# Intermediate Algebra/Printable version

The current, editable version of this book is available in Wikibooks, the open-content textbooks collection, at
https://en.wikibooks.org/wiki/Intermediate_Algebra

# Expressions and Formulas
## Expressions

Throughout your mathematical journey through Arithmetic, Pre-Algebra, and Geometry, you have been introduced to (and analyzing) equations, and perhaps even expressions, but may have ignored these aspects of math. Equations, of course, involve an equal sign ($=$) while an expression is merely the calculations involved.
When you simplify an expression, you are answering simple arithmetic problems until you result in the simplest answer, most likely a single number, but sometimes a fraction. When you evaluate an expression, you are utilizing variables to find a single number. Of course, you are familiar with all this from the Arithmetics classes you previously took and the Arithmetic review, correct? If your memory needs some refreshing and some practicing the Order Of Operations (Please Excuse My Dear Aunt Sally), here are some practice problems.

### Practice Problems
## Formulas

Now that we understand how to simplify and evaluate expressions, we can analyze formulas. Formulas are expressions made primarily of variables that are plugged in to evaluate the expression. Formulas are used for science and mathematics often, especially Geometry.

### Practice Problems
## Lesson Review

Unlike equations, expressions contain no equal sign. In equations, there are two separate expressions that are equal to each other, and you are trying to make both sides of the equal sign... equal. Well, with expressions, you are either simplifying or evaluating them. To simplify an expression, you do as many of the Order of Operations as you can to get to the simplest answer. Meanwhile, to evaluate, you plug in numbers for variables and simplify from there. Formulas are special kinds of expressions and/or equations that are used for science and mathematics.

## Lesson Quiz
# Solving Absolute Value Equations
## Absolute Values

Absolute Values represented using two vertical bars ($\vert$) are common in Algebra. They are meant to signify the number's distance from 0 on a number line. If the number is negative, it becomes positive. And if the number was positive, it remains positive:
$\left\vert 4\right\vert =4\,$
$\left\vert -4\right\vert =4\,$
For a formal definition:
$x\geq 0$
$\left\vert x\right\vert =x$
$x<0$
$\left\vert x\right\vert =-x$
The formal definition is simply a declaration of what the function represents at certain restrictions of the $x$-value. For any $x<0$, the output of the graph of the function on the $xy$ plane is that of $y=-x$.
Please note that the opposite (the negative, -) of a negative number is a positive. For example, the opposite of $-1$ is $1$.

