# CLEP College Algebra/Algebraic Operations

Addition, subtraction, multiplication, division, exponentiation, rationalization, simplification, etcetera. If you knew about four of these terms, you would have most likely understood the definitions of the first four. This is because you have most likely been trained very well to understand the first four terms in your elementary/primary school years. Some of these are simply extensions of those ideas.
Nevertheless, any good algebra textbook will cover the basic algebraic operations and properties of numbers. This is true because math is simply rules, in that you start with a basic set of rules so that you may operate and simplify — to be put into its most compact form — any expression — a written form of combined symbols that uses one of the two basic operations, addition and multiplication. To make sure you are not confused, here is an example:
* Example 0.1.a: Operate this expression: $20+10-5$.
* Algebra has never been more simple. This problem comes from a second grade class. In fact, many of you could probably this in your head (given the simple numbers). However, is it not strange to you that you know what you need to do once you see this expression? You add 10 to 20. Then, you subtract 5 from that resultant sum:
* $10+20-5$
* $=30-5$
* $=25$
* The truth is, you have been brainwashed, conditioned to follow the rules. We wonder how many would think to do this:
* $10+20+(-5)$
* $=10+(-5)+20$
* $=\left[10+(-5)\right]+20$
* $=5+20$
* $=25$
As should be obvious, you get the same answer. But why? That is what this chapter of the textbook seeks to explain.

## Basic Operations of Real Numbers
### Introduction to the Real Numbers

Along with that question, there is another: "what is a real number?" Before we get to that definition, we need to go ahead and go through all the systems that we worked with since primary school.
First, the natural numbers — the counting numbers or the numbers used when counting items of a set, not including zero. Then, the introduction of zero along with the natural numbers gives the full definition of the positive integers. It seems all is true until you learn about the negative integers $\left\{-1,\,-2,\,-3\,\ldots \right\}$. The full set of integers is both the positive and negative integers.
This is where the strange stuff really begins. Notice we have yet to talk about fractions, until now. The numbers in which a fraction ${\frac {m}{n}}$, where $m$ is or is not divisible by $n$, is considered rational. A rational number is simply the ratio between two numbers. By contrast, an irrational number is a number that cannot be denoted exactly by the ratio of two numbers. Later, after the introduction, you will get to see one proof of irrationality. However, a thorough grasp of algebra and real numbers is necessary to understand the proof. You will be able to do that later.
There are some restrictions on rational numbers however, some of which are situational, some of which are not, whereby the number is either an integer or results in a number that is undefined. On top of the numerator being divisible by the denominator, the following restrictions apply:
* The denominator is either $\mathbf {-1}$ or $\mathbf {1}$ when the numerator is an integer. This should be a fact everyone knows: dividing any integer by $1$ gives the same integer again. Dividing by a negative number gives you a negative number. In dividing by $-1$, given the numerator is an integer, you get the negative version of the numerator. Either way, so long as the numerator is an integer, when dividing by $1$ or $-1$, the result is an integer.
* The denominator is $\mathbf {0}$ for any rational numerator $\mathbf {m}$. For any rational numerator $m$, dividing by zero is never a good thing (in the "real" world). We will get into why after the first section. For now, keep in mind that this is simply impossible to define using numbers that we will talk about here.
* The numerator is $\mathbf {0}$ when rational denominator $\mathbf {n\neq 0}$. Like the above point, this small factoid will be proven rigorously later after section 1.
Most textbooks define a rational number as the quotient of two integers, and we are no different. Therefore, we cannot say that ${\frac {0.5}{0.25}}$ is not an integer until we make it a quotient of two integers: Here is how we would that problem:
* Example 1.1.a: Determine whether the following quotient is an integer:
* ${\frac {0.5}{0.25}}$ Most of you probably already know that ${\frac {1}{2}}=0.5$ and ${\frac {1}{4}}=0.25$. Therefore, let us put it into this form:
* ${\frac {0.5}{0.25}}={\frac {\frac {1}{2}}{\frac {1}{4}}}.$
* Dividing a fraction by another fraction can be expressed the same way like so:
* ${\frac {\frac {1}{2}}{\frac {1}{4}}}={\frac {1}{2}}\div {\frac {1}{4}}.$
* From here, most students know what to do: ${\text{K}}$eep the first fraction the same, ${\color {Red}{\text{C}}}$hange the division to multiplication, and ${\color {Green}{\text{F}}}$lip the second fraction (take the reciprocal).
* ${\frac {1}{2}}\div {\frac {1}{4}}={\frac {1}{2}}{\color {Red}\times }{\color {Green}{\frac {4}{1}}}$
* From here, multiply the two fractions (multiply the top bits together and do the same for the bottom bit) to find this:
* ${\frac {1}{2}}\times {\frac {4}{1}}={\frac {4}{2}}=2$
* The number $2$ belongs to the set of integers, so because it is a result of the quotient, ${\frac {0.5}{0.25}}$ belongs to the set of integers. $\blacksquare$
Note: In case you have not seen $\blacksquare$, this is a shorthand for "Quod Erot Demonstrandum," which literally translates from latin to "that which has been demonstrated" true. This is simply mathematician code-speak for "we have finished proving everything and left no unfinished business."
The real numbers include all the sets we talked about in this section. That means it includes the integers, the rational numbers, and the irrational numbers. The following image to the left denotes the sets that will be relevant to your studies for now.
Note the symbols used to denote the specific set.
* Integers: $\mathbb {Z}$. Putting a positive or negative superscript (e.g. positive superscript $\mathbb {Z} ^{+}$) denotes positive or negative integers. The Z comes from ganze Zahlen, which literally translates to whole numbers.
* Rational numbers: $\mathbb {Q}$, for quotient.
* Real Numbers: $\mathbb {R}$.
A quick note about notation: Normally, a mathematician will want to know what type of variables he is working with. On the CLEP exam, it is explicitly stated that the variable you are working with on specific problems are real unless otherwise denoted, as is written here: "Unless otherwise specified, the domain of any function is assumed to be the set of all real numbers for which is a real number." In symbolic terms, "belongs to the set of" is denoted by $\in$. Therefore, given $\mathbb {R}$ represents the real numbers, you can define $x$ such that $x\in \mathbb {R}$. This notation is not needed for the exam; however, it can be helpful to be quick when working with problems. To make sure you fully understand which numbers belong to which, let us look at an example:
* Example 1.1.b: Determine each of the following as either an integer or rational number:
* (a)$-1.597$
* (b)$-10$
* (c)${\frac {8}{2}}$
* (d)${\frac {5}{4}}$ (a) First, ask, is $-1.597$ an integer. Well, if you divide two integers only to suddenly get something beyond a decimal point, then it is not an integer and by definition becomes a rational number. Since the number is not an integer, and the number forms a decimal, then it must be the case that it was divided by two numbers and thus formulated a quotient: ${\frac {a}{b}}=-1.597$. We have sufficiently established this number as a rational number (although not rigourously): $-1.597\in \mathbb {Q}$.
* (b) $-10$ has no decimals, and can be written in the form ${\frac {a}{1}}=a$. This must be an integer: $-10\in \mathbb {Z}$.
* (c) Just because it is fraction does not automatically make it a rational number. Remember, if $2$ can divide $m$, the numerator of the fraction, it would automatically make it an integer. Notice we are not saying that the denominator has a factor to the numerator or that the denominator divides into the numerator. Rather, you can easily divide the numerator by the denominator. This is why it must be an integer: ${\frac {8}{2}}\in \mathbb {Z}$.
* (d) Because "4" cannot easily divide into the numerator, "5," the fraction is a rational number: ${\frac {5}{4}}\in \mathbb {Q}$.
Of course, each example above is defined as a real number and a rational number. Regardless, it is important to understand when a qualification limits the ability to define variables accordingly. Often times, the restrictions are out in the open so that no one is confused on what the number can or cannot be.
It is now your turn to try this out. This WikiBooks uses explorations to not only help cement concepts but also challenge readers to think like mathematicians. Problems in mathematics are not always simple, and at times, mathematicians need to generalize findings for everyone so that it can be made applicable in the abstract and the concrete as well. Nevertheless, it is now your turn, so try it out.
$x\in \mathbb {Z}$
$\mathbb {Q}$
$\mathbb {Z}$
(a) ${\tfrac {x}{1}}$ (b) ${\tfrac {x}{y}}$, where $y\in \mathbb {Q}$
${\tfrac {1}{x}}$
$x\in \mathbb {Z}$
${\frac {x}{1}}\in \mathbb {Z}$
(b) At the moment, you cannot determine ${\frac {x}{y}}$ to be part of the set of rational numbers until we make sure $y\notin \mathbb {Q}$. Because $y\in \mathbb {Q}$, set $y={\frac {a}{b}}$, where $a\in \mathbb {Z}$ and $b\in \mathbb {Z}$. Assume ${\frac {a}{b}}$ is coprime, meaning $a\nmid b$ ($a$ does not divide $b$). Knowing $y={\frac {a}{b}}$, the following is true: ${\dfrac {x}{\dfrac {a}{b}}}$ Simplifying the fraction to be in the form ${\frac {a}{b}}$, you get ${\frac {b\cdot x}{a}}$. There are two cases for this situation.
    CASE 1: $a\nmid x$, meaning ${\frac {b\cdot x}{a}}=b\cdot {\frac {x}{a}}\in \mathbb {Q}$.
    CASE 2: $a\mid x$. Knowing ${\frac {b\cdot x}{a}}=b\cdot {\frac {x}{a}}$, let ${\frac {x}{a}}=F$, where $F\in \mathbb {Z}$. Because $F\in \mathbb {Z}$, $b\cdot F\in \mathbb {Z}$.
If CASE 1 is true, then ${\frac {x}{y}}\in \mathbb {Q}$. Given this is a possibility, ${\frac {x}{y}}\in \mathbb {Q}$. If CASE 2 is true, then this definition for the fraction is still true because $\mathbb {Z}$ is a subset of $\mathbb {Q}$. Because the question is asking for the most specific one, this is the best answer one could give.
$1\in \mathbb {Z}$
$x\in \mathbb {Z}$
${\frac {1}{x}}\in \mathbb {Q}$
$x\nmid 1$
$x\in \mathbb {Z}$
${\frac {x}{1}}\in \mathbb {Z}$
(b) At the moment, you cannot determine ${\frac {x}{y}}$ to be part of the set of rational numbers until we make sure $y\notin \mathbb {Q}$. Because $y\in \mathbb {Q}$, set $y={\frac {a}{b}}$, where $a\in \mathbb {Z}$ and $b\in \mathbb {Z}$. Assume ${\frac {a}{b}}$ is coprime, meaning $a\nmid b$ ($a$ does not divide $b$). Knowing $y={\frac {a}{b}}$, the following is true: ${\dfrac {x}{\dfrac {a}{b}}}$ Simplifying the fraction to be in the form ${\frac {a}{b}}$, you get ${\frac {b\cdot x}{a}}$. There are two cases for this situation.
    CASE 1: $a\nmid x$, meaning ${\frac {b\cdot x}{a}}=b\cdot {\frac {x}{a}}\in \mathbb {Q}$.
    CASE 2: $a\mid x$. Knowing ${\frac {b\cdot x}{a}}=b\cdot {\frac {x}{a}}$, let ${\frac {x}{a}}=F$, where $F\in \mathbb {Z}$. Because $F\in \mathbb {Z}$, $b\cdot F\in \mathbb {Z}$.
If CASE 1 is true, then ${\frac {x}{y}}\in \mathbb {Q}$. Given this is a possibility, ${\frac {x}{y}}\in \mathbb {Q}$. If CASE 2 is true, then this definition for the fraction is still true because $\mathbb {Z}$ is a subset of $\mathbb {Q}$. Because the question is asking for the most specific one, this is the best answer one could give.
$1\in \mathbb {Z}$
$x\in \mathbb {Z}$
${\frac {1}{x}}\in \mathbb {Q}$
$x\nmid 1$
Now that we have thoroughly introduced the reader into the land of real numbers, it is now time to find its properties.

### The Most Fundamental Definitions of Real Numbers

Recall the introduction of this chapter in which we mention a fundamental principal of mathematics:
   [M]ath is simply rules, in that you start with a basic set of rules so that you may operate and simplify... any expression — a written form of combined symbols that uses one of the two basic operations, addition and multiplication.
How come there are only two basic operations? After all, there exists subtraction and division! Notice, however, what we did in Example 1.1.a. The expression has a minus sign in front of the "5," yet there is a way to rewrite the expression:$10+20-5=10+20+(-5)$. Our quandary at the beginning of the paragraph seems to now make sense. We defined these operations. If we say there are only two basic operations, then it is so. Subtraction and division are referred to as inverse operations in that they are the same as addition and multiplication with the only difference being a sign. To denote this, we will be using variables, which are symbols that denote a number that can be input such that it follows a restriction, if any exists. If you never got used to this idea in Algebra I and II, we will be introducing them slowly. Right now, get used to the idea that these are simply symbols that denote a possible real number.
* Inverse Definition of Addition: $a-b$ is materially equivalent to $a+(-b)$.
* Inverse Definition of Multiplication: ${\frac {a}{b}}$ is materially equivalent to $a\cdot {\frac {1}{b}}$, where $b\neq 0$.
Because the above symbols are variables, and the above operations are by definition true, any number can be input. An example should help cement these definitions in your mind. In the table below, inputs $a$ and $b$ are given on the respective column. The output of $a-b$ and $a+(-b)$ are given. Notice that $a-b$ and $a+(-b)$ are identical.
${\begin{array}{|c|c|}a&b&a-b&a+(-b)\\\hline 1&3&1-3=-2&1+(-3)=-2\\\hline 3&4&3-4=-1&3+(-4)=-1\\\hline 5&5&5-5=0&5+(-5)=0\\\hline 7&6&7-6=1&7+(-6)=1\\\hline \vdots &\vdots &\vdots &\vdots \\\hline 2a+1&b+3&2a+1-(b+3)&2a+\left(-\left[b+3\right]\right)\end{array}}\!$
This table should help the student unfamiliar with working with variables get an intuitive sense of the meaning. Note, the final row is a "rule" that gives the pattern for the number. Finally, one last definition should be established before moving.
If given expression ${\frac {a}{b}}=a\cdot {\frac {1}{b}}$, the term ${\frac {1}{b}}=b^{-1}$.
All that the above text says is that $b^{-1}$ is the same thing as ${\frac {1}{b}}$. For example, $8^{-1}={\frac {1}{8}}$. Now that you understand the most fundamental definitions of the real numbers, we can keep deriving new information from our definitions.

