# Basic 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/Basic_Algebra

# For Contributors

Welcome to the Basic Algebra project!
Each lesson has 5 sections:
* Vocabulary - simple explanations of new terms.
* Lesson - What does this new skill let us do?
* Example Problems - three example problems worked out in detail.
* Practice Games - links to outside games on the internet that reinforce skills needed for this new skill
* Practice Problems - in three sections (easy, medium, hard). Please list the question, then give the answer in parentheses. Example 2x=6: (x=3)
Remember: All textbooks have many, many problems in common. Don't worry if the problem you add is in a copyrighted textbook. No one owns 2x = 16, for example. Be sure that you do not copy lists of problems. Make sure all word problems you give are completely original as far as subject matter is concerned. Please be sure the subject of your word problem is a BE850 word.
We assume that these students are ready for algebra. We do include a review of basic mathematics ideas in order to direct the students towards integrating their prior math knowledge, especially math vocabulary. This is a standard practice in algebra textbooks.
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# Introduction to Basic Algebra Ideas/The Ideas of Algebra

Algebra is a way of working with numbers and signs to answer a mathematics problem (a question using numbers). You use Algebra in life.
It answers, for example, questions like these:
If I have 10 coins and need 25 coins to get a book, how many more coins do I need to get the book?
If you have 6 bottles of milk and you drink 2 bottles in a day, how many days will your milk last?
If you start running a race at a speed of 10 kilometers an hour, and your friend starts running 10 minutes later at a speed of 15 kilometers an hour, when will your friend catch up to you?
Algebra helps answer questions when building houses, making food, and doing science work.
This book will help you learn how to do mathematics using Algebra. It has chapters (parts of the book) with lessons (parts of the chapter about one idea).
A lesson has five parts:
* Vocabulary - lists special words you need for the lesson.
* Lesson - explains a new idea and how to use this idea.
* Example Problems - demonstrates the steps to complete problems using the new idea.
* Practice Games - provides entertaining challenges in which you complete math problems.
* Practice Problems - You practice your skills while completing problems.

# Introduction to Basic Algebra Ideas/Translating Words into Math Symbols
## Definitions

* The sum of two unknown numbers can be written as $x+y$.
* The difference of two unknown numbers can be written as $x-y$.
* The quotient of two unknown numbers can be written as $x\div y$.
* The product of two unknown numbers can be written as $x\times y$.
Keep in mind that $x+y$ is the same as $y+x$ and $x\times y$ is the same as $y\times x$.
However, $x-y$ is not the same as $y-x$, and $x\div y$ is not the same as $y\div x$.

## Practice Games

Put links here to games that reinforce these skills.
* http://www.quia.com/pa/19009.html
* http://www.quia.com/jg/278151.html

## Practice Problems
# Introduction to Basic Algebra Ideas/Simple Operations
## Lesson

An operation is a thing you do to numbers. You use signs like: +, –, ×, or ÷ for operations.
The equals sign is not an operation. You use the " = " sign for the equals sign. It is a special sign that you put between two things that are the same.

### Adding

Adding is a way to put two numbers together. You use the " + " sign for the adding operation. If you have 5 apples and a friend has 6 apples, you and your friend have 11 apples. This is 5 + 6 = 11 using the signs of Mathematics, and is read "five plus six equals eleven".
Order is not important when adding. 2 + 3 is the same as 3 + 2.

### Subtracting

Subtracting is a way of taking a number from another number. It is the opposite of adding. You use the " – " sign for the subtracting operation. If you have 5 apples and give 3 to a friend, you then have 2 apples left. This is 5 – 3 = 2 using the signs of Mathematics, and is read "five minus three equals two".
Order is important when subtracting. 3 – 2 is not the same as 2 – 3.

### Multiplying

Multiplying is a way of adding a number many times. You use the " × ", " • " or " * " sign for the multiplying operation. If you have 4 apple trees and every tree has 25 apples, you have 100 apples. This is 4 × 25 = 100 or 25 + 25 + 25 + 25 = 100 in the signs of Mathematics. The first way is shorter and easier to use.
Multiplying 25 by 4 is the same as adding 25, 4 times or adding 4, 25 times.
4 x 25 = 100 should be read "four times twenty-five equals one hundred".
Order is not important when multiplying. 2 × 3 is the same as 3 × 2.