### Practice Problems
## Absolute Value Equations

Now, let's say that we're given the equation $\left\vert k\right\vert =8$ and we are asked to solve for $k$. What number would satisfy the equation of $\left\vert k\right\vert =8$? 8 would work, but -8 would also work. That's why there can be two solutions to one equation (and later, even more solutions). (Answer this: why?)
* Example 1: Solve for $k$:$|2k+6|=8$
* Recall what the absolute value represents: it is the distance of that number to the left or right of the starting point, zero. This means that whatever the inside value represents, it must be either $8$ or $-8$. As such,
* $2k+6=8\quad {\text{OR}}\quad 2k+6=-8$.
* All that is left to do is to solve the two equations for $k$:
* $2k+6=8\qquad 2k+6=-8$
* $\Leftrightarrow 2k=2\qquad 2k=-14$
* $\Leftrightarrow k=1\qquad k=-7$
A basic principle of solving these absolute value equations is the need to keep the absolute value by itself. This should be enough for most people to understand, yet this phrasing can be a little ambiguous to some students. As such, a lot of practice problems may be in order here.
Example 2: Solve for $k$:$3|2k+6|=12$
We will show you two ways to solve this equation. The first is the standard way, the second will show you something incredible.
Standard way: Multiply the constant multiple by its inverse.
We'd have to divide both sides by $3$ to get the absolute value by itself. We would set up the two different equations using similar reasoning as in the first example:
$2k+6=4\quad {\text{OR}}\quad 2k+6=-4$.
Then, we'd solve, by subtracting the 6 from both sides and dividing both sides by 2 to get the $k$ by itself, resulting in $k=-5,-1$. We will leave the solving part as an exercise to the reader.
Other way: "Distribute" the three into the absolute value.
Play close attention to the steps and reasoning laid out herein, for the reasoning for why this works is just as important as the person using the trick, if not moreso. Let us first generalize the problem. Let there be a positive, non-zero constant multiple $c$ multiplied to the absolute value equation $|2k+6|$:
$c\cdot |2k+6|=|c|\cdot |2k+6|\quad {\text{OR}}\quad c\cdot |2k+6|=|-c|\cdot |2k+6|$.
Let us assume both are true. If both statements are true, then you are allowed to distribute the positive constant $c$ inside the absolute value. Otherwise, this method is invalid!
${\begin{aligned}|c|\cdot |2k+6|&=|c(2k+6)|&\qquad |-c|\cdot |2k+6|&=|-c(2k+6)|\\&=|2ck+6c|&\qquad &=|-2ck-6c|=|-(2ck+6c)|\\&=|1|\cdot |2ck+6c|={\color {red}1\cdot |2ck+6c|}&\qquad &=|-1|\cdot |2ck+6c|={\color {red}1\cdot |2ck+6c|}\end{aligned}}$
Notice the two equations have the same highlighted answer in red, meaning so long as the value of the constant multiple $c$ is positive, you are allowed to distribute the $c$ inside the absolute value bars. However, this "distributive property" needed the property that multiplying two absolute values is the same as the absolute value of the product. We need to prove this is true first before one can use this in their proof. For the student that spotted this mistake, you may have a good logical mind on one's shoulder, or a good eye for detail.
Prove:$|b|\cdot |c|=|bc|$
We will first look at the simplest case: the two values $b{\text{ and }}c$ are constants. We know the following properties are true:
$|-b|=|b|=b$
$|-c|=|c|=c$
From this, we can conclude that $|b|\cdot |c|=bc$ for any $b,c\in \mathbb {R}$. We also know the following is true:
$bc=d<0\Leftrightarrow |d|=-d>0$. This simply means that for some product $bc$ that equals a negative number $d$, the absolute value of that is $-d$, or the distance from zero. Because $d<0$, multiplying the two sides by $-1$ will change the less than to a greater than, or $d<0\Leftrightarrow -d>0$.
$bc=d=0\Leftrightarrow |d|=d=0$. For some product $bc$ that equals a number $d=0$, the absolute value of that is $0$.
$bc=d>0\Leftrightarrow |d|=d>0$. For some product $bc$ that equals a positive number $d$, the absolute of the product is $d$.
Because we multiplied two absolute values, the product is either positive or zero. We can therefore use the second and third bullet point to conclude that
$|b|\cdot |c|=bc\geq 0\Leftrightarrow |d|=d=bc=|bc|\geq 0$.
We can use all five bullet points to show that for any constants $b,c\in \mathbb {R}$, $|b|\cdot |c|=|bc|\blacksquare$. This was the simplest case. However, we have already proven the hardest case, where both $b{\text{ and }}c$ are variables. The five bullet points we have is enough to demonstrate this fact. As a result, this proof will be left as a trivial exercise for the reader.
By confirming the general case, we may be employ this trick when we see it again. Let us apply this property to the original problem (this gives us the green result below):
$3|2k+6|={\color {green}|6k+18|=12}$
This all implies that
$6k+18=12\quad {\text{OR}}\quad 6k+18=-12$.
From there, a simple use of algebra will show that the answer to the original problem is again $k=-5,-1$. :
Let us change the previous problem a little so that the constant multiple is now negative. Without changing much else, what will be true as a result? Let us find out.
* Example 3: Solve for $k$:$-4|2k+6|=8$
* We will attempt to the problem in two different ways: the standard way and the other way, which we will explain later.
* Standard way: Multiply the constant multiple by its inverse.
* Divide like the previous problem, so the equation would look like this: $|2k+6|=-2$. Recall what the absolute value represents: it is the distance of that number to the left or right of the starting point, zero. With this, do you notice anything strange? When you evaluate an absolute value, you will always get a positive number because the distance must always be positive. Because this is means a logically impossible situation, there are no real solutions. Notice how we specifically mentioned "real" solutions. This is because we are certain that the solutions in the real set, $\mathbb {R}$, do not exist. However, there might be some set out there which would have solutions for this type of equation. Because of this posibility, we need to be mathematically rigorous and specifically state "no real solutions."