### Closed Door Policy of Real Numbers

Let us go back to the set of integers. If we add any two integers, $p+q$, you get an integer as a result. Similar, multiplying any two integers, $p\cdot q$, gives another integer. However, if you divide two unique integers, ${\frac {p}{q}}=p\cdot q^{-1}$, you will not always get an integer. For example, ${\frac {3}{4}}=0.75$, which is not an integer. We would say the following is true:
    Under addition, subtraction, and multiplication, the set of integers are closed.
By closed, we mean to say that the set of integers using that specific operation will always give another integer. Similar to integers, real numbers are also closed. The following definition below tells us the most common form:
Under the operation of addition, subtraction, multiplication, and division (excluding denominator 0), the use of the operation to any real numbers is closed.
This is a simple but powerful tool to remember in the mathematical lexicon. This fact is so fundamental, in fact, that it can be used to prove a number is or is not an integer, or rational number, or a real number. To prove a number does or does not belong to a given set, you must show that it can or cannot belong to that set using those rules. We defined those operations, so you must show that a number is not defined under those axioms (definitions).
This idea is key to doing proofs in mathematics, so it is best we introduce the concept now. Also, proving a number is an integer is a very important concept in number theory. You thought you knew everything about integers? Try number theory (after learning as much as you can about Algebra) to find out how wrong you are!

### The Laws of Real Numbers (Under Equality)

Numbers are a society. They follow rules that were dictated by a ruler. In this instance, the ruler is humans. We did not discover "4," we only said that there are 4 things. Numbers are adjectives, not nouns, so by definition, they cannot be discovered. As such, everything we have talked about so far is only definitions. Keep in mind that some things are extensions of the definitions we made. However, humans have been applying these definitions for so long that we were able to "discover" new facts about mathematics. We will see the discoveries in the second section. However, for now, let us define the rules that humans have used for hundreds of years now!
For all real numbers $a$, $b$, and $c$, $(a+b)+c=a+(b+c)$. That is, the grouping of real numbers does not change the sum.
It turns out proving this is true takes a lot more than basic algebra. Unfortunately, for the curious student, it is hard to explain these simply. If you want to prove these are true, you will want to do abstract algebra. In fact, the three definitions below cannot be simply proven (or at least shown) with either algebra or geometry, so simply accept these as true for now.
For all real numbers $a$ and $b$, $a+b=b+a$. That is, the order of addition does not matter.
For all real numbers $a$, $b$, and $c$, $(a\cdot b)\cdot c=a\cdot (b\cdot c)$. That is, the grouping by which three numbers are multiplied does not change the product.
For all real numbers $a$ and $b$, $a\cdot b=b\cdot a$.
For all real numbers $a$, $b$, and $c$, $c\cdot (a+b)=c(a+b)=a\cdot c+b\cdot c=ac+bc$.
As a convention in this book, we will outline all proofs that we can present in the green boxes below. This should hopefully make the later examples more manageable to students (since a large part of this textbook will be to help students see that math is not about memorization but discovery). Plus, for students that have no interest in them (even though they should), it can be skipped. However, we do not recommend this approach unless you are only here to memorize facts, which may not make math fun for you. hould make proofs easy to find. Plus, for students that have no interest in them (even though they should), it can be skipped. We hope that the language we will use will allow students to engage in these proofs with little to no hesitation!s
* Proof 1: For any positive real numbers $a,b,c$, the distributive law is true.
* A proof of the commutative law of multiplication involves understanding a little bit of geometry. A rectangle $ABCD$ has dimensions $c\cdot m$. Form an altitude at some point $P$ along the line segment ${\overline {AC}}=m$ (i.e. arbitrarily form a vertical line at some point along line segment ${\overline {AC}}$). Let $m=a+b$. Here, ${\overline {AP}}=a$ and ${\overline {PC}}=b$. See the image to the left for reference.
* By the definition of area, the length times the width gives the number of unit squares in a rectangle. The area of the entire rectangle is $cm=c\left(a+b\right)$. Forming a vertical altitude line parallel to the sides of the rectangle means ${\overline {PQ}}=c$. As such, the sum of the area of the two segments must be equivalent to entire rectangle. Therefore, $ac+bc=c\left(a+b\right)$. This makes the distributive law true for any $a>0$, $b>0$, and $c>0$. $\blacksquare$
* Notice that this proof required another definition, the area of a rectangle. It is very hard to find a proof that does not involve a definition of some kind. Perhaps it may be impossible. Also, notice that we were not able to prove this is the case for all real numbers. To do that, we need to find a way to say that a given distance is negative. We can do that later when we learn about graphing.
For now, we will accept that the distributive law is true for all real numbers.
We have two more properties to define before we move on. Notice how they are in different sections. There is a reason for that, as you will see with the zero factor property and the identity elements.
For all real numbers $a$ and $b$, $a\cdot b=0$ if and only if $a=0$ or $b=0$ or $a=0$ and $b=0$. In other words, for all real numbers $a$ and $b$, $a\cdot b=0$ if and only if at least one of the factors is zero.
Say that we have found out that the following equation below is true:
    $(x-5)\cdot \left({\cfrac {{\dfrac {x+1}{4}}-{\dfrac {3}{2}}}{5}}\right)=0$
The scary looking equation aside, use some basic logic so that the answer shall be easy to find. For the expression to be true, it needs to be the case that either $x-5=0$ or ${\cfrac {{\dfrac {x+1}{4}}-{\dfrac {3}{2}}}{5}}=0$ or both. If at least one of the factors is zero, then the expression will always be zero. This is simply another definition of mathematics that we humans have used for many years now. If you want to solve such an equation, wait for it. If you can already, go ahead. Otherwise, keep reading.
For all real numbers $a$, the following are always true:
* $a+0=a$. When adding $a$ by $0$, you get the same real number $a$.
* $1\cdot a=a$. When multiplying $a$ by $1$, you get the same real number $a$.
When you hear "identity," you think of a person's name. In math, however, identity simply means a number that does not change another number. What number makes that true? Let us some basic logic to figure this out.
* What number makes it so that by adding it to another you get the same number? If a result does not change, that logically means the difference is zero because if there exists absolutely no change, then the difference between the overall output and the initial input is zero. As such, for any real number $a$, the identity element must be $a-a$, or $0$.
* What number makes it so that by multiplying it to another you get the same number? We can use the exact same logic to find the number we are looking for. If we are multiplying a result, we want some way to find the overall change when multiplying. In this instance, division would work. Excluding $a=0$ from the real numbers, ${\frac {a}{a}}=1$. Can we say that multiplying $a=0$ by 1 gives the same number? Yes, $0\cdot 1=0$, so the identity element is true for all real numbers.
If for the real numbers $a$, $b$, and $c$, $a=b$, and $b=c$, then $a=c$.
This is simply true by definition. For example, if $2+2=4$, and $2\cdot 2=4$, then $2+2=2\cdot 2$.

### Check Your Understanding

It is always a good idea to test your knowledge, to see if you truly understand. However, it can often be difficult to test yourself when you cannot think of any problems to do. As such, here are some problems that relate all the knowledge you learned.

#### Part A

Instructions: without using a calculator, find a way to calculate these expressions below in under 1 minute and 30 seconds. For each "text box," write the number that makes the value true. Do not write commas in the boxes because those are interpreted as decimals.

#### Part B

Instructions: In under 2 minutes, without using a calculator, answer each question below.
More will be added later

## Introduction to Algebra

You may have saw the algebra required part of the check your understanding and may have wanted to give up. "When did math introduce letters anyway; it is already hard enough as it is!" Well, there should not be any need to worry because is easy once you understand the core principles behind it (that is, until you get to modern [abstract] algebra). The most fundamental of those rules is outlined below:
When given an equation – expressions set to equality or inequality – the opposing sides must be equal to each other, or at least one of the expressions is either lesser than or greater than the other.
Is $2+2=4$? Yes, it is. We know the operation must be an equation, because it is true to what it states, that adding two things with two things is the same as four things. However, what happens when you introduce variables – objects that introduce a source of change to an equation. If one side is equal to the other, then the variable of interest must be equal to some expression so that it operates with the other variables or non-variables such that it equals the other side. Look at the example below to see what we mean:
    $5T+2=17$
The value of $T$ must equal some constant number to make the above equation true; otherwise, the expression cannot be equal. If put into a sentence, the above equation is asking us to "find the value of $T$ that makes it so that multiplying it by 5 and adding 2 gives 17." In the very early days of math, it used to be the case that questions were asked in such a fashion. Algebra makes these problems easier (both in the statement of it and in the evaluation of it). In this section, we will discuss how to solve single-variable equations and multi-variable equations. Then, we will determine how to find the value of variables when given a system of equations.

### Simplifying the Complicated

Before we do that cool stuff, however, it is important the student has a grasp on the anatomy of equations. The first idea students learn is "term":
The term is a single mathematical expression that is either a single number (a constant), a single variable, multiplied variables, or variable(s) multiplied by constants, each separated by a plus ($+$) or fraction sign ($/$).
For example, look at the following expression: $14l+16w$. There are two terms according to the definition above: $14l$ and $16w$. $14l$ is a variable multiplied by a constant, which is also similar to $16w$. The two terms are separated by a plus sign. Combining terms make up the expression. The equation is equality or inequality of two expressions. This forms the basic anatomy of the equation and the expression. Sometimes, however, there are more terms than needed in the expression. The procedure during such times is to simplify the equation or expression.
The act of simplifying equations (or expressions) is one whereby "like-terms" are condensed and combined, without changing the equation's equality or inequality illegally.
Before we keep refining the above definition (e.g. what are "like-terms"?), it is important to get an example of this. Some equations or expressions that are not simplified are provided below:
* $1+2+3+4+5$
* ${\frac {1}{3}}(3x+4x)+4y=30$
* $2xy+8yx+10y-10x=20$
* $2^{16}+2^{16}+2^{16}+2^{16}+2^{8}+2^{8}+2^{8}+2^{8}-2^{10}$
The bullet points below correspond to the equivalent unsimplified expressions above, in the order given.
* $15$
* ${\tfrac {7}{3}}x+4y=30$
* $xy+y-x=2$
* $2^{18}$
Making things simpler is usually what you want to do in math problems, and this is especially true in College Algebra, because the act of making things simpler is inherent to finding the one value of the variable of interest to make the equation true. Despite the very simple act of simplifying equations, students may make some common errors because there is no definite idea of "like terms." It is important to get this idea correct before moving on to like-terms. As such, the definition and examples are provided below.
Like-terms are two terms that are either constants or have the same variables with the same exponents (values that denote the number of times a term is multiplied by the same number).
The above definition is one of the better ones; however, it can be a little confusing to students. Therefore, an exploration is in order. Imagine the following expression below:$2x^{3}+3x^{2}y+3xy^{2}+y^{3}-x^{3}+5x^{2}+10-100-10$
This expression above is quite messy, so let us see if we can simplify the above expression. First, let us look for terms that have the same variables with exponents within them. Those that are "alike" will be highlighted in their corresponding color.${\color {Red}2x^{3}}+3x^{2}y+3xy^{2}+y^{3}{\color {Red}-x^{3}}+5x^{2}+{\color {blue}10-100-10}$
Notice how negatives are highlighted. This is because it is separated by the plus sign, for $a-b=a+(-b)$. As such, those are different terms. Notice how $3x^{2}y$ and $5x^{2}$ are "unlike" terms. This is because $5x^{2}$ is missing the variable $y$. Because it does not have the same variable, those two are unlike terms. The next procedure would be to rewrite the above equation like so:$={\color {Red}2x^{3}}{\color {Red}-x^{3}}+3x^{2}y+3xy^{2}+y^{3}+5x^{2}+{\color {blue}10-100-10}$
To combine the above terms, simply combine the constant values or the constant values corresponding to the like terms. This is true because the constant values next to those corresponding constant values can be factored. For example, $2{\color {green}x}-5{\color {green}x}=(2-5){\color {green}x}=-3x$, where the variable $x$ is multiplied to the corresponding constant values. From this property, the idea of combining like-terms make sense.
   As long as the variables multiplied to the constants are exactly alike, where each variable is identical along with the exponent value, the two terms can be combined due to factoring allowing the corresponding constants to be added.
This is the most important principle behind factoring, and should be rigorous enough such that any mathematician (and student) can agree that the definition of "like-term" makes sense. Enough fo that, it is time to finish the problem we have stated. Keeping the colors for the like terms, we yield:$={\color {Red}x^{3}}+3x^{2}y+3xy^{2}+y^{3}+5x^{2}{\color {blue}-100}$
This answer above is the simplest form available for this above expression. Keep in mind that the simplest form is a little subjective, and it is based on what the question is asking, so there is no true definitive way to determine "simplest form." The best one can do is infer whether that form is the easiest one can use. For the above expression, it could be the case that this may not be the simplest, but this is due to notation (which we will get into more detail in Binomial Theorem). If you are curious to see the simplest form, which you may not understand at this point in time, here it is:$=\sum _{k=0}^{3}\left({\binom {3}{k}}x^{k}y^{3-k}\right)+5\left(x-2{\sqrt {5}}\right)\left(x+2{\sqrt {5}}\right)$
One other thing about simplifying equations is a little warning. An expression is provided, and a few students got different answers. Your job is to determine who is correct (and yes, these are your exploration).
$-2xy+4yx-10y+4x-9$ However, each person got a different answer. Determine which person got the correct answer, and explain your choice. (A) $-9$ (B) $6xy+4x-10y-9$ (C) $-2xy+4yx-10y+4x-9$ (D) $-2xy-2yx-9$
$2xy+4x-10y-9$
$ab=ba$
$a=x$
$b=y$
$xy=yx$
$-2xy+4yx=-2xy+4xy$
$2xy$
$xy$
$2xy+4x-10y-9.$
$-2xy+4yx$
$x$
$y$
$ab=ba$
$a=x$
$b=y$
$xy=yx$
$-2xy+4yx=-2xy+4xy$
$2xy$
$xy$
$2xy+4x-10y-9.$
$-2xy+4yx$
$x$
$y$
Should you get the above exploration, you basically understand the idea of simplifying equations and maybe the associate property of multiplication. Either way, good job if you got that one correct.
$x^{2}+3x-5-2(4x+5)$
However, each person got a different answer. Determine which person got the correct answer, and explain your choice. (A) $x^{2}-5x+5$ (B) $x^{2}-5x-15$ (C) $x^{2}-5x$ (D) $x^{2}-5x-10$
$x^{2}+11x+5$
$c(a+b)=ac+bc$
$a=4x$
$b=5$
$c=-2$
${\begin{aligned}-2(4x+5)&=(-2)(4)+(-2)(5)\\&=-8x-10\\\end{aligned}}$.
As such, $x^{2}+3x-5-8x-10$ is the result. Combining all "like terms," you get
$x^{2}-5x-15.$
$c(a+b)=ac+bc$
$a=4x$
$b=5$
$c=-2$
${\begin{aligned}-2(4x+5)&=(-2)(4)+(-2)(5)\\&=-8x-10\\\end{aligned}}$.
As such, $x^{2}+3x-5-8x-10$ is the result. Combining all "like terms," you get
$x^{2}-5x-15.$
The above explorations may be how the CLEP exam may try to trick you into picking the wrong answer. It is because each answer choice may be a common mistake a student will make when trying to simplify an expression. These are usually the first questions of the test, so it may be best not to get those wrong because of an embarrassing mistake like not distributing the negative 4 to all terms or forgetting about the negative number in the first place.