### Dividing

Dividing is a way of subtracting a number many times. It is the opposite of multiplying. You use the " ÷ " or " / " sign for the dividing operation. If 3 friends take an equal part of a bag of 12 apples, every friend gets 4 apples. Or, you may take away (subtract) 3 apples from the bag of 12 apples, exactly 4 times until the bag is empty. This is 12 ÷ 3 = 4 or 12 – 3 – 3 – 3 – 3 = 0 in the signs of Mathematics.
12 ÷ 3 = 4 should be read "twelve divided by three equals 4".
Order is important when dividing. 3 ÷ 2 is not the same as 2 ÷ 3.

## Example Problems

* 3 + 4 = 7
* 5 – 3 = 2
* 6 × 2 = 12
* 13 + 14 = 27
* 21 – 9 = 12
* 8 × 3 = 24

## Practice Games

* Arcademics Grand Prix (Multiplication)
* Arcademics Jet Ski (Addition)
* Arcademics Sailboat (Subtraction)
* Arcademics Drag Race (Division)

## Practice Problems
# Introduction to Basic Algebra Ideas/Exponents and Powers
## Vocabulary
### Exponent

    A number written in superscript that denotes how many times the base will be multiplied by itself.

### Base (or radix)

    The number to be multiplied by itself.
Example: $5^{2}=25$
In this example, the base is 5 and the exponent is 2.

## Lesson

We use exponents to show when we're multiplying the same number more than one time.

### $3\times 3=3^{2}$

    Three times three equals three to the second power (or three squared)

### $3\times 3\times 3=3^{3}$

    Three times three times three equals three to the third power (or three cubed)

### $3\times 3\times 3\times 3=3^{4}$

    Three times three times three times three equal three to the fourth power

### $2\times 2\times 2=2^{3}$

    Two times two times two equals two to the third power
Note that any nonzero number raised to the 0 power is always equal to 1.

### $2^{0}=1$

    Two to the zero power equals one
We can also raise any number to a negative exponent. This is called the inverse exponent and places the number on the bottom of a fraction with a 1 on top:

### $2^{-2}={\frac {1}{2^{2}}}={\frac {1}{4}}$

    Two to the negative two equals one over two to the second power

## Example Problems

Let's evaluate these expressions.
* $7^{2}$
| $7\times 7$ | Seven to the second power, or seven squared, means seven times seven. |
| $49$ | Seven times seven is forty-nine. |
    Seven to the second power equals forty-nine.
* What is the area of a square with a side of 3 meters length?
| Area = (length of the side)$^2$ | The area, or space inside, of a square is equal to the length of the side of the square to the second power. |
| (3 meters)$^2$ | The length of the side is 3 meters, so the area is (3 meters) squared. |
| $3\times 3$ meters$^2$ | 3 squared is the same as 3 times 3. |
| 9 square meters | Three times three is nine. |
    So, the area of a square with a side length of 3 meters is 9 square meters.
* $c^{2}$ where $c=6$
| $6^{2}$ | First, we replace the variable "c" in the expression with 6, which is what it equals. |
| $6\times 6$ | 6 squared equals 6 times 6. |
| $36$ | 6 times 6 equals 36. |
    So, c squared is 36.
* $x^{3}$ where $x=10$.
| $10^{3}$ | First, we replace the variable "x" in the expression with 10, which is what it equals. |
| $10\times 10\times 10$ | 10 to the third power, or 10 cubed, is equal to 10 times 10 times 10. |
| $100\times 10$ | 10 times 10 equals 100. |
| $1000$ | 100 times 10 equals 1000. |
    So, x to the third power is 1000.
* $y^{4}$ where $y=2$
| $2^{4}$ | First, we replace the variable "y" in the expression with 2, which is what it equals. |
| $2\times 2\times 2\times 2$ | 2 to the fourth power is equal to 2 times 2 times 2 times 2. |
| $4\times 2\times 2$ | 2 times 2 equals 4. |
| $8\times 2$ | 4 times 2 equals 8. |
| $16$ | And 8 times 2 equals 16. |
    So, y to the fourth is 16.
* $3^{-3}$
| ${\frac {1}{3^{3}}}$ | Three to the negative third power, which can be expressed as 1 over three cubed. |
| ${\frac {1}{27}}$ | Three cubed equals 3 times 3 times 3 which equals 27. |
    So, three to the negative third power equals one twenty-seventh.