* Other way: "Distribute" the constant multiple into the absolute value.
* Here, we notice that the constant multiple $c<0$. The problem with that is there is no $g$ such that $|g|<0$. The only way this would be true is for $-|g|<0$ because
* $-|g|<0\qquad {\text{Divide both sides by }}-1$
* $|g|>0$
* With this property, we may therefore only distribute the constant multiple as $|c|$ with a negative $-1$ as a factor outside the absolute value. As such,
* $-4|2k+6|=-|8k+24|=8\qquad {\text{Divide both sides by }}-1$
* $|8k+24|=-8$
* It seems the other way has us multiply a constant by its inverse to both sides. Either way, this "other method" still gave us the same answer: there is no real solution.
The problem this time will be a little different. Keep in mind the principle we had in mind throughout all the examples so far, and be careful because a trap is set in this problem.
* Example 4: Solve for $x$:$|3x-3|-3=2x-10$
* There are many we ways can attempt to find solutions to this problem. We will do this the standard and allow any student to do it however they so desire.
* $|3x-3|-3=2x-10\qquad {\text{Add the }}3{\text{ to both sides.}}$
* $|3x-3|=2x-7$
* Because the absolute value is isolated, we can begin with our generalized procedure. Assuming $2x-7>0$, we may begin by denoting these two equations:
* (1)
* $3x-3=2x-7$
* (2)
* $3x-3=-(2x-7)$
* These are only true if $2x-7>0$. For now, assume this condition is true. Let us solve for $x$ with each respective equation:
* Equation (1)
* $3x-3=2x-7\qquad {\text{Add }}3{\text{ and subtract }}2x{\text{ on both sides.}}$
* $x=-4$
* Equation (2)
* $3x-3=-(2x-7)\qquad {\text{Distribute }}-1{\text{.}}$
* $3x-3=-2x+7\qquad \quad {\text{Add }}3{\text{ and add }}2x{\text{ on both sides.}}$
* $5x=10\qquad \qquad \qquad \quad {\text{Divide }}5{\text{ on both sides.}}$
* $x=2$
* We have two potential solutions to the equation. Try to answer why we said potential here based on what you know so far about this problem.
* Why did we state we had two potential solutions?
* Because we had to assume that $2x-7>0$ and $|3x-3|=2x-7$ is true for the provided $x$.Because we had to assume that $2x-7>0$ and $|3x-3|=2x-7$ is true for the provided $x$.
* Because of this, we have to verify the solutions to this equation exist. Therefore, let us substitute those values into the equation:
* $|3(-4)-3|=2(-4)-7$. Notice that the right-hand side is negative. Also, the left-hand side and the right-hand side are not equivalent. Therefore, this is not a solution.
* $|3(2)-3|=2(2)-7$. Notice the right-hand side is negative, again. Also, the left-hand side and the right-hand side are not equivalent. Therefore, this cannot be a solution.
* This equations has no real solutions. More specifically, it has two extraneous solutions (i.e. the solutions we found do not satisfy the equality property when we substitute them back in).
Despite doing the procedure outlined since the first problem, you obtain two extraneous solutions. This is not the fault of the procedure but a simple result of the equation itself. Because the left-hand side must always be positive, it means the right-hand side must be positive as well. Along with that restriction is the fact that the two sides may not equal the other for the values whereby only positive values are given. This is all a matter of properties of functions.
* Example 5: Solve for $a$:$6\left\vert 5{\frac {a}{6}}+{\frac {1}{12}}\right\vert ={\frac {3}{5}}|15a+15|$
* All the properties learned will be needed here, so let us hope you did not skip anything here. It will certainly make our lives easier if we know the properties we are about to employ in this problem.
* $6\left\vert 5{\frac {a}{6}}+{\frac {1}{12}}\right\vert ={\frac {3}{5}}|15a+15|\qquad {\text{Distribute, so to speak, the constant terms.}}$
* $\left\vert 5a+{\frac {1}{2}}\right\vert =|9a+9|$
* Looking at the second equation might be the first declaration of absurdity. However, an application of the fundamental properties of absolute values is enough to do this problem.
* (3)
* $5a+{\frac {1}{2}}=|9a+9|$
* (4)
* $5a+{\frac {1}{2}}=-|9a+9|$
* Peel the problem one layer at a time. For this one, we will categorize equations based on where they come from; this should hopefully explain the dashes: 3-1 is first equation formulated from (3), for example.
* (3-1)
* $9a+9=5a+{\frac {1}{2}}$
* (3-2)
* $9a+9=-\left(5a+{\frac {1}{2}}\right)$
* (4-1)
* $-(9a+9)=5a+{\frac {1}{2}}$
* (4-2)
* $-(9a+9)=-\left(5a+{\frac {1}{2}}\right)$
* We can demonstrate that some equations are equivalents of the other. For example, (3-1) and (4-2) are equivalent, since dividing both sides of (4-2) by $-1$ gives (3-1). After determining all the equations that are equivalent, distribute $-1$ to the corresponding parentheses.
* (5)
* $9a+9=5a+{\frac {1}{2}}$
* (6)
* $9a+9=-5a-{\frac {1}{2}}$
* (7)
* $-9a-9=5a+{\frac {1}{2}}$
* Now all that is left to do is solve the equations. We will leave this step as an exercise for the reader. You will find that two of three potential solutions are identical, so that means there are two potential solutions: $a=-{\frac {19}{28}},-{\frac {17}{8}}$. All that is left to do is verify that the equation in the question is true when looking at these specific values of $a$:
* $a=-{\frac {19}{28}}$
* $\left\vert 5\left(-{\frac {19}{28}}\right)+{\frac {1}{2}}\right\vert =\left\vert 9\left(-{\frac {19}{28}}\right)+9\right\vert$ is true. The two sides give the same value: ${\frac {81}{8}}=10.125$.
* $a=-{\frac {17}{8}}$
* $\left\vert 5\left(-{\frac {17}{8}}\right)+{\frac {1}{2}}\right\vert =\left\vert 9\left(-{\frac {17}{8}}\right)+9\right\vert$ is true. The two sides give the same value: ${\frac {81}{28}}\approx 2.893$.
* Because both solutions are true, the two solutions are $a=-{\frac {19}{28}},-{\frac {17}{8}}\blacksquare$.