### Solving for One Variable in An Equation

The level of detail we will go into here may seem excessive. However, this magnification will be necessary to explain why we are certain some things work they way they work. Plus, some students appreciate when questions are answered that they would likely ask. As such, bear with us as we demolish through these simple problems. As another bonus, you will understand algebra more than current students taking the College Algebra class in university, so you should look at this as a blessing. The blessing comes in this encompassing principle:
   For an equation that has a variable, the variable is only true when going backwards. That is, using inverse operations, the operation on the other side is now transferred to the other side without the variable.
This basic property is what allows us to solve for equations. After all, since both sides change by the same amount, there should be no discrepancy between the equality signs. This gives us the basic definitions below:
If $a,b,c\in \mathbb {R}$, and $a=b$, then
    $a+c=b+c$
If $a,b,c\in \mathbb {R}$, and $a=b$, then
    $a-c=b-c$
If $a,b,c\in \mathbb {R}$, and $a=b$, then
    $ac=bc$
If $a,b,c\in \mathbb {R}$, and $a=b$, then
    ${\frac {a}{c}}={\frac {b}{c}}$
These properties above, along with the properties of real numbers listed in the previous section, will be very helpful when trying to solve equations. We will be stating these properties as we work through them in the examples that follow. Therefore, it is a good idea to understand how we present our work and incorporate that style into your work for the exploration. Although this may be a multiple choice test, if a student moves on to a higher math, then it is important to know how to communicate mathematically, especially on paper.
The process of isolating the variable of interest is such that on one side only, the variable of interest has no constants added or multiplied to it, except for 0 and 1, respectively, and the variable is equal to a constant or expression with the variable of interest not on the other side.
The above statement can be a little confusing if one has not seen an example of this procedure, so two examples are given below, one with only one variable and one with multiple variables. From there, the examples will present a more concise procedure.
* Example 2.2.a: Find the value of $x$ that makes the equation true: $10+5x=210$.
* From here on out, because whatever is true on one side is also true of the other, we will no longer be stating that the two sides are equal. However, keep this in mind every time you are solving equations, because if both sides must equal, then what is done one side must be done to the other to make them equal! We will step you through the process so that one can see how to isolate the variable of interest. The equation in the problem is rewritten below:
* $10+5x=210$
* Add $-10$ to both sides of the equation. The effect of adding $-10$ is to make it so that the left side of the equation is adding the identity element to $+5x$.
* $10+({\color {Red}-10})+5x=210+({\color {Red}-10})$
* ${\color {Red}0}+5x=200$
* Because the identity element does not change the left side, we may safely ignore it. Because $5$ is multiplied to the variable $x$, to make it so that $x$ is not changed, multiply ${\frac {1}{5}}$ to both sides of the equation. The effect of multiplying ${\frac {1}{5}}$ is to make it so that the left side of the equation is multiplying the identity element to $x$.
* $\left({\color {Red}{\frac {1}{5}}}\right)\cdot 5x=\left({\color {Red}{\frac {1}{5}}}\right)\cdot 200$
* ${\color {Red}1}x=40$
* Because the identity element does not change the left side, we may safely ignore it. From this, we have successfully isolated the variable of interest, $x$ by itself. We learned that the value of $x$ that makes the equation true is $x=40$.
We know this makes the equation true because when isolating the variable, we made sure the equation always remained the same. This process ensured that the value of $x$ will always give the right number. Nevertheless, if you do not feel confident about the work you did, a good way to double check your answer is to "plug it in" and see if it holds true. Here is how one would do it:
${\begin{aligned}210&=10+5(40)\\&=10+200\\&=210\end{aligned}}$
If the expression on one side, after being operated, gives the same result as the expression on the other side, then the value for the variable, $x$, you found must be true so that it satisfies the equation. Unfortunately, sometimes, on the CLEP exam, you may not always be given one-variable equations. There may be times when you are given more than one variable. The same procedure applies for what we did above, except this time, you must be extra sure of yourself that the answer you have is definitely correct, for although you may substitute the answer you got into the equation, this one may be a bit more tedious.
It is given that we will keep stepping through the process for you. After all, the student that is not familiar with working with equations tends to get lost if not stepped through. Plus, for the student that understands the algebra, it helps to see problems so that one can see how to work them through step-by-step. Mind you, these steps are not meant to be efficient. For now, think of these as little problems that set up the big bad later on.
* Example 2.2.b: Find the value of $\Delta x$ that makes the equation true: $\Delta F={\frac {\Delta x}{\Delta t}}m-{\frac {1}{2}}$.
* The first step when doing any problem is to not panic. Do not get too bogged down with the details; simply fight for your answer. Of course, the CLEP exam is multiple choice, although it is a good idea to learn how to do these problems without relying heavily on multiple choice tricks. Tricks should only be a last resort for problems that you have no idea how to work out during the short amount of time you have to think. This problem is way easier than it may seem at first, and using tricks on it may be slower than not using them and doing it the straight way.
* Realize that the same philosophy we had when working with problems is the same: whatever is done to one side, you do on the other as well. Before we begin, let us write down the equation again:
* $\Delta F={\frac {\Delta x}{\Delta t}}m-{\frac {1}{2}}$
* Let us begin by adding ${\frac {1}{2}}$ to both sides of the equation. This will eliminate the constant term on the right hand side and move it over to the other side. The effect of adding ${\frac {1}{2}}$ is to make it so that the left side of the equation is adding the identity element to ${\frac {\Delta x}{\Delta t}}m$
* $\Rightarrow \Delta F+\left({\color {Red}{\frac {1}{2}}}\right)={\frac {\Delta x}{\Delta t}}m-{\frac {1}{2}}+\left({\color {Red}{\frac {1}{2}}}\right)$
* $\Rightarrow \Delta F+\left({\color {Red}{\frac {1}{2}}}\right)={\frac {\Delta x}{\Delta t}}m+\left({\color {Red}0}\right)$
* $\Rightarrow \Delta F+{\frac {1}{2}}={\frac {\Delta x}{\Delta t}}m$
* Next, eliminate the fractional term multiplied to the variable of interest. Here, we will be a little slow and work it through one at at time before we introduce a neat consequence of the slow working we have done here. For now, think it through this way. If we want to get rid of something that is divided, we want to work with the inverse operation. Therefore, we need to multiply the fractional term. First, let us rewrite what we see so far into something simple:
* ${\frac {\Delta x}{\Delta t}}m=\left(m\Delta x\right)\left(\Delta t\right)^{-1}$.
* Doing this is equivalent because we know that ${\tfrac {\Delta x}{\Delta t}}m=\left(\Delta x\left(\Delta t\right)^{-1}\right)m$. Because multiplication is commutative, we understand that the terms can be multiplied in any way we want. Plus, because we know that multiplication is also associative, we know grouping does not affect multiplication. Therefore, we may rewrite the following expression like so:
* ${\begin{aligned}\left(\Delta x\left(\Delta t\right)^{-1}\right)m&=\Delta x\left(\Delta t\right)^{-1}m\\&=m\Delta x\left(\Delta t\right)^{-1}\\&=\left(m\Delta x\right)\left(\Delta t\right)^{-1}\\&={\frac {m\Delta x}{\Delta t}}\end{aligned}}$.
* It is worth your time understanding all the steps underlying the above indented line. All we did is simply eliminate the parentheses that are unnecessary, moved some terms around, then added the parentheses back in. You can think of these parentheses as denoting the numerator of the fraction.
* Either way, we know the above expression is equal to $\Delta F+{\frac {1}{2}}$, so what we do on one side must be done to the other. To get rid of the division, we must multiply, as shown below.
* $\Rightarrow \left(\Delta F+{\frac {1}{2}}\right)\cdot \left({\color {Red}\Delta t}\right)=\left(m\Delta x\right)\left(\Delta t\right)^{-1}\cdot \left({\color {Red}\Delta t}\right)$
* $\Rightarrow \left(\Delta F+{\frac {1}{2}}\right)\cdot \left({\color {Red}\Delta t}\right)=\left(m\Delta x\right)\left({\color {Red}1}\right)$
* $\Rightarrow \left(\Delta F+{\frac {1}{2}}\right)\cdot \Delta t=\left(m\Delta x\right)$.
* Notice the parenthesis around $F+{\frac {1}{2}}$. This is important, because we are multiplying this entire term by $\left(\Delta t\right)$. We will explain in full detail why this is important later, but for now. Let us continue.
* Notice we are nearly done isolating the variable $\Delta x$. To complete the step, we must transfer $m$ to the other side. As such, do the inverse operation. The inverse operation of multiplication is division. As such, multiply both sides by $m^{-1}$:
* $\Rightarrow \left(\Delta F+{\frac {1}{2}}\right)\Delta t\cdot \left({\color {Red}m^{-1}}\right)=m\cdot \left({\color {Red}m^{-1}}\right)\cdot \Delta x$
* $\Rightarrow \left(\Delta F+{\frac {1}{2}}\right)\Delta tm^{-1}=\Delta x$.
* Readers may also be wondering why we did not put parentheses around $\left(\Delta F+{\frac {1}{2}}\right)\Delta t$. Again, we will get to that one soon. Realize this instead: we are finished with this problem.
We know some readers might be asking why we did not divide this weird symbol $\Delta$ (capital Delta) from $x$. For one, the question is asking for $\Delta x$, but also, $\Delta$ may be an important abbreviation for something: the change in $x$. The equation may look a little bit more messy having to define to different $x$ and $t$ and $F$. In the interest of not confusing a lot of scientists, specifically, you write delta to denote a change in the variable of interest. This is how it works, so it may as well be a good idea to be exposed to this notation. If we were to redefine the above equation without any $\Delta$, then we would write the following:
    $F-F_{0}={\frac {x-x_{0}}{t-t_{0}}}m-{\frac {1}{2}}$, which simplifies to $\left(F-F_{0}+{\frac {1}{2}}\right)\left(t-t_{0}\right)\left(m\right)^{-1}=x-x_{0}$. You could see why $\Delta$ is very popular in science.
One of the more strange aspects of the problem come with the parentheses. We asked why we put parentheses around items $\Delta F+{\frac {1}{2}}$ but not around $\left(\Delta F+{\frac {1}{2}}\right)\Delta t$. The first one is easier to explain than the second, so we will begin with the easier one.
* Any term that is finite will effectively equal something. Wen adding terms, such as $x+y$, it must equal some other term, say $A$. When multiplying the entire expression by a term $C$, you are effectively multiplying the end result, $A$, by $C$. If $x+y=C$, and you are multiplying $A$ by $C$, then $AC=C(x+y)$. This is why, when adding terms, you must put parentheses around what you are multiplying. There is no law stating that multiplying and adding are commutative or associative (recall that it is distributive, however).
* By extension of the first premise, if $\left(\Delta F+{\frac {1}{2}}\right)\Delta t$ is multiplied by some term, say $m^{-1}$, then if $\Delta F+{\frac {1}{2}}=B$, then $B\cdot \Delta t\cdot m^{-1}$. Because multiplying is commutative and associative, there is no need to put parentheses are around $B\Delta t$. As such, the expression did not need parentheses.
There is trick that we can learn from the above problem. Recall how we stated that we are complicating the problem a bit with what we are doing, specifically with the fractional term. This is because we are not utilizing a common trick with fractions multiplied to a term of interest. However, the understanding of this "trick" should be necessary before we ever utilize it. Tricks are fine, but that is not what math is all about. This is what makes math boring for so many students. What is this trick?
Definition: A reciprocal is a fraction ${\frac {a}{b}}$ that swaps the location of the numerator and the denominator such that it forms the fraction ${\frac {b}{a}}$.
Trick: When a variable of interest, say $x$, is multiplied by a fractional term, say ${\frac {a}{b}}$, such that it gives $c$, multiplying by the reciprocal allows the variable of interest to be "isolated."
Let us unpack this slowly. A reciprocal of a fraction is effectively swapping the location of the numerator and the denominator. For example, the reciprocal of ${\frac {3}{4}}$ is ${\frac {4}{3}}$. When you multiply by the reciprocal, what you effectively do is make it so that the following is true:
${\begin{aligned}{\frac {a}{b}}\cdot {\frac {b}{a}}&=ab^{-1}\cdot ba^{-1}\\&=\left(aa^{-1}\right)\left(b^{-1}b\right)\\&=1\cdot 1\\&=1\end{aligned}}$.
As such, multiplying by the reciprocal will shorten the time it takes to finish the problem. Let us see how this would have worked with the previous problem. If the following is true,
${\begin{aligned}\Delta F+{\frac {1}{2}}&={\frac {\Delta x}{\Delta t}}m\\&=\Delta x\left(\Delta t\right)^{-1}m\\&=m\left(\Delta t\right)^{-1}\Delta x\\&={\frac {m}{\Delta t}}\Delta x\end{aligned}}$,
then multiplying by the reciprocal ${\frac {m}{\Delta t}}$ to both sides gives
    $\left(\Delta F+{\frac {1}{2}}\right){\frac {\Delta t}{m}}=\Delta x$.
We got the same answer with very little delay. The only thing is that the answer does not look the same. However, they effectively are. If $\left(\Delta F+{\frac {1}{2}}\right)\Delta tm^{-1}=\Delta x$, then because of the associate property,
    $\left(\Delta F+{\frac {1}{2}}\right)\left(\Delta tm^{-1}\right)=\Delta x$
    $\Rightarrow \left(\Delta F+{\frac {1}{2}}\right)\left({\frac {\Delta t}{m}}\right)=\Delta x$
the same answer from the above paragraph. The long convoluted steps we took gave us the same answer! This is where the true beauty of math is found. Hiding the beauty loses out on the intricacies that took place in the above problem.
We are not saying to not remember the trick. However, we are saying to lock this reasoning in your heart. If you cannot do that, then just remember the trick because any way you can shave time of the test, the better.
The color below indicates the concise argument of the full explanation given in the answer of the example above.
* Example 2.2.b (REPEAT): Find the value of $\Delta x$ that makes the equation true: $\Delta F={\frac {\Delta x}{\Delta t}}m-{\frac {1}{2}}$.
* This example is a concise argument of the above example. Concise explanations must be given for everyone to understand what you are doing. Following along a huge line algebra can actually confuse people, so this is necessary.
* ${\begin{aligned}\Delta F&={\frac {\Delta x}{\Delta t}}m-{\frac {1}{2}}\\\Delta F&=\Delta x{\frac {m}{\Delta t}}-{\frac {1}{2}}&\left(\Delta x{\frac {m}{\Delta t}}={\frac {\Delta x}{\Delta t}}m\right)\\\Delta F+{\frac {1}{2}}&=\Delta x{\frac {m}{\Delta t}}&{\text{(Additive property of equality.)}}\\{\frac {\Delta t}{m}}\left(\Delta F+{\frac {1}{2}}\right)&=\Delta x&{\text{(Multiplicative inverse property of equality.)}}\\\Delta x&={\frac {\Delta t}{m}}\left(\Delta F+{\frac {1}{2}}\right)&{\text{(Symmetric property of equality.)}}\end{aligned}}$
Notice how we used different letters. Always keep in mind what the question wants you to find.
Example 2.2.c: Solve for $a$ in the following equation given below:
${\frac {1}{5}}+2a=15+{\frac {5}{2}}\cdot a$.
This example will be explaining every step. The repeat of this example will demonstrate the more concise argument. Play close attention to the repeat so that you may be able to digest the information. At the end of each example (except the word problems and long problems which require explanations), there will be a concise solving of the multistep equations presented.
Apply the subtraction property of equality so that the following is true:
$\Leftrightarrow {\frac {1}{5}}-15={\frac {5}{2}}a-2a$
Simplify and combine like terms:
$\Leftrightarrow -{\frac {224}{5}}={\frac {1}{2}}a$
Apply the multiplication property of equality by multiplying both sides by $2$:
$\Leftrightarrow -{\frac {448}{5}}=a$
Because $a=b\Leftrightarrow b=a$ by the symmetric property of equality, $a=-{\frac {448}{5}}\blacksquare$
Concise Argument
${\begin{aligned}{\frac {1}{5}}+2a&=15+{\frac {5}{2}}a\\{\frac {1}{5}}-15&={\frac {5}{2}}a-2a&{\text{(Addition property of equality.)}}\\-{\frac {224}{5}}&={\frac {1}{2}}a&{\text{(Simplify and combine like-terms.)}}\\-{\frac {448}{5}}&=a&{\text{(Multiplication property of equality.)}}\\a&=-{\frac {448}{5}}\blacksquare &{\text{(Symmetric property of equality.)}}\end{aligned}}$ :
The next problem will bring us back to the single-variable equation before introducing a harder variation of the single variable equation problem.
Example 2.2.d: Solve for $z$ in the following equation given below:
$z\cdot {\dfrac {-{\frac {1}{2z^{2}}}}{2+{\frac {1}{2z}}}}=-2$.
Fractions in algebra can be very tricky. That is why this is taught in a college algebra course. Knowing how to manipulate fractions will be very important when discussing rational functions. Although not its own section on the CLEP College Algebra exam, it will be very important to be able to simplify these functions so that one can easily determine its graph.
Apply the property of multiplication (proven in Intro to Algebraic Proofs) that $c\cdot {\frac {a}{b}}={\frac {ac}{b}}$:
${\dfrac {-z\cdot {\frac {1}{2z^{2}}}}{2+{\frac {1}{2x}}}}={\dfrac {\frac {1}{2z}}{2+{\frac {1}{2z}}}}=-2$