## Practice Games

* http://www.math.com/school/subject2/practice/S2U2L2/S2U2L2Pract.html
* http://www.quia.com/pop/50485.html (scientific notation)
* http://www.softschools.com/math/games/exponents_practice.jsp
* http://www.quia.com/quiz/358716.html (King Kong Scientific Notation)
* http://www.shodor.org/interactivate/activities/OrderOfOperationsFou/ (order of operations including exponents)

## Practice Problems

Use / as the fraction line!

# Introduction to Basic Algebra Ideas/Order of Operations
## Vocabulary
### Order of Operations

    The order of operations is the rule at which you apply operations within a mathematical formula. There are two common mnemonics.
In mathematics, we use BODMAS:
* Brackets
* Orders (e.g. exponents)
* Division
* Multiplication
* Addition
* Subtraction
In the United States, you may also see PEMDAS:
* Parentheses
* Exponents
* Multiplication
* Division
* Addition
* Subtraction
Other variations exist, but the rules for order of operations remain the same.

## Lesson

Evaluate the expression $3+4\times 5$.
1st solution - If you multiply first, it is $3+20$ and evaluates to 23.
2nd solution - If you add first, it is $7\times 5$ and evaluates to 35.
Is the first or second answer correct?
With no order of operations, both answers would be expected, but if an expression evaluates to more than one answer, math becomes ambiguous and does not work. For math to work there is only one order of operations to evaluate a mathematical expression.
The order of operations is Parenthesis, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). This can be remembered in two ways: "Please Excuse My Dear Aunt Sally" or PEMDAS.
Please note that for Multiplication and Division and Addition and Subtraction you do whichever one comes first going from left to right, and that is why both PEMDAS and BODMAS work. The following list, from top to bottom, is the order of operations in Algebra. Operations at the top of the list are completed first, and operations on the same line are completed from left to right.
* Parenthesis ( )
* Exponent ^
* Multiply $\times$, Divide $\div$
* Add $+$, Subtract $-$
Parenthesis is a special operation that has the most precedence. You use the ( and ) signs to make a separate expression from a group of terms. You evaluate an expression in parenthesis first. You use parenthesis if you need to do an operation with less precedence first. If the term in parenthesis is juxtaposed to a variable with no multiplication, then you treat this implicit multiplication the same as any other multiplication. (example: In $1/2x$, the $2x$ is a juxtaposed, implicit multiplication, so it means the same as $1/2\times x$, and multiplication does not take precedence over the division in the Order of Operations. If you want $2x$ to be a proper term, you should write $1/(2x)$)

## Example problems

Let's evaluate these expressions.
| $5+x^{3}$ where $x=2$ $=5+2^{3}$ $=5+(2\times 2\times 2)$ $=5+8$ $=13$ | $(5+x)^{3}$ where $x=2$ $=(5+2)^{3}$ $=7^{3}$ $=343$ |
| ${\frac {6}{x}}+3$ where $x=3$ $={\frac {6}{3}}+3$ $=2+3$ $=5$ | ${\frac {6}{x+3}}$ where $x=3$ $={\frac {6}{3+3}}$ $={\frac {6}{6}}$ $=1$ |
| $8-x+2$ where $x=3$ $=8-3+2$ (evaluate the – operation first. – and + have the same precedence but is left-to-right) $=5+2$ $=7$ | $8-(x+2)$ where $x=3$ $=8-(3+2)$ $=8-5$ $=3$ |
Back to the first problem: Evaluate the expression $3+4\times 5$.
There is only one answer, 23, because we multiply first.
If we want to add first, we can use parentheses.
If we write $(3+4)\times 5$, then we add first, and get 35.

## Practice games

* http://www.quia.com/cm/11715.html
* http://regentsprep.org/regents/math/orderop/orderPrac.htm