### Practice Problems
## Lesson Review

An absolute value (represented with |'s) stands for the number's distance from 0 on the number line. This essentially makes a negative number positive although a positive number remains the same. To solve an equation involving absolute values, you must get the absolute value by itself on one side and set it equal to the positive and negative version of the other side, because those are the two solutions the absolute value can output. However, check the solutions you get in the end; some might produce negative numbers on the right side, which are impossible because all outputs of an absolute value symbol are positive!

## Lesson Quiz
# Solving Inequalities
## Inequalities

As opposed to an equation, an inequality is an expression that states that two quantities are unequal or not equivalent to one another. In most cases we use inequalities in real life more than equations (i.e. this shirt costs $2 more than that one).
Given that a and b are real numbers, there are four basic inequalities:

### a < b

    a is "less than" b:
    Example: 2 < 4 ; -3 < 0; etc.

### a > b

    a is "greater than" b:
    Example: -2 > -4 ; 3 > 0 ; etc.

### $a\leq b$

    a is "less than or equal to" b:
    Example: If we know that $x\leq 7$, then we can conclude that x is equal to any value less than 7, including 7 itself.

### $a\geq b$

    a is "greater than or equal to" b:
    Example: Conversely, if $x\geq 7$, then x is equal to any value greater than 7, including 7 itself.

## Properties of Inequalities

Just as there are four properties of equality, there are also four properties of inequality:
Addition Property of Inequality
If a, b, and c are real numbers such that a > b, then a + c > b + c. Conversely, if a < b, then a + c < b + c.
Subtraction Property of Inequality
If a, b, and c are real numbers such that a > b, then a - c > b - c. Conversely, if a < b, then a - c < b - c.
Multiplication Property of Inequality
If a, b, and c are real numbers such that a > b and c > 0, then ac (or a * c) > bc (or b * c). Conversely, if a < b and c > 0, then ac < bc. (Note that if c = 0, then both sides of the inequality are in fact equal.) We will also review cases where c is less than zero later in the lesson.
Division Property of Inequality
If a, b, and c are real numbers such that a > b, and c > 0, then ${\frac {a}{c}}>{\frac {b}{c}}$. Conversely under the same conditions, if a < b, then ${\frac {a}{c}}<{\frac {b}{c}}$. As with the Multiplication Property, there are special cases that will be discussed later when c < 0.
Note that all four properties also work with inequalities where $a\leq b$ or $a\geq b$.