For the next step to make sense, we need to prove one property.
Proof: ${\dfrac {\frac {a}{b}}{c}}={\frac {a}{bc}}$
One theorem we will take as given for this proof is that $m^{-1}\cdot n^{-1}=(mn)^{-1}$ (referred to as multiples of base). This theorem will be proven in the Exponents chapters.
First, we rewrite ${\dfrac {\frac {a}{b}}{c}}$ as ${\frac {a}{b}}\div c$.
${\dfrac {\frac {a}{b}}{c}}={\frac {a}{b}}\div c$
After, we apply the definition of division:
$\Leftrightarrow {\frac {a}{b}}\div c=a(b)^{-1}(c)^{-1}$
We apply the multiples of base theorem
$\Leftrightarrow a(b)^{-1}(c)^{-1}=a(bc)^{-1}$
to conclude that
$\Leftrightarrow {\dfrac {\frac {a}{b}}{c}}={\frac {a}{bc}}\Box$
This can be given in the concise argument below:
${\begin{aligned}{\dfrac {\frac {a}{b}}{c}}&={\frac {a}{b}}\div c&{\text{(Definition of division.)}}\\&=a(b)^{-1}(c)^{-1}&{\text{(Definition of division.)}}\\&=a(bc)^{-1}&{\text{(Multiples of base theorem.)}}\\&={\frac {a}{bc}}\Box &{\text{(Definition of division.)}}\end{aligned}}$
From the above proof, we can finally use the multiples of base lemma (a "little" theorem used to prove what we wanted to show):
$\Leftrightarrow -{\dfrac {1}{2z\left(2+{\frac {1}{2z}}\right)}}=-2$
Apply the distributive property within the denominator of the fraction:
$\Leftrightarrow -{\frac {1}{4z+1}}=-2$
Apply the division property of equality by dividing both sides by $-1$:
$\Leftrightarrow {\frac {1}{4z+1}}=2$
The multiplicative property of equality will be used here. Another way to do the problem will be presented after:
$\Leftrightarrow 1=2(4z+1)$
Distribute the $2$ to $4z+1$:
$\Leftrightarrow 1=8z+2$
Subtract the $2$ and apply the subtractive property of equality:
$\Leftrightarrow -1=8z$
Finally, divide the $8$ and apply the division property of equality:
$\Leftrightarrow -{\frac {1}{8}}=z$
Because $-{\frac {1}{8}}=z$, by the symmetric property of equality, $z=-{\frac {1}{8}}\blacksquare$
Concise Explanation
${\begin{aligned}z\cdot {\dfrac {-{\frac {1}{2z^{2}}}}{2+{\frac {1}{2z}}}}&=-2\\{\dfrac {\frac {1}{2z}}{2+{\frac {1}{2z}}}}&=-2&\left(c\cdot {\frac {a}{b}}={\frac {ac}{b}}\right)\\-{\dfrac {1}{2z\left(2+{\frac {1}{2z}}\right)}}&=-2&\left({\dfrac {\frac {a}{b}}{c}}={\frac {a}{bc}}\right)\\-{\frac {1}{4z+1}}&=-2&{\text{(Distributive prop.)}}\\{\frac {1}{4z+1}}&=2&{\text{(Division prop. of equality)}}\\1&=2(4z+1)&{\text{(Multiplicative prop. of equality)}}\\1&=8z+2&{\text{(Distributive prop.)}}\\-1&=8z&{\text{(Subtraction prop. of equality.)}}\\-{\frac {1}{8}}&=z&{\text{(Division prop. of equality.)}}\\x&=-{\frac {1}{8}}\blacksquare &{\text{(Symmetric prop. of equality.)}}\end{aligned}}$ :
This example also has another trick that can be applied. In actuality, it is another property of equality not often in college algebra classes, and that is the following.
New property added later after new edit.
* Example 2.2.e: There is an object moving with a positively increasing velocity every $t$ seconds. The object that is $400{\text{ kg}}$ is constantly running at an acceleration of $5{\frac {\text{m}}{{\text{s}}^{2}}}$ along a linear path. If a photo gate located $T$ seconds from the car's starting position gives the runner's speed as $17{\frac {\text{m}}{\text{s}}}$, find the amount of time it took the car to reach that final speed, ${\overrightarrow {v}}$, given the initial photo gate recorded the car's initial speed at $2{\frac {\text{m}}{\text{s}}}$.
* Quite the scary looking problem, especially with all the units and numbers. Breathe in, and chunk the information you see above. Read one sentence first until you understand. Usually, information that is given should be read carefully. Another technique is to draw a sketch, as shown below.
* A car with the velocity-time graph and corresponding vector diagram is drawn so that it models the situation described in the word problem. The vector diagram is vertical so that space is saved. You may see a common theme with physics in College Algebra. Get used to that, because Physics loves Algebra.
* First, we need to ask, "How do we model equations to situations?" A good start is with the numbers and variables we have to work with. A good technique is to write down information that will be useful for the problem. Here, we will list ALL numbers and information that are in the problem which will help when calculating.
* Positively increasing velocity every $t$ seconds
* $400{\text{ kg}}$ object.
* Constant $5{\frac {\text{m}}{{\text{s}}^{2}}}$ acceleration.
* $17{\frac {\text{m}}{\text{s}}}$ at time $T$ seconds.
* Car's initial speed is at $2{\frac {\text{m}}{\text{s}}}$.
* All of the above information can be used when writing equations. Next, find out what the question is asking. According to the text above, it is asking us to "find the amount of time it took the car to reach that final speed, ${\overrightarrow {v}}$..." As such, since we are looking for the time it took to equal $17{\frac {\text{m}}{\text{s}}}$, we may use the third, fourth, and fifth bullet points to reach this equation:
* $17=5T+2$
* Let us make sure everyone understands what is going on here.
* Notice we have decided not to keep the units. This is to not confuse students; plus, we know what the units are meant to be at the end (seconds), so there is no need to show the work using units. Plus, it helps us to not be bogged down with so much detail. Always remember, however, to put down units at the end!
* We decided to put the variable of interest "next to" $5$. This is because $5$ is representative of the constantly increasing speed per second, $t$. The only way one would understand this is if one read the problem carefully (or listed the information like we did).
* The reason why we didn't use $t$ is because it is an interstitial variable. What we will be doing is basically "replacing" the variable with $T$ because that is what we want to determine. This idea of "replacing" will be formalized after Example 2.2.d.
* One other bit of information that is missing from this equation is the "$400{\text{ kg}}$". This one is quite simple: the problem is not asking about the force$^{[Footnote1]}$ on the car. Instead, it is asking for the time it took the car to reach the velocity at time $T$.
* Because the car started at $2{\frac {\text{m}}{\text{s}}}$, it would be impossible to change that speed per second. Similarly, the final speed cannot change in the end, so it would be impossible to change that per second. This is by definition.
* Either way, we are finally ready to begin isolating the variable. Do the same procedure as before. Here, let us add $-2$ to both sides of the equation to add the identity element to $5T$:
* $\Rightarrow 17+({\color {Red}-2})=5T+2+({\color {Red}-2})$
* $\Rightarrow 15=5T+({\color {Red}0})$
* $\Rightarrow 15=5T$.
* Finally, multiply both sides by the inverse operation of multiplying $5$ to $T$, ${\frac {1}{5}}$, so that way the identity element is multiplied to $T$:
* $\Rightarrow 15\cdot \left({\color {Red}{\frac {1}{5}}}\right)=5T\cdot \left({\color {Red}{\frac {1}{5}}}\right)$
* $\Rightarrow 3=({\color {Red}1})T$
* $\Rightarrow T=3$.
* The car took 3 seconds to reach the final velocity, ${\overrightarrow {v}}$
It turns out this special car can reach $17{\tfrac {\text{m}}{\text{s}}}$ in 3 seconds. If you are a bit confused as to how we reached the answer, well you should be. This is a preview into modeling equations, specifically linear equations. You will see another problem like this later. However, for right now, you are finished. The most confusing part about this problem is the units. It is certainly possible many students are not exposed to physics problems like these in math class, so these students may not get this problem correct simply because of the nature of the word problem.
On a brighter note, you may be noticing a pattern with how we are solving these problems. What we are effectively doing is a "backwards order of operations." The plan was to show three examples of this in action before simply giving the answer. Simply put, the person simplifying the equation will be doing so in a manner consistent with reversing the order by which operations are to be done systematically. Before doing this procedure, make sure to simplify the equation as much as possible. The ultimate demonstration of this procedure is shown in Example 2.2.g. For now, follow along with one more word problem.
* Example 2.2.f: Assume two blocks with masses $M$ and $m$ are attached to a pulley, where $M$ is on a wooden table with a coefficient of kinetic friction $\mu _{k}=0.5$, and $m$ is suspended in the air. Assuming the system has no change in speed or direction, the following equation models the situation:
* ${\overrightarrow {F_{f}}}-{\overrightarrow {F_{w}}}=0$
* where ${\overrightarrow {F_{f}}}=-gM\mu _{k}$ and ${\overrightarrow {F_{w}}}=-gm$. Based on the following situation, what is true about the mass of the object suspended in the air? Woah, what a scary looking problem! Most of the problem's "scare factor" comes from the situation in which the equation it models comes from. Most students will want to understand what the problem is talking about first before they do the problem. However, this is not Physics – this is College Algebra, so there is no need to worry too much. We will step you through the process so that one can see how to isolate the variable of interest. The equation in the problem is rewritten below:
* (1.2.2.1)
* ${\overrightarrow {F_{f}}}-{\overrightarrow {F_{w}}}=0$
* In case you never read a math textbook before, (1.2.2.1) denotes an equation we will be referring to later in this textbook. The decimals denote "chapter/section/subsection/order of equation reference."
* First things first, whatever we see is equal to another object, we will use the transitive property to make our lives easier. Case in point, ${\overrightarrow {F_{f}}}=-gM\mu _{k}$ and ${\overrightarrow {F_{w}}}=-gm$, so "plug those in" to equation 1.2.1.
* (1.2.2.2)
* $-gM\mu _{k}-\left(-gm\right)=0$
* Because of the associative and commutative property of multiplication, $-\left(-gm\right)=-1\left(-1\cdot gm\right)=-1\cdot -1\cdot g\cdot m=gm$, so the equation must be equivalent to
* (1.2.2.3)
* $-gM\mu _{k}+gm=0$
* The problem is looking for a truth about the mass suspended in the air, so isolate the variable $m$. Before we do that, notice that $g$ is found in both terms, $-gM\mu _{k}$ and $mg$, so based on the fact that both terms have a factor of $g$, $-gM\mu _{k}+gm=g\left(-M\mu _{k}+m\right)$. Have the factored form rewritten below:
* (1.2.2.4)
* $g\left(-M\mu _{k}+m\right)=0$
* Because of the zero factor property, $g=0$ or $-M\mu _{k}+m=0$.$^{[Footnote2]}$ Because of the given situation, we will ignore $g$ to find out more about $m$. Because of how $-M\mu _{k}+m=0$ is written, move $-M\mu _{k}$ to the other side by adding $M\mu _{k}$ to both sides of the equation. This will make it so that the identity element on the left side is added to $m$:
* (1.2.2.5)
* $-M\mu _{k}+\left({\color {Red}M\mu _{k}}\right)+m=0+\left({\color {Red}M\mu _{k}}\right)$
* (1.2.2.6)
* ${\color {Red}0}+m={\color {Red}M\mu _{k}}$
* Because the identity element does not change the left side, we may safely ignore it. From here, we are done. We know that the kinetic friction $\mu _{k}=0.5={\frac {1}{2}}$, so the mass of the block suspended in the air is half of the mass of the block on the wooden table. This would most likely be the answer you would find on the multiple choice question.
Here, we were able to use algebra to find out a truth about the system when there is no change in direction or speed. You can see why physics loves algebra, and it is for this precise reason. In fact, other fields of science love algebra as well. However, physics is perhaps the biggest lover. From our experience with equations, you may have noticed we used the properties we talked about in the beginning. If you skipped that section, or do not understand it, you are screwing yourself over. Algebra requires those fundamentals, so if you don't understand them, you must understand it now! After you finally understand those fundamental concepts, it is time to learn more about new definitions.
A single-variable equation is an equation in which only one variable of interest is shown throughout.
Examples of these equations are shown below:
* $2x+5=5x-2$
* ${\frac {7x}{10}}+{\frac {7}{10x}}=100$
Notice how the variable $x$ can be on both sides of the equation and yet still be called a single-variable equation.
A literal equation is one in which at least two unique variables are shown throughout.
Examples of these equations are shown below:
* ${\frac {\mathbf {F} }{m}}=\mathbf {a}$
* $\Delta \mathbf {x} =\mathbf {v} _{0}t+{\frac {1}{2}}\mathbf {a} t^{2}$
* $\Psi _{s}=-iCRT$, where pressure constant $R=0.0831$ liter/mole Kelvin.
Sometimes, the CLEP exam may ask you to interpret an equation. Information is given and you will have to solve for or interpret new information based on that equation. These would be the hardest problems of the exam.
Substituting is the act of replacing a variable with a value or other expression. This is informally referred to as "plugging in" for a variable.
In Example 2.2.e, because ${\overrightarrow {F_{f}}}-{\overrightarrow {F_{w}}}=0$, and ${\overrightarrow {F_{f}}}=-gM\mu _{k}$ and ${\overrightarrow {F_{w}}}=-gm$, we may substitute knowing what we know. This is how we got (1.2.2.3).
Because we have yet to introduce an example of a single variable equation with the variable of interest on both sides, let us do the problem below.
Example 2.2.g: Find the value for $x$ that makes the equation true:
${\frac {20(1+x)}{x}}=10$ This equation is something that is surprisingly simple to solve. We recommend you try this problem yourself before we explain how to do it, both as a test of your knowledge and as a way to check your answer.
One common student misconception that should be addressed swiftly is this: do not divide the $x$ within the parentheses. There are two reasons for this:
$x\nmid (1+x)$. The parentheses denotes them as two different numbers.
Even after distributing the 20 to each term in the parentheses, ${\frac {20+20x}{x}}\neq {\frac {40}{x}}$. Notice that ${\frac {20+20x}{x}}=(20+20x)\left(x^{-1}\right)=20x^{-1}+20x\left(x^{-1}\right)$, so in general, ${\frac {a}{b}}+{\frac {c}{b}}={\frac {a+c}{b}}$. Because of this truth, ${\frac {20+20x}{x}}={\frac {20}{x}}+{\frac {20x}{x}}={\frac {20}{x}}+20$. This is the only way to truly divide the expression by $x$. However, this method may actually be slower than the one we are about to show you.
If dividing by $x$ is slow, then how do we do this quickly? Notice how we are dividing $x$ with another $x$. A good way to eliminate them by moving the bottom $x$ to the other side. Because the inverse operation of division is multiplication, do that so that way the identity element is multiplied to the left hand side:
${\frac {20(1+x)}{x}}\cdot {\color {red}x}=10\cdot {\color {red}x}$
$\Rightarrow 20(1+x)=10{\color {red}x}$.
Some students may want to distribute the 20 to each term in the parentheses. While this may be fine when working with expressions, since we are working with an equation, let us make this process go faster by multiplying both sides by ${\frac {1}{10}}$. This can be done because $20(1-x)$ is the same as multiplying $20$ to some variable $1-x=C$, so ${\frac {1}{10}}\cdot 20C={\frac {20C}{10}}=2C$. As such,
$\Rightarrow 20(1+x)\cdot {\color {red}{\frac {1}{10}}}=10\cdot {\color {red}{\frac {1}{10}}}\cdot x$
$\Rightarrow {\color {red}2}(1+x)=x$.
Then, you can distribute the 2 to each term in the parentheses.
$\Rightarrow 2+2x=x$.
Now, add both sides with $-2x$. This is to isolate one side with $x$.
$\Rightarrow 2+2x+({\color {Red}-2x})=x+({\color {Red}-2x})$
$\Rightarrow 2={\color {Red}-x}$.
Because the $x$ is not isolated (you are multiplying $-1$ to the $x$), multiply both sides by ${\frac {1}{-1}}$.
$\Rightarrow 2\cdot ({\color {Red}{\frac {1}{-1}}})=-x\cdot ({\color {Red}{\frac {1}{-1}}})$
$\Rightarrow {\color {Red}-}2=x$.
Concise Explanation
${\begin{aligned}{\frac {20(1+x)}{x}}&=10\\20(1+x)&=10x&{\text{(Multiplication prop. of equality.)}}\\2(1+x)&=x&{\text{(Division prop. of equality.)}}\\2+2x&=x&{\text{(Distributive prop.)}}\\2&=-x&{\text{(Subtraction prop. of equality.)}}\\-2&=x&{\text{(Division prop. of equality.)}}\\x&=-2&{\text{(Symmetric prop. of equality.)}}\blacksquare \end{aligned}}$ :
In the above example, we mentioned that it may be faster to multiply both sides by $x$ to easily solve for $x$. Let us explore how solving for $x$ would be difficult for the below equation:
    ${\frac {20}{x}}+20=10$
First, notice that the side with $x$ is added by $20$, so it would be best to get rid of that constant. Add $-20$ to both sides:
    ${\frac {20}{x}}+20+({\color {red}-20})=-10+({\color {red}-20})$
    ${\frac {20}{x}}+{\color {red}0}={\color {red}-10}$
Notice that the above expression can be rewritten to this form: $20x^{-1}=-10$. Because every problem with a fraction can be thought of as simply multiplying, so by this very idea, one can isolate the $x$ by multiplying both sides by ${\frac {1}{20}}$.
    $\left({\color {red}{\frac {1}{20}}}\right)\cdot 20x^{-1}=-10\cdot \left({\color {red}{\frac {1}{20}}}\right)$
    $\left({\color {red}1}\right)\cdot x^{-1}=-\left({\color {red}{\frac {1}{2}}}\right)$
Because the reciprocal of $x$ is equal to $-{\frac {1}{2}}$, one could take the reciprocal of the reciprocal to both sides. That is to say, take the reciprocal of ${\frac {1}{x}}$ and $-{\frac {1}{2}}$. This would mean the solution set would be
    $x=-2$
You may not be certain doing one or the other is faster because of how the above example was framed. However, the way we did the example (boxed) is the faster way in terms of thought processing. A student intimately familiar with algebraic operations may do the above method (not in box), and may perhaps do it quickly. With the introduction of these algebraic operations, however, and presuming no intimate familiarity with algebra for the reader, we supposed it would be faster to do the boxed method.