## Practice problems
# Introduction to Basic Algebra Ideas/Variables and Expressions
## Lesson

A variable is a letter or symbol that takes place of a number in Algebra. Common symbols used are $a$, $x$, $y$, $\theta$ and $\lambda$. The letters x and y are commonly used, but remember that any other symbols would work just as well.
Variables are used in algebra as placeholders for unknown numbers. If you see "3 + x", don't panic! All this means is that we are adding a number who's value we don't yet know.
Some examples of variables in use:
* $3x$ -- three times of $x$.
* $5-y$ -- five minus $y$
* $2\div s$ or ${\frac {2}{s}}$-- 2 divided by $s$
A term is a number or a variable or a cluster of numbers and variables multiplied and or divided separated by addition and subtraction.
Examples of terms:
* $3+5$ The terms are 3 and 5.
* ${\frac {6}{x}}$ The term is $6/x$, 6 over $x$ is one term, because the operation is division.
* $6x+5$ The terms are 6$x$ and 5, 6$x$ and 5 are separate terms because they are separated by a addition or subtraction.
An operation is a thing you do to numbers, like add, subtract, multiply, or divide. You use signs like +, –, *, or / for operations.
An expression is one or more terms, with operations between all terms.
Examples of expressions:
* $3\div 6$
* $8\times x$
* $x\times 6+y$
* $a\times b\times c\times d$
To evaluate an expression, you do the operations to the terms of an expression.
Examples of evaluating expressions:
* $3+4$ evaluates to 7.
* $18\div 3$ evaluates to 6.
* $4\times 5-3$ evaluates to 17.
To evaluate an expression with variables, you substitute (put a thing in the place of an other thing) numbers for the variables.
Examples of substituting: (Substitute 3 for x in these examples.)
* $x+4$ is $3+4$.
* $18\div x$ is $18\div 3$.
* $4\times 5-x$ is $4\times 5-3$.

## Example Problems

Evaluate the following expressions
| $5\times x$ | When $x=2$ |
| $5\times 2$ | Substitute 2 for $x$. |
| $10$ | Evaluate $5\times 2$ to get the answer. |
| ${\frac {x}{3}}+y$ | When $x=9$ and $y=4$ |
| ${\frac {9}{3}}+4$ | Substitute 9 for $x$ and substitute 4 for $y$. |
| $7$ | Evaluate ${\frac {9}{3}}+4$ to get the answer. |

## Practice Games

* Evaluating expressions
* Advanced Expression Evaluator

## Practice Problems

remember order of operations

# Introduction to Basic Algebra Ideas/Working With Negative Numbers
## Lesson
### Negative Numbers

A positive number is a number greater than zero.
A negative number is a number less than zero. You make a negative number by doing the negative operation on a positive number. You use the " – " sign for the negative operation. This sign is the same you use for subtracting.

### Adding and Subtracting

Adding a negative number is the same as subtracting a positive number.
* $7+(-4)=7-4$
* $x+(-y)=x-y$
Subtracting a negative number is the same as adding a positive number.
* $7-(-4)=7+4$
* $x-(-y)=x+y$

### Multiplying and Dividing

Multiplying a negative number by a positive number, or a positive number by a negative number makes the result negative.
* $(-2)\times 3=-6$
* $2\times (-3)=-6$
Multiplying a negative number by a negative number makes the result positive.
* $(-2)\times (-3)=6$
You do the same for dividing.
* $(-6)\div 3=-2$
* $6\div (-3)=-2$
* $(-6)\div (-3)=2$

### Exponentiating

Exponentiating a negative number to an even (a number you can divide by two) power makes the result positive.
* $(-3)^{2}=9$
* $(-x)^{2}=(-x)\times (-x)=x^{2}$
Exponentiating a negative number to an odd (a number you can not divide by two) power makes the result negative.
* $(-2)^{3}=-8$
* $(-x)^{3}=(-x)\times (-x)\times (-x)=x^{2}\times (-x)=-x^{3}$

### Order of Operations

The negative operation has the same precedence as multiplying and dividing.
* $3+8\div 4=3+2=5$
* $-3^{2}=-(3\times 3)=-9$
* $(-3)^{2}=(-3)\times (-3)=9$

## Example Problems

* $4+(-4)=0$
* $4+(-7)=-3$
* $0+(-2)=-2$
* $-5+7-2\times (-4)=10$

## Practice Problems
# Introduction to Basic Algebra Ideas/Solving Equations Using Properties of Mathematics
## Lesson

It is very important to show math in the simplest way. For example, ${\frac {5}{10}}$ is the same as ${\frac {1}{2}}$, but ${\frac {1}{2}}$ is better because it is easier to understand. The simplest answer is usually the best.
Associative Properties mean that you can do the problem in any order, and the answer will always be the same.
Commutative Properties mean that you can change the order of numbers around, and the answers will always be the same.
Equality Properties state that if two numbers on either side of the equation are equal, and the operation processed is the same, as well having the same variables in the operations, the result will be the same for both sides of the equation. Read the section devoted to the Addition and Subtraction Properties of Equality for more information.