### Trichotomy Property

The statements above form the basis of Trichotomy Property:
* a < b
* a = b
* a > b
So, if we are given any two unknown real-number values, then any one of the three statements will hold true.

## Solving Inequalities

Solving algebraic inequalities is more or less identical to solving algebraic equations. Consider the following example:
Although it may be an inequality, we can use the Properties of Inequality stated above to solve. Start by subtracting 2x from both sides:
Finish by adding 7 to both sides.
This can be rewritten as $x\geq 11$. To check, substitute any value greater than or equal to 11. However, in order to satisfy the Trichotomy Property, we'll substitute three different values: 10, 11, and 12.
| $2x+4\leq 3x-7$ | $2x+4\leq 3x-7$ | $2x+4\leq 3x-7$ |
| $2(10)+4\leq 3(10)-7$ | $2(11)+4\leq 3(11)-7$ | $2(12)+4\leq 3(12)-7$ |
| $24\leq 23$ | $26\leq 26$ | $28\leq 29$ |
Ten is incorrect, whereas eleven and twelve satisfy the solution. Therefore, the solution set - the set of all answers which satisfy the original inequality - is $x\geq 11$. Written in set notation, the answer is {$x|x\geq 11$}. This is read as "the set of all x such that x is greater than or equal to 11".

## Special Cases - A variable in the denominator

For example, consider the inequality
    ${\frac {2}{x-1}}<2\,$
In this case one cannot multiply the right hand side by (x-1) because the value of x is unknown. Since x may be either positive or negative, you can't know whether to leave the inequality sign as <, or reverse it to >. The method for solving this kind of inequality involves four steps:
* Find out when the denominator is equal to 0. In this case it's when $x=1$.
* Pretend the inequality sign is an = sign and solve it as such: ${\frac {2}{x-1}}=2\,$, so $x=2$.
* Plot the points $x=1$ and $x=2$ on a number line with an unfilled circle because the original equation included < (it would have been a filled circle if the original equation included <= or >=). You now have three regions: $x<1$, $1<x<2$, and $x>2$.
* Test each region independently. in this case test if the inequality is true for 1<x<2 by picking a point in this region (e.g. x=1.5) and trying it in the original inequation. For x=1.5 the original inequation doesn't hold. So then try for 1>x>2 (e.g. x=3). In this case the original inequation holds, and so the solution for the original inequation is 1>x>2.

### Special Cases

Let us suppose that we were given this inequality to solve:
Using the same steps as above, start by subtracting 8 from both sides.
Now divide both sides by -3.
Check this solution by substituting three numbers. We'll use -7, -6, and -5.
| $-3x+8\geq 26$ | $-3x+8\geq 26$ | $-3x+8\geq 26$ |
| $-3(-7)+8\geq 26$ | $-3(-6)+8\geq 26$ | $-3(-5)+8\geq 26$ |
| $29\geq 26$ | $26\geq 26$ | $23\geq 26$ |
Wait, what happened? -7 and -6 satisfy the inequality, yet -7 is non-inclusive in the solution set! And -5, which is greater than -6, does not satisfy the inequality!
This is a special case involved in solving inequalities. Because the coefficient of the x-term was negative, the constant on the other side (26, which became 18 and then -6) switched signs. In order to attain a valid solution if a negative number is divided, we need to switch the sign in order to make sure that the solution set is correct. Thus, $x\geq -6$ becomes $x\leq -6$, or more specifically, {$x|x\leq -6$}.

#### Practice Problems
## Graphing Solutions

Because inequalities have multiple solutions, we need to be able to represent them graphically. In order to do so, we use the number line, depicted below:
There are two ways of graphing solutions; however, each one is unique depending on the nature of the inequality. If, for example, the inequality contains the < or > sign, we use an open circle ("O") and place it on (or above) the corresponding position on the number line, then draw another line either left or right of the solution (based on the sign) to indicate the infinite number of solutions in the set. For example, if we were to graph x < 4:
The open circle indicates that 4 is not included in the solution set; however, all values less than 4 satisfy the solution.
If an inequality contains a $\leq$ or $\geq$, then a closed circle (it will be depicted here with a *) is placed on or above the corresponding position on the number line. Hence, the solution $x\geq -2$ is graphed as follows.
The asterisk (*) indicates that -2 is inclusive in the solution set, as are all values greater than -2.
Practice Problems
Graph the following inequalities on a number line:
1. x > 4
2. $x\leq 1$
3. -1 > x
Solutions
1.
2.
(Note that your actual graph should have a closed circle in place of the dot. If in doubt, simply draw a circle and color it black.)
3. Be careful here; you need to rearrange the inequality first before graphing. When rewritten, the inequality becomes x < -1:

## Lesson Review

An inequality is a statement justifying that two quantities are not equal to each other. There are four cases of inequalities, two of which allow for equality ($a\leq b$ and $a\geq b$). The four properties of inequality, which are more or less parallel to the properties of equality, can be used to solve simple inequalities. The only exceptions are in multiplication and division, where the signs must be reversed if both sides are multiplied or divided by a negative number. Finally, the solution set of an inequality can be graphed on the number line with either an open circle ("O") or a closed circle ("*"), depending on the original sign used.

## Lesson Quiz

1. What is an inequality? List the four possible cases of inequalities.
2. State the Trichotomy Property in your own words.
3. Solve the following and graph all solutions:

### Quiz Answers

1. An inequality states that two quantities are not equal. The four cases of inequalities: a < b, a > b, $a\leq b$, $a\geq b$.
2. Answers will vary. Just make sure you didn't copy and paste the actual definition.
3a. x > 1
3b. $x\leq {\frac {5}{3}}$
3c. $x\leq 6$
3d. x < 1

# Absolute Value Inequalities
## Compound Inequalities

In the previous section we dealt with inequalities with one specific constraint - that is, only one solution set. However, there is a good chance that real-world situations will require multiple constraints. For example, computer manufacturers will want their products to be sold within certain ranges so that customers will continue to buy their computers.
Such situations are also bound to happen in algebra, and can be represented using compound inequalities. A compound inequality is a statement that puts two constraints on a constant, variable, or expression. For example, the inequality -4 < x < 3 indicates that x has two constraints: it must be (a) greater than -4 and (b) less than 3. These compound inequalities are often referred to as "and" inequalities because they require that the expression in the center meets the two restraints.
Using the graphing methods covered in the last lesson, we can show the solution set of -4 < x < 3 on the number line:
Notice that the solution is bounded by the two values.
On the other hand, a compound inequality can have one out of two constraints. These inequalities are called "or" inequalities - the expression can satisfy either one of the conditions. If, for example, x < 3 or x $\geq$ 7, then x is equal to any number less than 3 or greater than or equal to 7. Any number between 3 and 7 will not satisfy either condition.
Represented graphically, the inequality x < 3 or x $\geq$ 7 looks like this:
Notice that, as opposed to the inequality in the previous example, the two constraints travel in opposite directions. This is a key principle of compound inequalities, and one that should be kept in mind when graphing solutions.

### Solving Compound Inequalities

Because compound inequalities are two separate constraints combined into one, all operations must be performed on all sides of the inequality.
Example 1: Solve for x: -7 < 2x + 4 $\leq$ 12.
Stop for a second and pretend that the statement above was two separate inequalities. The logical first step would be to subtract 4 from both sides. Because the two are now put into one, however, we must subtract 4 from all sides. Therefore:
$\leq$
$\leq$
$\leq$
Now divide all sides by 2.
$\leq$
${\frac {-11}{2}}<{\frac {2x}{2}}\leq {\frac {8}{2}}$
$\leq$
Example 2: Solve for x: -3 > -x + 7 or 2x - 6 $\leq$ -4.
Because the two inequalities act as separate entities, solve each one separately. Start by subtracting 7 from both sides in the first inequality, and add 6 to both sides in the second.
Finish by dividing by respective coefficients (-1 and 2).

# Linear Equations
## Linear Equations

A linear equation is an equation that forms a line on a graph.

### Slope-Intercept form

A linear equation in slope-intercept form is one in the form $y=mx+b$ such that $m$ is the slope, and $b$ is the y-intercept. An example of such an equation is: $y=3x-1$

#### Finding y-intercept, given slope and a point

The y-intercept of an equation is the point at which the line produced touches the y-axis, or the point where $x=0$ This can be very useful. If we know the slope, and a point which the line passes through, we can find the y-intercept. Consider:
$y=3x+b$ Which passes through $(1,2)$ $2=3(1)+b$ Substitute $2$ and $1$ for $x$ and $y$, respectively $2=3+b$ Simplify. $-1=b$ $y=3x-1$ Put into slope-intercept form.