### Solving for Multiple Variables

Notice how in Example 2.2.d, we were able to get rid of many variables by substituting information into what we know. This gave us a taste of solving multiple variables. You would be hard-pressed to find situations in which you do not need to solve for two variables. An example will perhaps be a good place to start with knowing about situations. Realize, however, that the solution will not be provided. We will be exploring the solution.
* Example 2.3.a: A couple has a shared fund where either John or Mary will set aside part of their income to contribute to necessary payments. John is willing to set aside 52% of his income to pay for the income while Mary is willing to set aside 35% of her income to pay for the necessary expenses. If the necessary expenses make up $800 per month, while the total overall income earned by the couple is $2,150 per month, what is the minimum amount Mary must set aside so that she fulfills her promise?
* This is quite an interesting problem. First, we need to know how much Mary and John are making. Of course, we could ask them for more details, but let us have some fun, instead. First, let us write down what we know is true:
* Mary and John make a combined income of $2,150 per month.
* Mary wants to set aside 35% of her income.
* John wants to set aside 52% of his income.
* The necessary expense is a total of $800. Mary and John want to set aside their percentage so they could pay for the expense as a whole.
* Since we do not know how much each is making, let us define some variables. Let $m$ represent Mary's monthly income and let $j$ represent John's monthly income. Since we know Mary and John make a combined income of $3,500 per month, the following equation below is true:
* (1.2.3.1)
* $m+j=2150$
* Since we know Mary and John set aside a certain percentage of their income towards monthly expenses, we know the following equation is also true:
* (1.2.3.2)
* $0.35m+0.52j=800$
* This is all the information we have. However, how do we determine the income of Mary and John?
This is quite the situation, and we will return to it later. However, as of right now, look at this as the challenge questions we will soon be able to solve. For now, let us work with an abstract space as opposed to the real world. Why do that? Because the real-world is messy while the abstract world follows our rules and our logic, although we must be precise with it.
* Example 2.3.b: If $2xy-4=-10$ and $y=-1$ are part of the system of equations, find $x$. Back into familiar territory once again. This problem seems a little similar to Example 2.2.d in terms of things for which we can substitute. This technique seems to work especially well in a situation like this and can make our lives especially easier. Let us bullet point the information we see in front of us into what is called a system of equations. We denote these systems using brackets on the left-hand side (usually). Sometimes, these systems can simply be denoted by writing the two equations on top of another.
* ${\begin{cases}2xy-4=-10\\y=-1\end{cases}}$
* Notice that this system has defined $y=-1$, so because it is a part of this system, we may use this information on $2xy-4=-10$. Before we do that, let us solve for $x$ on the first equation. We do the procedure as always. By now, you should be familiar enough with this algebraic manipulation such that you may not need us anymore. However, we will still explain some manipulations we believe need explaining. Let us begin using algebra on the first equation in the system:
* $2xy-4=-10$
* $\Rightarrow 2xy=-6$
* Recall that multiplication is commutative. Since we are trying to isolate for $x$, the variable of interest, and $2y$ is multiplied to $x$, it is possible to divide both sides by $2y$. The justification for this is simply a matter of associativity and commutativity, which we have explained thoroughly in Example 2.2.b. Keep in mind that this associativity and commutativity allows the student to divide like the following:
* ${\begin{aligned}\Rightarrow x&=-{\frac {6}{2y}}\\&=-{\frac {3}{y}}\end{aligned}}$
* This type of analysis will be explored further on the next page. Regardless, this helped us solve for $x$. Using what we know, $y=-1$, we may do the following to find the answer to the question:
* ${\begin{aligned}x&=-{\frac {3}{(-1)}}\\&=3\,\blacksquare \end{aligned}}$
What we have done is solve by substitution, which is a technique whereby an isolated variable on one of the equations of the system can be used to solve for the variable of interest on the other. Keep in mind, this can be done for any system of equation, and this also gives us information about what we can do with systems:
When given two equations, it is always possible to solve for a variable in one of the systems of equations and then substitute what you found into the other.
The word system suggests something ingrained, a mechanism with different parts that work together. If all of these parts work together to help find a solution for each variable within, then it may be possible to use those individual equations to help solve for another. Sometimes, the equations need not have two different variables. Take a look at the next problem.
* Example 2.3.c: If ${\frac {5}{x}}=200$, what is $54000-{\frac {200}{x}}$?
* It may seem unusual to use systems of equations to work these types of problems out. We believe most students will instinctively know how to work this problem out completely. Nevertheless, the relationship between the system and this problem is self-evident. There are two equations within this question; and we know to use both to solve this problem. Therefore, let the following be true:
* (1.2.3.3)
* ${\frac {5}{x}}=200$
* (1.2.3.4)
* $54000-{\frac {200}{x}}=y$
* We have two unknowns, $(x,y)$, and we want to find $y$, implicitly. To do that, we need to know what $x$ is. The only equation that can help us here is (1.2.3.3). First, we need to solve for $x$.
* ${\begin{aligned}{\frac {5}{x}}&=200\\&\Rightarrow 5x^{-1}=200\\&\Rightarrow x^{-1}=40\\&\Rightarrow x=40^{-1}={\frac {1}{40}}\end{aligned}}$
* Now that we know what $x$ equals to, we can finally find the solution set $(x,y)$. The only equation where we can find $y$ is (1.2.3.4). Therefore:
* ${\begin{aligned}54000-{\frac {200}{x}}&=54000-{\frac {200}{40^{-1}}}\\&=54000-200\left(40^{-1}\right)^{-1}\\&=54000-200\cdot 40&{\text{This step will be justified in the next chapter.}}\\&=54000-8000\\&=46000\blacksquare \end{aligned}}$
At times, however, solving by substitution is not always the most efficient. For example, in this situation, it would perhaps be better to use the fact that ${\frac {5}{x}}=200$ since(1.2.3.4) has the term ${\frac {200}{x}}$. The terms are similar by a constant multiple, $k$. By this logic, $k\cdot {\frac {5}{x}}={\frac {200}{x}}$ implies that $k={\frac {200}{5}}=40$. By the transitive property, ${\frac {5}{x}}=200$ implies that $40\cdot 200={\frac {200}{x}}=8000$. As such, $54000-{\frac {200}{x}}=54000-8000=46000$.
Most of the time, substitution will work the fastest. However, situations may appear where elimination of a variable may improve the situation dramatically. This option has a natural reason for it working, and it all has to do with operations.
* Example 2.3.d: If $mv={\frac {\sqrt {3}}{2}}mv_{1}+{\sqrt {2}}mv_{2}$ and $0={\frac {1}{2}}mv_{1}-{\sqrt {2}}mv_{2}$ are part of the system of equations, find $v_{1}$ and $v_{2}$ in terms of $v$.
* One could take the effort of substituting each information needed to find $v$ and $v_{1}$. However, the amount of information needed to answer this question is not enough with substitution alone. This is because of $v_{2}$, which requires more information in order to make it way easier to manage. We will have to look at these equations together. But first, we will label them.
* (1.2.3.5)
* $mv={\frac {\sqrt {3}}{2}}mv_{1}+{\sqrt {2}}mv_{2}$
* (1.2.3.6)
* $0={\frac {1}{2}}mv_{1}-{\sqrt {2}}mv_{2}$
* These two equations are part of a system. Whatever is true for one is true for the other. However, if these are part of a system, then when combining these two different relations, whatever action is done to one side must be done to the other side. This fundamental principle of algebra can be extended beyond just one system. We can sufficiently demonstrate this as such: let $A=mv$ and $B=0$. By the transitive property, $mv={\frac {\sqrt {3}}{2}}mv_{1}+{\sqrt {2}}mv_{2}$ or (1.2.3.5), and $0={\frac {1}{2}}mv_{1}-{\sqrt {2}}mv_{2}$ or (1.2.3.6), meaning:
* (1.2.3.7)
* $A=mv={\frac {\sqrt {3}}{2}}mv_{1}+{\sqrt {2}}mv_{2}$
* (1.2.3.8)
* $B=0={\frac {1}{2}}mv_{1}-{\sqrt {2}}mv_{2}$
* If we let $A$ and $B$ stand for different equations, because those equations each have their own relation to another side by the transitive property, then combining equations is a valid action. $\blacksquare$ (We learned an important fact within our problem that required demonstrating the fact to be true before we could use it. This is called a lemma, and it is often sneaky in where it shows up. Keep an eye out for lemmas.) From what we learned above, we can combine two different equations and it will not break anything. Therefore, let us find $A+B$:
* ${\begin{aligned}A+B&=mv+0\\&=\left({\frac {\sqrt {3}}{2}}mv_{1}+{\sqrt {2}}mv_{2}\right)+\left({\frac {1}{2}}mv_{1}-{\sqrt {2}}mv_{2}\right)\\&={\frac {\sqrt {3}}{2}}mv_{1}+{\sqrt {2}}mv_{2}+{\frac {1}{2}}mv_{1}-{\sqrt {2}}mv_{2}\\&=\left({\frac {\sqrt {3}}{2}}+{\frac {1}{2}}\right)mv_{1}+{\color {Red}{\sqrt {2}}mv_{2}-{\sqrt {2}}mv_{2}}\\&=\left({\frac {\sqrt {3}}{2}}+{\frac {1}{2}}\right)mv_{1}+{\color {Red}0}\end{aligned}}$
* Hopefully, by now, you can successfully follow along with the algebra without any need for guidance. Either way, we found the equation that results from adding the two relations, (1.2.3.7) and (1.2.3.8).
* (1.2.3.9)
* $A+B=mv=\left({\frac {\sqrt {3}}{2}}+{\frac {1}{2}}\right)mv_{1}$
* If we look back at the question, we want to find $v_{1}$ in terms of $v$, so let us do that. First simplify the equation. Notice that $m$ is a multiple on both sides of the equation, so it can safely be divided, assuming $m\neq 0$.
* $mv=\left({\frac {\sqrt {3}}{2}}+{\frac {1}{2}}\right)mv_{1}$
* $\Rightarrow v=\left({\frac {\sqrt {3}}{2}}+{\frac {1}{2}}\right)v_{1}$
* $\Rightarrow {\frac {v}{\left({\frac {\sqrt {3}}{2}}+{\frac {1}{2}}\right)}}=v_{1}$
* This step will be justified in the next chapter. We are done with the first part after this:
* $\Rightarrow {\frac {2v}{\left({\sqrt {3}}+1\right)}}=v_{1}$
* We have answered the first half of the question. However, we need to find $v_{2}$. We can do this easily with (1.2.3.8), since we can use the zero factor property. Let us rewrite the equation below.
* ${\begin{aligned}0&={\frac {1}{2}}mv_{1}-mv_{2}{\sqrt {2}}\\&=m\left({\frac {1}{2}}v_{1}-v_{2}{\sqrt {2}}\right)\end{aligned}}$
* $\Rightarrow m=0{\text{ or }}{\frac {1}{2}}v_{1}-v_{2}{\sqrt {2}}=0$
* We may safely ignore $m$ and work with the other factor.
* $\Rightarrow {\frac {1}{2}}\left({\frac {2v}{\left({\sqrt {3}}+1\right)}}\right)-v_{2}{\sqrt {2}}=0$
* Now solve for $v_{2}$ in terms of $v$:
* $\Rightarrow v_{2}{\sqrt {2}}={\frac {1}{2}}\left({\frac {2v}{\left({\sqrt {3}}+1\right)}}\right)$
* ${\begin{aligned}\Rightarrow v_{2}&={\frac {1}{\sqrt {2}}}\cdot {\frac {1}{2}}\left({\frac {2v}{\left({\sqrt {3}}+1\right)}}\right)\\&={\frac {1}{2{\sqrt {2}}}}\left({\frac {2v}{\left({\sqrt {3}}+1\right)}}\right)\\&={\frac {2v}{2{\sqrt {2}}\left({\sqrt {3}}+1\right)}}\\&={\frac {v}{{\sqrt {2}}\left({\sqrt {3}}+1\right)}}\end{aligned}}$
* We have found the two solutions to the problem: $v_{1}={\frac {2v}{\left({\sqrt {3}}+1\right)}}$ and $v_{2}={\frac {v}{{\sqrt {2}}\left({\sqrt {3}}+1\right)}}\blacksquare$.
The above problem demonstrates the next principle of a system of equations, which we proved to be true in the problem above as well. The usefulness of systems of equations is the ability to combine these two principles of systems of equations to help solve problems. The conveniences of using algorithms are one of the principles by which mathematicians have lived by because it can help easily solve problems that would be nearly inconvenient or even impossible to solve without them.
When given two equations, it is always possible to combine the two equations in a system.
Systems of equations can be combined or solved individually, and these two actions can be combined in any number of ways to easily and verily solve for the two variables for which we seek. If possible, it may be best to solve by elimination, which is to combine the two equations in such a way as to eliminate one variable and solve for the other through the combined equations. This was demonstrated in the above problem, and it is how we found a useful reason for using The Second Reality of Systems.
From what you learned above, you should be able to solve any problem involving two systems of equations. Before we get back to our original problem, let us revisit a problem; however, let us add more information and change the situation a little so that we may learn how physics has found itself in this book again:
* Example 2.3.e: Assume two blocks with non-zero masses $M$ and $m$ are attached to a pulley, where $M={\frac {3}{2}}m$ is on a wooden table with a coefficient of kinetic friction $\mu _{k}=0.5$, and $m$ is suspended in the air. The following equation models the situation:
* ${\begin{cases}\mathbf {F} _{T}+\mathbf {F} _{f}=Ma\\\mathbf {F} _{T}+\mathbf {F} _{w}=-ma\end{cases}}$
* where $\mathbf {F} _{f}=-gM\mu _{k}$, and $\mathbf {F} _{w}=-gm$, where $g$ is a constant approximately equal to $10{\tfrac {\text{m}}{{\text{s}}^{2}}}$. Based on the following situation, Which of the following must be true? Select all that apply.
* The force of friction, $\mathbf {F} _{f}$, of the block on the table would be nine times the mass $m$ if it were on the table.
* The force of tension, $\mathbf {F} _{T}$, of the block on the table is nine times the mass suspended in the air.
* The magnitude of the acceleration, $a$, of the system is $1{\tfrac {\text{m}}{{\text{s}}^{2}}}$.
* We are back again with the scary problem, only this time, with choices. The choices are a disguise of a certain problem: true-false. We want to determine which statement(s) is/are true, and only the true ones. The choices delimit our focus, and so we will focus intently on what it states. Start with the ones we can easily determine based on the information we have.
* True or false? The force of friction, $\mathbf {F} _{f}$, of the block on the table would be nine times the mass $m$ if it were on the table. False!
* This one is the easiest to evaluate because we have all the necessary information.
* $\mathbf {F} _{f}=-gM\mu _{k}$
* $M={\frac {3}{2}}m$
* $g\approx 10{\tfrac {\text{m}}{{\text{s}}^{2}}}$
* $\mu _{k}=0.5$
* This is just a simple substitution exercise. If this is true, then we have found one of our answers easily. If not, then we did not waste too much time.
* ${\begin{aligned}\mathbf {F} _{f}&=-gM\mu _{k}\\&=-{\frac {3}{2}}\cdot {\frac {1}{2}}\cdot 10m&{\text{[Clearly false, but let us continue.]}}\\&=-{\frac {3}{2}}\cdot 5m\\&=-{\frac {15}{2}}m\\\end{aligned}}$
* It was pretty clear that the force of friction would not have been 9 times the mass suspended in the air because neither 10, 2, or 5 were a multiple of 9. Nevertheless, the final answer simply cemented this fate.
* True or false?The acceleration, $a$, of the system is $1{\frac {\text{m}}{{\text{s}}^{2}}}$. True!
* Look at the system of equations above. Let $A={\color {Green}\mathbf {F} _{T}+\mathbf {F} _{f}=Ma}$ and $B={\color {Red}\mathbf {F} _{T}+\mathbf {F} _{w}=-ma}$. Is there any term in the equation that can be eliminated through the combination of the two equations? Yes, one can subtract $A$ from $B$ to get the following:
* ${\begin{aligned}A-B&=Ma-[-ma]=(M+m)a\\&=\left(\mathbf {F} _{T}+\mathbf {F} _{f}\right)-\left(\mathbf {F} _{T}+\mathbf {F} _{w}\right)\\&={\color {Red}\mathbf {F} _{T}}+\mathbf {F} _{f}-{\color {Red}\mathbf {F} _{T}}-\mathbf {F} _{w}\\&={\color {Red}\mathbf {F} _{T}}-{\color {Red}\mathbf {F} _{T}}+\mathbf {F} _{f}-\mathbf {F} _{w}\\&={\color {Red}0}+\mathbf {F} _{f}-\mathbf {F} _{w}\end{aligned}}$
* As we now know, the following equation models the situation:
* (1.3.3.10)
* $\mathbf {F} _{f}-\mathbf {F} _{w}=(M+m)a$
* As in Example 2.2.d, we can substitute information we see familiar in the problem above. In this instance,
* $\mathbf {F} _{f}=-gM\mu _{k}$ and $\mathbf {F} _{w}=-gm$ will be needed here. As such our next equation shows itself:
* (1.3.3.11)
* $-gM\mu _{k}-(-gm)=(M+m)a$
* Because of the associative and commutative property of multiplication, $-\left(-gm\right)=gm$ (see Example 2.2.d in case a student needs to remember why), so the equation must be equivalent to:
* (1.3.3.12)
* $-gM\mu _{k}+gm=(M+m)a$
* Notice that $g$ is a factor of both terms on the left-hand side, so $-gM\mu _{k}+gm=g\left(-M\mu _{k}+m\right)$. Have the factored form rewritten below:
* (1.3.3.13)
* $g\left(-M\mu _{k}+m\right)=(M+m)a$
* Recall that we are trying to solve for $a$, so let us go ahead and divide both sides by $M+m$:
* (1.3.3.14)
* ${\frac {g\left(-M\mu _{k}+m\right)}{M+m}}=a$
* At this point in time, we have all the information necessary to determine if the above statement is true. We know the following:
* $g=10{\tfrac {\text{m}}{{\text{s}}^{2}}}$
* $\mu _{k}=0.5$
* $M={\frac {3}{2}}m$
* These points above are information that can help us determine whether the acceleration is positive. This should be a good step in the right direction, because if the acceleration is negative, then we know that we are done with this problem and can move on to the other choice to evaluate if they are true. Either way, let us substitute information we know:
* ${\begin{aligned}a&={\frac {10\left(-{\frac {3}{2}}m\cdot {\frac {1}{2}}+m\right)}{{\frac {3}{2}}m+m}}\\&={\frac {10m\left(1-{\frac {3}{2}}\cdot {\frac {1}{2}}\right)}{m\left({\frac {3}{2}}+1\right)}}&{\text{(Eliminate the }}m{\text{.)}}\\&={\frac {10\left(1-{\frac {3}{4}}\right)}{{\frac {3}{2}}+1}}\\&={\frac {10\cdot {\frac {1}{4}}}{\frac {5}{2}}}\\&={\frac {\frac {5}{2}}{\frac {5}{2}}}\\&=1{\tfrac {\text{m}}{{\text{s}}^{2}}}\end{aligned}}$
* This is the final piece of information needed to find the answer. We wanted to evaluate whether the acceleration is indeed $1{\tfrac {\text{m}}{{\text{s}}^{2}}}$, and sure enough we found information that stated this is a possible answer.
* True or false?The force of tension, $\mathbf {F} _{T}$, of the block on the table is nine times the mass suspended in the air. True!
* This is a strange statement. Based on the system of equations, if $\mathbf {F} _{T}$ is not different for one or the other, then how is it possible for the tension to be different for each block. However, a cursory glance is not an analysis, and it is indeed certain that our conclusion is too hasty to be true. Look back at the system of equations:
* ${\begin{cases}\mathbf {F} _{T}+\mathbf {F} _{f}=Ma\\\mathbf {F} _{T}+\mathbf {F} _{w}=-ma\end{cases}}$
* Now that we know what the acceleration equals, we can use that information to answer the question involving the force of tension for the block of mass $M$ (meaning we have to use the first equation to keep it in terms of $M$). However, in physics, it is often best to keep the solution general to make sure we learn something new with what we already know. Keeping in mind $\mathbf {F} _{T}$ is unknown, $\mathbf {F} _{f}=-gM\mu _{k}$, and $a={\frac {g\left(-M\mu _{k}+m\right)}{M+m}}=1$. By what we know:
* (1.3.3.15)
* $\mathbf {F} _{T}-gM\mu _{k}=M{\frac {g\left(-M\mu _{k}+m\right)}{M+m}}$
* Here, $g$ is a factor on the right side of the equation, because
* ${\begin{aligned}{\frac {g\left(-M\mu _{k}+m\right)}{M+m}}&=\left(g\left(-M\mu _{k}+m\right)\right)\left(M+m\right)^{-1}\\&=g\left(-M\mu _{k}+m\right)\left(M+m\right)^{-1}&{\text{}g{\text{ is on the very left here.]}}\\&=g{\frac {-M\mu _{k}+m}{M+m}}\\&=g{\frac {m-M\mu _{k}}{M+m}}&\left[-M\mu _{k}+m=m-M\mu _{k}\right]\end{aligned}}$
* We can take the opportunity to solve for $\mathbf {F} _{T}$ found in (1.3.3.15).
* $\Rightarrow \mathbf {F} _{T}=Mg{\frac {\left(-M\mu _{k}+m\right)}{M+m}}+Mg\mu _{k}$
* $\Rightarrow \mathbf {F} _{T}=Mg\left({\frac {-M\mu _{k}+m}{M+m}}+\mu _{k}\right)$
* Notice how Equation (1.3.3.14) shows up again on the right hand-side, only this time missing a factor of $g$. Because ${\frac {g\left(-M\mu _{k}+m\right)}{M+m}}=a$, we may easily rewrite the above equation like so:
* $\Rightarrow \mathbf {F} _{T}=Mg\left({\frac {a}{g}}+\mu _{k}\right)$
* Recall how $M={\frac {3}{2}}m$, $g=10{\tfrac {\text{m}}{{\text{s}}^{2}}}$, $\mu _{k}=0.5$, and $a=1{\tfrac {\text{m}}{{\text{s}}^{2}}}$. Using what we know, we may safely conclude that
* ${\begin{aligned}\mathbf {F} _{T}&=Mg\left({\frac {a}{g}}+\mu _{k}\right)\\&=Ma+M\mu _{k}g\\&={\frac {3}{2}}ma+{\frac {3}{2}}mg\mu _{k}\\&={\frac {3}{2}}m\left(a+g\mu _{k}\right)\\&={\frac {3}{2}}m\left(1+10\cdot {\frac {1}{2}}\right)\\&={\frac {3}{2}}m\left(6\right)\\&=9m{\text{ Newtons}}\end{aligned}}$
* According to the above calculations, $\mathbf {F} _{T}$ in terms of $m$ is nine times the mass suspended in the air. Taken literally, this is true!
We have found useful applications of systems of equations in both real-life scenarios, abstract algebraic problems, and proofs. Nevertheless, we have still yet to learn one more tool to work with. This abstract problem below will be the last demonstration of it before we get to the two explorations.
* Example 2.3.f: Let an initial positive number be multiplied by the mixed number $1{\tfrac {1}{3}}$ with that result being divided by the sum of the initial number and $1{\tfrac {1}{3}}$. Then, subtract the result of that by the ratio of another number to the original. This operation results in $50$. Given twice the reciprocal of twice the sum between the initial number and the mixed number $1{\tfrac {1}{3}}$ minus the result of that to half the other number will equal $50$ times half the original number, find the greater.
* This type of question is hard enough to answer for the average student since most students have experience with answering questions that are usually straightforward with some context. However, this question is something one would see if one were in a math competition (despite the confusing language).
* Here is a nice tip to keep in mind. Anytime the problem seems to be wordy, make it into an equation! This makes problems so much easier, as you hopefully figured out by now. Let us go ahead and decode this problem. Let the initial number be $x$ and the other number be $y$. Let us go ahead figure out what the problem says.
* Note: a mixed fraction of $1{\tfrac {1}{3}}$ is the same as $1+{\frac {1}{3}}={\frac {3}{3}}+{\frac {1}{3}}={\frac {4}{3}}$, so we will be using this from here on.
* Let an initial positive number be multiplied by the mixed number $1{\tfrac {1}{3}}$ with that result being divided by the sum of the initial number and $1{\tfrac {1}{3}}$.
* In simple terms, $x$ is multiplied by ${\frac {4}{3}}$. Afterwards, that product is then divided by the sum of those two numbers. Therefore, the expression is
* ${\frac {{\tfrac {4}{3}}x}{x+{\tfrac {4}{3}}}}$
* Then, subtract the result of that by the ratio of another number to the original.
* Subtract the previous expression with ${\frac {y}{x}}$, leaving you with a continuation:
* ${\frac {{\tfrac {4}{3}}x}{x+{\tfrac {4}{3}}}}-{\frac {y}{x}}$
* This operation results in $50$.
* Make this expression equal to $50$, thereby giving you the first of many equations in this problem.
* (1.2.3.16)
* ${\frac {{\tfrac {4}{3}}x}{x+{\tfrac {4}{3}}}}-{\frac {y}{x}}=50$
* Given twice the reciprocal of twice the sum between the initial number and the mixed number $1{\tfrac {1}{3}}$ minus the result of that to half the other number will equal $50$ times half the original number...
* That is a little too much, so let us further chunk that information.
* "Given twice the reciprocal of twice the sum between the initial number and the mixed number $1{\tfrac {1}{3}}$..."
* Multiply the reciprocal of two times $x+{\tfrac {4}{3}}$ by two. Meaning multiply ${\frac {1}{2\left(x+{\tfrac {4}{3}}\right)}}$ by two, thereby giving you the partial expression
* $2\cdot {\frac {1}{2\left(x+{\tfrac {4}{3}}\right)}}$
* Note: although some of you may want to eliminate the two multiplied to the fraction by the two in the denominator, it may be a good idea to keep the factor of two, at least until we see the entire equation.
* "...minus the result of that to half the other number..."
* Subtract the previous expression by ${\frac {y}{2}}$.
* $2\cdot {\frac {1}{2\left(x+{\tfrac {4}{3}}\right)}}-{\frac {y}{2}}$
* "...will equal $50$ times half the original number..."
* Make the expression we have now equal to $50\cdot {\frac {x}{2}}$, thereby giving us our next equation:
* (1.2.3.17)
* $2\cdot {\frac {1}{2\cdot \left(x+{\tfrac {4}{3}}\right)}}-{\frac {y}{2}}=50\cdot {\frac {x}{2}}$
* Despite these scary-looking equations, we have made our lives much easier. Now we simply have to find one of the numbers, whether the initial or the other. In this instance, it would be much easier to find the initial number because we can find a way to eliminate the other.
* Recall how we simply just allowed the two equations to be added upon the other. We justified this through the transitive property. Plus, we introduced equations through the principle that whatever is done to one side must be done to the other due to the property of equality. Well, since adding and multiplying are definite candidates, then one can say that multiplying the entire equation would work out in our favor. Therefore, multiplying the two equations by any factor is okay.
* Because this principle works in isolation, we need to make sure it works when combined with what we already know. Let the following be true:
* (1.2.3.18)
* $A={\frac {{\tfrac {4}{3}}x}{x+{\tfrac {4}{3}}}}{\color {Red}-{\frac {y}{x}}}=50$
* (1.2.3.19)
* $S_{1}=2\cdot {\frac {1}{2\cdot \left(x+{\tfrac {4}{3}}\right)}}-{\frac {y}{2}}=50\cdot {\frac {x}{2}}$
* Our goal here is to transform $-{\frac {y}{2}}$ to ${\frac {y}{x}}$, so that we may do a version of the following operation: $A+S_{1}$. This will eliminate ${\frac {y}{x}}$, highlighted in red in (1.2.3.18).
* First, notice there is a common factor of ${\frac {1}{2}}$ in $S_{1}$. Since ${\frac {y}{x}}$ does not have a factor of ${\frac {1}{2}}$, we may multiply $S_{1}$ by $2$ to have a common factor of $1$ between $-y$ and ${\frac {y}{x}}$. The operation below will color the thing of interest we are trying to eliminate in red (to the best of our abilities). This gives us the following:
* ${\begin{aligned}2S_{1}&=2\left[2\cdot {\frac {1}{2\cdot \left(x+{\tfrac {4}{3}}\right)}}-{\color {Red}{\frac {y}{2}}}\right]&\left[{\text{Recall }}C{\tfrac {1}{a}}={\tfrac {C}{a}}{\text{.}}\right]\\&=2\left[{\frac {2}{2\cdot \left(x+{\tfrac {4}{3}}\right)}}-{\color {Red}{\frac {y}{2}}}\right]&{\text{[Recall the distributive property.]}}\\&={\color {Green}2}\cdot \left[{\frac {2}{{\color {Green}2}\cdot \left(x+{\tfrac {4}{3}}\right)}}\right]-{\color {Green}2}\left({\frac {\color {Red}y}{\color {Green}2}}\right)&{\text{[Green colors will be eleminated.]}}\\&={\frac {2}{x+{\tfrac {4}{3}}}}-{\color {Red}y}\end{aligned}}$
* This same principle is applied to the right-hand side of Equation (1.2.3.17), and since the exercise for that is trivial, we will leave all the calculations in between the final answer to the reader as an exercise. From here, this gives us out next equation.
* (1.2.3.20)
* $2S_{1}={\frac {2}{x+{\tfrac {4}{3}}}}{\color {Red}-y}=50x$
* The highlighted variable of interest in red has a factor of $-1$ that is not found in the goal, ${\frac {y}{x}}$, so one will need to divide $2S_{1}$ by $-1$ so that it cancels out and gets us closer to ${\frac {y}{x}}$. (Alternatively, one could multiply by $-1$ because of a basic principle of multiplication — a negative times a negative equals a positive.) We will leave this critical operation as an exercise for the reader since the difficulty of this step is trivial.
* One factor missing as a result of this manipulation is ${\frac {1}{x}}$ since the only difference between $y$ and ${\frac {y}{x}}$ is the factor of ${\frac {1}{x}}$. As such, this needs to be multiplied onto $-2S_{1}$. As such, the following operation is true:
* ${\begin{aligned}{\frac {1}{x}}\cdot \left(-2S_{1}\right)&={\frac {1}{x}}\left[-{\frac {2}{x+{\tfrac {4}{3}}}}+{\color {Red}y}\right]\\&={\frac {1}{x}}\cdot \left[-{\frac {2}{x+{\tfrac {4}{3}}}}\right]+{\frac {1}{x}}\cdot {\color {Red}y}\\&=-{\frac {2}{x\left(x+{\tfrac {4}{3}}\right)}}+{\color {Red}{\frac {y}{x}}}\end{aligned}}$
* Our plan has finally been revealed — the goal is accomplished. Let $-{\frac {2}{x}}S_{1}=S_{2}$. By what we have proven in Example 2.3.d, the lemma shows that we can combine to equations through addition or subtraction. However, based on what was stated, when doing the operation $A+S_{2}$, by the transitive property, $A+S_{2}=A-{\frac {2}{x}}S_{1}$, meaning we can combine two equations when multiplying by either a constant or variable. This is our lemma!
* The operation is true. First, we will write the result of our work through the problem. (Also, the other side of the equation is also left as another trivial exercise for the reader.)
* (1.2.3.21)
* $-{\frac {2}{x}}S_{1}=-{\frac {2}{x\left(x+{\tfrac {4}{3}}\right)}}+{\color {Red}{\frac {y}{x}}}=-50$
* All that is left to do is do the operation we have been waiting for: $A-{\frac {2}{x}}S_{1}$
* ${\begin{aligned}A-{\frac {2}{x}}S_{1}&={\frac {{\tfrac {4}{3}}x}{x+{\tfrac {4}{3}}}}{\color {Red}-{\frac {y}{x}}}-{\frac {2}{x\left(x+{\tfrac {4}{3}}\right)}}+{\color {Red}{\frac {y}{x}}}=50-50\\&={\frac {{\tfrac {4}{3}}x}{x+{\tfrac {4}{3}}}}-{\frac {2}{x\left(x+{\tfrac {4}{3}}\right)}}=0\end{aligned}}$
* (1.2.3.22)
* $A-{\frac {2}{x}}S_{1}={\frac {{\tfrac {4}{3}}x}{x+{\tfrac {4}{3}}}}-{\frac {2}{x\left(x+{\tfrac {4}{3}}\right)}}=0$
* This is all but the first phase of our plan. Now we need to solve for $x$. It would be nice to evaluate the expression with a common denominator because we can make use of the zero factor theorem. After all, the denominator cannot be set equal to zero, yet the fractional term is zero. Therefore, the numerator must be zero. It would be easy to evaluate if the fraction was singular.
* Notice how the denominator term has a common factor of $x+{\frac {4}{3}}$ (in red). Because of the way the problem was designed, the missing factor is $x$. As such, it would be best to multiply ${\frac {x}{x}}$ to the first fraction. If you don't understand, it may be best to follow along below:
* ${\begin{aligned}{\frac {{\tfrac {4}{3}}x}{\color {Red}x+{\tfrac {4}{3}}}}-{\frac {2}{x\left({\color {Red}x+{\tfrac {4}{3}}}\right)}}&={\frac {x}{x}}\cdot {\frac {{\tfrac {4}{3}}x}{\color {Red}x+{\tfrac {4}{3}}}}-{\frac {2}{x\left({\color {Red}x+{\tfrac {4}{3}}}\right)}}\\&={\frac {{\tfrac {4}{3}}x^{2}}{\color {Green}x\left(x+{\tfrac {4}{3}}\right)}}-{\frac {2}{\color {Green}x\left(x+{\tfrac {4}{3}}\right)}}&{\text{(The common denominator is highlighted in green.)}}\\&={\frac {{\tfrac {4}{3}}x^{2}-2}{x\left(x+{\tfrac {4}{3}}\right)}}=0\end{aligned}}$
* (1.2.3.23)
* ${\frac {{\tfrac {4}{3}}x^{2}-2}{x\left(x+{\tfrac {4}{3}}\right)}}=0$
* Because the fractional term is equal to zero, this must mean the numerator must equal zero. Therefore, we will only look at the numerator to make our life easier:
* (1.2.3.24)
* ${\frac {4}{3}}x^{2}-2=0$
* The following factorized form may appear to come out of nowhere. However, this will be explained in the Polynomials chapter of this Wikibooks. For now, simply pretend you understand and follow along:
* ${\begin{aligned}{\frac {4}{3}}x^{2}-2&={\frac {4}{3}}\left(x^{2}-{\frac {3}{2}}\right)\\&={\frac {4}{3}}\left(x-{\frac {\sqrt {3}}{\sqrt {2}}}\right)\left(x+{\frac {\sqrt {3}}{\sqrt {2}}}\right)\\&={\frac {4}{3}}\left(x-{\frac {\sqrt {6}}{2}}\right)\left(x+{\frac {\sqrt {6}}{2}}\right)=0\end{aligned}}$
* (1.2.3.25)
* $x-{\frac {\sqrt {6}}{2}}=0$ or $\Rightarrow x+{\frac {\sqrt {6}}{2}}=0$
* The problem states we started with an initial positive number. Therefore, we will only look at the equation that will give a positive solution to this equation.
* $x-{\frac {\sqrt {6}}{2}}=0$
* $\Rightarrow x={\frac {\sqrt {6}}{2}}$
* Before declare we are done with the problem, let us make sure we are being mathematically valid. We need to make sure we are not falling into the trap of making (1.2.3.23) a ${\frac {0}{0}}$ situation. To do that, we may use the zero factor theorem to our advantage. Because we must not allow the denominator to equal $0$, if we set the denominator to zero, we know what $x$ cannot equal. If our solution $x$ turns out to make the denominator equal to zero, then we have no solution to the original problem of the question asked.
* (1.2.3.26)
* $x\left(x+{\tfrac {4}{3}}\right)\neq 0$
* Per the zero factor theorem, $x\neq 0$ and $x+{\tfrac {4}{3}}\neq 0$. "Solving" for $x$ tells us that our solution does not make the denominator equal to zero. Although we will not explicitly prove it, it is obvious that $x={\frac {\sqrt {6}}{2}}$ is a valid solution.
* Some of our readers might think the problem is over. However, we are barely in phase two of our plan to find the solution. The problem has asked us to find the greater or the number that is the greatest and to declare the magnitude of that number. As such, we need to solve for $y$ and compare it to $x$ before we say we are finished with the problem. This may waste time if we found the greater number, but it will make us lose points if we find the wrong number, so it is better to determine the answer.
* Since (1.2.3.18) and (1.2.3.19) have some $y$, we will use one of those two equations. If a person is clever, one might try to work with one of the simplified equations. We will allow that student to be clever. We will choose to go with (1.2.3.18) since we will not have to substitute for $x$ a lot, which will make calculations easier as a result. No matter what equation the student chooses (that has a $y$), the calculations will be ugly.
* (1.2.3.27)
* ${\frac {{\tfrac {4}{3}}\left({\tfrac {\sqrt {6}}{2}}\right)}{\left({\tfrac {\sqrt {6}}{2}}\right)+{\tfrac {4}{3}}}}-{\frac {y}{\left({\tfrac {\sqrt {6}}{2}}\right)}}=50$
* Let us hope the marking scheme we have devised below is easy to follow. Any numbers that can be eliminated will be highlighted in Red while terms that can be factored will be in Green. Also, if you recall, the square root function has some strange property whereby multiplying the number to itself will result in the number inside the square root (such operations are colored in Orange). With this information in mind, let us apply some math:
* ${\begin{aligned}{\frac {{\tfrac {\color {Green}4}{3}}\left({\tfrac {\sqrt {6}}{\color {Green}2}}\right)}{\left({\tfrac {\sqrt {6}}{2}}\right)+{\tfrac {4}{3}}}}-{\frac {y}{\left({\tfrac {\sqrt {6}}{2}}\right)}}&={\frac {\tfrac {2{\sqrt {6}}}{3}}{\tfrac {3{\sqrt {6}}+8}{6}}}-{\frac {2y}{\sqrt {6}}}\\&={\frac {{\color {Green}6}\cdot 2{\sqrt {6}}}{{\color {Green}3}(3{\sqrt {6}}+8)}}-{\frac {2y}{\color {Orange}{\sqrt {6}}}}\cdot {\frac {\sqrt {6}}{\color {Orange}{\sqrt {6}}}}\\&={\frac {2\cdot 2{\sqrt {6}}}{3{\sqrt {6}}+8}}-{\frac {{\color {Green}2}y{\sqrt {6}}}{\color {Green}6}}\\&={\frac {4{\sqrt {6}}}{3{\sqrt {6}}+8}}-{\frac {y{\sqrt {6}}}{3}}\\&={\frac {3(4{\sqrt {6}})-y{\color {Orange}{\sqrt {6}}}(3{\color {Orange}{\sqrt {6}}}+8)}{3(3{\sqrt {6}}+8)}}\\&={\frac {3\cdot 4{\sqrt {6}}-y\cdot 6\cdot 3-y{\sqrt {6}}\cdot 8}{3\cdot 3{\sqrt {6}}+3\cdot 8}}\\&={\frac {12{\sqrt {6}}-18y-8y{\sqrt {6}}}{9{\sqrt {6}}+24}}\\&={\frac {12{\sqrt {6}}-(18+8{\sqrt {6}})y}{9{\sqrt {6}}+24}}=50\\\end{aligned}}$
* (1.2.3.28)
* ${\frac {12{\sqrt {6}}-(18+8{\sqrt {6}})y}{9{\sqrt {6}}+24}}=50$
* We are nearly finished with our plan. We simply need to solve for $y$. The easy way to do it is by multiplying both sides by the denominator of the left side of Equation (1.2.3.28). From there, it is easy to show that $y<x$ without needing to complete the calculation fully. We will leave that step as a non-trivial exercise for the reader.
* With the information we have provided, it is safe to say that $x>y$ and so the greater is $x$. In terms of magnitude, the greater is
* $x={\frac {\sqrt {6}}{2}}\blacksquare$
This example was used to introduce the important proof of concept, so to speak, involving the manipulations allowed over a system of equations. If one knows about the induction hypothesis, one can easily prove the manipulations we have formulated regarding the system of equations applies for any system involving $n$ equations. Either way, our final reality of systems reveals itself:
When given two equations, it is always possible to combine the two equations in a system such that at least one equation in that system was multiplied by a constant.
The three realities described herein will work for any problem involving systems of equations in some way. We will explore these ideas more in the second set of chapters in this Wikibooks, including the idea of more than two equations. For now, let us explore this concept to its fullest.
$m$
$m$
$j$
$-0.52$
$m$
$-0.17m=-318$
$1170.59$
$598.59$
$m$
$m$
$j$
$-0.52$
$m$
$-0.17m=-318$
$1170.59$
$598.59$
Note: required for the next exploration is the difference of squares formula: $x^{2}-y^{2}=(x+y)(x-y)$. An explanation for the formula can be found in our Polynomials chapter.
$s$
$a$
$b$
${\frac {a+b}{2}}={\frac {2\left(a+b\right)}{a-b}}{\text{ where }}a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)=4$ Since there are two equations that have the same variables, put them into a systems of equations: ${\begin{cases}{\frac {1}{2}}\left(a+b\right)=2{\frac {\left(a+b\right)}{a-b}}\\\left(a-b\right)\left(a+b\right)=4\end{cases}}$ The following statement is also true due to cross multiplying: $\left(a+b\right)\left(a-b\right)=4\left(a+b\right)$ From there, solve:
    $\left(a+b\right)\left(a-b\right)=4=4\left(a+b\right)$
    $\Rightarrow 1=\left(a+b\right)$
Substitute the value into the system of equation for the second equation and solve for $a-b$:
    $\left(a-b\right)\left(a+b\right)=4$
    $\Rightarrow \left(a-b\right)\cdot 1=4$
    $\Rightarrow \left(a-b\right)=4$
After wards, put the two derived equations into another systems of equation and solve down for $a$ and $b$: ${\begin{cases}\left(a-b\right)=4\\\left(a+b\right)=1\end{cases}}$ Add the two equations and solve for $a$:
    $2a=5\Rightarrow a={\frac {5}{2}}=2{\frac {1}{2}}$.
Substitute into either equation and solve for $b$: $a+b=1\Rightarrow {\frac {5}{2}}+b=1\Rightarrow b=-{\frac {3}{2}}=-1{\frac {1}{2}}$.
$b<a$
$b=s=-1{\frac {1}{2}}$
$a$
$b$
${\frac {a+b}{2}}={\frac {2\left(a+b\right)}{a-b}}{\text{ where }}a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)=4$ Since there are two equations that have the same variables, put them into a systems of equations: ${\begin{cases}{\frac {1}{2}}\left(a+b\right)=2{\frac {\left(a+b\right)}{a-b}}\\\left(a-b\right)\left(a+b\right)=4\end{cases}}$ The following statement is also true due to cross multiplying: $\left(a+b\right)\left(a-b\right)=4\left(a+b\right)$ From there, solve:
    $\left(a+b\right)\left(a-b\right)=4=4\left(a+b\right)$
    $\Rightarrow 1=\left(a+b\right)$
Substitute the value into the system of equation for the second equation and solve for $a-b$:
    $\left(a-b\right)\left(a+b\right)=4$
    $\Rightarrow \left(a-b\right)\cdot 1=4$
    $\Rightarrow \left(a-b\right)=4$
After wards, put the two derived equations into another systems of equation and solve down for $a$ and $b$: ${\begin{cases}\left(a-b\right)=4\\\left(a+b\right)=1\end{cases}}$ Add the two equations and solve for $a$:
    $2a=5\Rightarrow a={\frac {5}{2}}=2{\frac {1}{2}}$.
Substitute into either equation and solve for $b$: $a+b=1\Rightarrow {\frac {5}{2}}+b=1\Rightarrow b=-{\frac {3}{2}}=-1{\frac {1}{2}}$.
$b<a$
$b=s=-1{\frac {1}{2}}$
More explorations will be added later.