#### Finding slope, given y-intercept and a point

The slope of a line is defined as the amount of change in x and y between two points on the line.
If we know the y-intercept of the line, and a point on the line, we can easily find the slope. Consider:
$y=mx+4$ which passes through the point $(2,1)$
$y=mx+4$
$1=2m+4$ Replace $x$ and $y$ with $1$ and $2$, respectively. $-3=2m$ Simplify. $-3/2=m$ $y=-3/2x+4$ Put into slope-intercept form.

### Standard form

The Standard form of a line is the form of a linear equation in the form of $Ax+By=C$ such that $A$ and $B$ are integers, and $A>0$.

#### Converting from slope-intercept form to standard form

Slope-intercept equations can easily be changed to standard form. Consider the equation: $y=3x-1$ $-3x+y=-1$ Subtract -3x from each side, satisfying $Ax+By=C$ $3x-y=1$ Multiply the entire equation by $-1$, satisfying $A>0$ $A$ and $B$ are already integers, so we don't have to worry about changing them.

#### Finding the slope of an equation in standard form

In the standard form of an equation, the slope is always equal to ${\frac {-A}{B}}$

# The Coordinate (Cartesian) Plane
## Coordinate Plane

    A Coordinate Plane (Also referred to as a Cartesian Plane, after René Descartes) is a 2-dimensional plane with a horizontal axis and a vertical axis. Both axes extend to infinity, but in graphs only segments of them are drawn (and sometimes arrows are used to indicate the infinite length, such as in the picture to the right). The horizontal axis is known as the x-axis and the vertical axis is known as the y-axis. The point where they intersect is known as the origin.

### Coordinates of a Point

    When a point is plotted on a coordinate plane, its coordinates can be found. The x-coordinate is the point's distance from the y axis, and the y-coordinate is the distance from the x-axis. When the coordinates are integer numbers, they can be easily found on a graph by looking at the numbers on the axes. Together, the coordinates can be expressed as an ordered pair . The ordered pair (4,3) represents a point for squares to the right of the origin and three squares upward from it. On the x-axis, numbers increase toward the right and decrease toward the left. On the y-axis, numbers decrease going downward and increase going upward.

### Plotting an Equation on a Coordinate Plane

    It is also possible to represent an equation on a coordinate plane by plotting multiple ordered pairs. In the case of a linear equation in the form $y=mx+b$, the x-coordinate for a given point will be equal to the x variable at that point and likewise the y-coordinate will be equal to the y variable at that point.
    The simplest way to produce a graph of a linear equation is to solve for y at arbitrary values of x and plot the results. In the case of the equation $y=2x+4$, if you were graphing for x-coordinates from 0 to 5, you would find the value of y in that equation for all integer values of x from 0 to 5.
       $(2*0)+4=4$
       $(2*1)+4=6$
       $(2*2)+4=8$
       $(2*3)+4=10$
       $(2*4)+4=12$
       $(2*5)+4=14$
    The resulting ordered pairs are (0,4),(1,6),(2,8),(3,10),(4,12) and (5,14). All linear equations have an infinite number of possible ordered pairs, but for graphing, a finite number will suffice. Plot these coordinate pairs on a sheet of graph paper (Count off the length of the x-coordinate to to right then count the length of the y-coordinate upward if need be). Once all of the points are plotted, draw a line connecting them. This line represents all the values of the equation $y=2x+4$ from $x=0$ to $x=5$.