### Solving for Variables in Inequalities

There are many times where we need to know about what values for some variables could tell us about the restriction of the overall operation. For example, we may need to know about the restriction of the variable so that we may be allowed to meet some certain criteria.
An inequality is an equality where one side may not precisely equal the other. That is to say that the one side may be greater than ($>$) or less than ($<$) the other. From these, truths can extend. This possibility gives us our first property.
For any two real numbers $a$ and $b$, there exists only one of these possibilities:
* $a<b$
* $a=b$
* $a>b$
This property states that no two numbers can ever be less than and equal to. This is impossible. This is what is referred to as a strict inequality.
There exists a greater than or equal to ($\geq$) and a less than or equal to ($\leq$). However, for a greater than or equal to, it is simply saying that $a\geq b$, where either $a$ is greater than $b$ or $a=b$. It is not saying it is both, but a possibility of both exists. The same reasoning applies for a less than or equal to. As such, a number can only ever be in one of the states above.
For any three real numbers $a$, $b$, and $c$, there exists only one of these possibilities:
* If $a\leq b\leq c$, then $a\leq c$.
* If $a\geq b\geq c$, then $a\geq c$.
For when either premise is a strict inequality, then the conclusion is a strict inequality:
* If $a\leq b<c$, then $a<c$.
* If $a<b\leq c$, then $a<c$.
* If $a\geq b>c$, then $a>c$.
* If $a>b\geq c$, then $a>c$.
This property should make sense. Suppose $a<b<c$. Because $b<c$ and $a<b$, we can safely conclude $a<c$. This property is very important when it comes to finding inequalities that are similar, and especially when wanting to show that some inequality may be impossible. Notice how $a<b<c$ is equivalent to "$a<b$and $b<c$." This is called chained notation.
For any two real numbers $a$ and $b$, we can conclude that:
* If $a<b$, then $b>a$.
* If $a>b$, then $b<a$.
This should make sense. If $-3<10$, then obviously $10>-3$. Because this is a property, it is defined true.
For any three real numbers $a$, $b$, and $c$, adding any real number $c$ to both sides of an inequality give the following result:
* For $a<b$, $a+c<b+c$.
* For $a>b$, $a+c>b+c$.
Let us say that a number exists in the state whereby $a<b$. If adding another real number $c$ to both sides of the inequality, then $a+c<b+c$. This makes sense because if $a<b$, then adding another constant to the both sides will not in any effect change the inequality because both grew by the same amount. As such, $a$ is still less than $b$.
There is similar idea to this for multiplying the two sides by the same real number. However, it is a little more complex than that. We need to look at the cases for $c>0$ and $c<0$. Let there be two real numbers $a$ and $b$ such that $a<b$. If multiplying by $c>0$ to both sides, then $ac<bc$ because both sides are multiplied by the same number, so the product must still be $ac<bc$.
For any three real numbers $a$, $b$, and $c\neq 0$, multiplying real number $c$ to both sides of an inequality give the following result:
* If $a<b$ and $c>0$, then $ac<bc$.
* If $a<b$ and $c<0$, then $ac>bc$.
* If $a>b$ and $c>0$, then $ac>bc$.
* If $a>b$ and $c<0$, then $ac<bc$.
A similar idea for division exists for any $c\neq 0$
For any three real numbers $a$, $b$, and $c\neq 0$, dividing by the real number $c$ to both sides of an inequality give the following result:
* If $a<b$ and $c>0$, then ${\frac {a}{b}}<{\frac {b}{c}}$.
* If $a<b$ and $c<0$, then ${\frac {a}{b}}>{\frac {b}{c}}$.
* If $a>b$ and $c>0$, then ${\frac {a}{b}}>{\frac {b}{c}}$.
* If $a>b$ and $c<0$, then ${\frac {a}{b}}<{\frac {b}{c}}$.
These properties have their special cases that are very useful in some problems, especially with systems of inequalities.
For any two real numbers $a$ and $b$, multiplying both sides by $-1$ to both sides of an inequality give the following result:
* If $a<b$, then $-a>-b$.
* If $a\leq b$, then $-a\geq -b$.
* If $a>b$, then $-a<-b$.
* If $a\geq b$, then $-a\leq -b$.
For any two real non-zero numbers $a$ and $b$, making both sides its additive inverse makes it so that:
* If $a<b$, then ${\frac {1}{a}}>{\frac {1}{b}}$.
* If $a\leq b$, then ${\frac {1}{a}}\geq {\frac {1}{a}}$.
* If $a>b$, then ${\frac {1}{a}}<{\frac {1}{a}}$.
* If $a\geq b$, then ${\frac {1}{a}}\leq {\frac {1}{a}}$.
While these properties are much more plentiful than the properties of equalities, it is important to understand these properties, especially when working with systems of equations.
More will be added later.

### Exercises

Directions: Some questions will require you to select from among five choices. For these questions, select the BEST of the choices given. Some questions will require you to type a numerical answer in the box provided. Some questions will require you to select one or more answer choices.
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## Footnotes

* The "force" on the car would be referring to the net force, which is $\mathbf {F} _{\text{net}}=m\mathbf {a}$. The acceleration $\mathbf {a} ={\frac {\mathbf {v} -\mathbf {v} _{0}}{T}}$, which is given in the problem as $5{\frac {\text{m}}{{\text{s}}^{2}}}$. The mass $m=400{\text{ kg}}$. From here, we would determine the net force to be $\mathbf {F} _{\text{net}}=\left(400{\text{ kg}}\right)\left(5{\frac {\text{m}}{{\text{s}}^{2}}}\right)=2000{\text{ kg}}\cdot {\frac {\text{m}}{{\text{s}}^{2}}}=2,000{\text{ Newtons}}$.
* Notice how we mentioned that $g=0$ or $-M\mu _{k}+m=0$. This is the case because it is not always the true that both must equal zero. We only need at least one to be true. Here, $g\approx -9.81$, which is not zero! However, in the context of the algebra, we are saying that if we were to have the factored form with $g\left(-M\mu _{k}+m\right)=0$, then we may state that the factor $g=0$ or the factor $-M\mu _{k}+m=0$, so that we may ignore $g$ to find out more about $m$. After learning about this, we now know for sure that $-M\mu _{k}+m=0$. Otherwise, the equation makes no sense. Alternatively, if one did not see the factor of $g$ in both terms, then one could have simply done the same procedure but then multiplied by ${\frac {1}{g}}$. Here is precisely what we mean: $gm-gM\mu _{k}=0\Rightarrow gm=gM\mu _{k}\Rightarrow m=M\mu _{k}$