# Slope

To find the slope of a line you have to use the graph:
With two points (x1,y1) and (x2,y2),
$m=(y2-y1)/(x2-x1)$
An example of this is to find the slope of the two points (4,3) and (8,5).
$m=y2(5)-y1(3)/x2(8)-x1(4)$ $m=5-3/8-4$ $m=2/4$ $m=1/2$
Therefore, the slope of those two points is 1/2. Now to graph the slope of a line you use the formula:
$y=mx+b$
Where b = the Y-Intercept and m = the slope of the line.
But first we must find what the Y-Intercept is, to do this we choose one of the points that we are given, and input it into the formula. For this example, I'm going to use the point (4,3):
$y(3)=m(1/2)x(4)+b$ $3=1/2(4)+b$ $3(+2)=2(-2)+b$ $5=B$
We now have the Y-Intercept, this can then be input into the formula to find the various points to graph. Using a simple T-Chart, pick various numbers and substitute them in for X, and you will find Y. Write these down, and graph them.
x = 0 $y=m(1/2)x(0)+5$ $y=1/2(0)+5$ $y=5$
x = -2 $y=m(1/2)x(-2)+5$ $y=1/2(-2)+5$ $y=-1+5$ $y=4$
x = 2 $y=m(1/2)x(2)+5$ $y=1/2(2)+5$ $y=1+5$ $y=6$
X - Y 0 - 5 -2 - 4 2 - 6
Plot the points, and connect the dots to form your line.
Now a shortcut to doing this is to use the rise over run (Rise/Run) formula. Which is simply that whatever the slope is, (In the case of our example, 1/2) is the rise and run of our slope.
The rise is how many up from our Y-Intercept we go before we plot our point. The run is how many left or right (X-Axis) we move before we plot our point. So if we have a slope of 1/2, our Rise is 1 and our Run is 2. This means that we move 1 (one) up from our Y-Intercept and 2 (two) to the right (since 2 is a positive number, if it were -2 we would move to the left on the X-Axis). So our plot would be at (6,2), and the next point after that would be (7,4) and so on.
Rise/Run format is much easier and faster, but you should still understand how to use a T-Chart.

# Writing Linear Equations

Linear equations have two variables: $x$ and $y$.

## Examples

$y=5+7x$
$y=2-4x+4$
$y=7/2-4x$
If one were doing a walkathon and wanted to raise $5 for every mile walked, an equation for the total money raised, $y$ as a function of the number of miles walked, $x$ would be $y=5x$. Would donors, however, promise to not only pay $5 per mile but give an additional $30 just for taking part, the equation would be $y=5x+30$. Note that this example is limited as negative miles don't make sense, but the line defined by the equation is infinite in either direction.

# Systems of Equations By Algebra
## Solving Systems of Linear Equations by Using Algebra

Generally, you're not going to want to solve a system using graphs, simply because it takes too much time. There are two algebraic methods for solving systems of linear equations.

### Addition

The ideal situation for the Addition method (also known as Elimination method) is one in which a variable in the two equations has opposite coefficients. For instance: $6x+3y=42$ $2x-3y=22$ We would simply add up the values in the two equations, canceling out $y$ in the process. $8x=64$ This is the result of the initial addition. $x=8$ Simplify. Now, all we have to do is substitute $8$ for each occurrence of $x$,and solve for $y$. $6(8)+3y=42$ Substitute the value of $x$. $48+3y=42$ Simplify. $3y=-6$ Subtract 48 from each side. $y=-2$ Divide each side by 3.
However, even if the variables don't easily cancel out, simply just try with constant multiplications and so on. $3x+8y=48$ $x-4y=22$ We would simply multiply the second equation throughout by 2 and get: $2x-8y=44$ Then add up: $x=4$ Substitute: $(8)-4y=22$ $-4y=18$ $y=-{\frac {9}{2}}$
In some occasions, you may need to multiply both sides. For example:
$3y+2x=5$ $4y+3x=10$
In this case, we will multiply the first equation by three and the second equation by two.
$9y+6x=15$ $8y+6x=20$
$y=-5$
$9\times -5+6x=20$ $-45+6x=20$ $6x=85$ $x=14{\frac {1}{6}}$

### Substitution

This is another method to solve a system of linear equations. This is ideal if one of the equations is laid out where one variable has a coefficient of one or negative one. $y=3x+1$ $x+2y=16$ Here you can simply substitute the first algebraic expression that y equals in to the second. $x+2(3x+1)=16$ Now simply slove the problem $x+6x+2=16$ $7x+2=16$ $7x+2-2=16-2$ ${\frac {7x}{7}}={\frac {14}{7}}$ $x=2$ Then plug it into the equation you substituted earlier. $y=3(2)+1$ $y=6+1$ $y=7$ To check your work simply plug both x and y into one part of your system. $x+2y=16$ $(2)+2(7)=16$ $16=16$ check.
Example where variable is not on one side:
$x+y=9$ $3x+5y=25$
Switch first equation so x is on one side
$x=9-y$
Substitute
$3(9-y)+5y=25$
Distribute and solve
$27-3y+5y=25$ $2y=-2$
$y=-1$ $x=10$