#### Associative Property of Addition

The associative property of addition shows us that when adding multiple values together, the outcome will always be the same. You can group numbers together in parenthesis, and it will still end up having the same outcome. For example, $3+(4+5)=(3+4)+5$. The order remains the same, but the grouping has changed. The result, however, is consistent.

#### Associative Property of Multiplication

The associative properties work for both addition and multiplication. Think of grouping $3\times 3$ together. You end up with 9. What about $3\times (3\times 4)$? If we change the grouping here, the result will be the same. Try to visualize why this is. When multiplying, you're often building objects in rows and columns.
For example, a 2 inch $\times$ 4 inch block will be 2 inches across, and 4 downward. If you had a block 4 inches across and 2 inches downward, it would be the same size overall. When you put objects in parenthesis, remember to do those operations first. You may end up with certain "sides" of the operation or object being larger or smaller, but the total area will always have the same outcome.

#### Commutative Property of Addition

The commutative property of addition shows that no matter what order numbers are in when we add them, the result will always be the same. For example, $x+y=y+x$, just how $3+4=4+3$. Even though the order in which we added has shifted, the result doesn't change, and the statements on both sides of the equal sign remain true.

#### Addition Property of Equality

The addition property of equality states that if two variables or numbers are equal to each other on each side of the equation, and the operation they go through is alike, the resulting sum will be the same. If for example, both $x$ and y are 6, and you add $z$ to each of them, that is,
if $x=y$ then $x+z=y+z$
or if $6=6$ then $6+z=6+z$

#### Subtraction Property of Equality

The subtraction property of equality states that if two variables or numbers are equal to each other on each side of the equation, and the operation they go through is alike, the resulting difference will be the same. If for example, both $x$ and y are 6, and you subtract $z$ from each of them, that is,
if $x=y$ then $x-z=y-z$
or if $6=6$ then $6-z=6-z$

## Example Problems

Find $x$ where $y=6$.
* ${\frac {x+y}{2}}=14$
| ${\frac {x+6}{2}}=14$ | Substituting 6 for y using the given y |
| $2\left({\frac {x+6}{2}}\right)=2(14)$ | By multiplying the denominator on either side simplifies it |
| $x+6=28$ | Taking down the parentheses |
| $x+6-6=28-6$ | Subtracting 6 on both side does not affect the property of the given formula but again simplifies it |
| $x=22$ | To arrive at the answer |
* $(x-3y)+2x=15$
| $(x-3\times 6)+2x=15$ | Again taking the hints as given, $y=6$ |
| $(x-18)+2x=15$ | Dealing with the parentheses |
| $x-18+2x=15$ | Taking down the parentheses is the most natural next move using PEMDAS |
| $3x-18=15$ | Adding $x+2x$ |
| $3x-18+18=15+18$ | Adding 18 to either side |
| $3x=33$ | Taking down the parentheses |
| ${\frac {3x}{3}}={\frac {33}{3}}$ | Divide the multiplier to either side to emerge the solution |
| $x=11$ | And simplify to get the answer |
* ${\frac {15-y}{x}}=y^{2}$
| ${\frac {15-6}{x}}=6^{2}$ | Substitute $y=6$ |
| ${\frac {9}{x}}=36$ | Parentheses |
| $9=36x$ | Multiply by the denominator (x) on either side |
| $x={\frac {9}{36}}$ | Divide the multiplier on either side |
| $x={\frac {1}{4}}$ or $0.25$ | Simplify or divide fraction |

## Practice Games

* http://www.math.com/school/subject2/S2U3Quiz.html
* http://www.quia.com/rr/4096.html

## Practice Problems

Answer should be correct to two decimals - round up or down accordingly until the answer is correct to two decimals

# Introduction to Basic Algebra Ideas/Chapter Review
## Lesson 1. Simple Operations

An operation is a thing you do to numbers. You use signs like: +, –, ×, or ÷ for operations. The Equals Sign is not an operation.

### Adding

    Adding is a way to put two numbers together.
* 1 + 2 = 3

### Subtracting

    Subtracting is a way of taking a number out from another number.
* 2 – 1 = 1

### Multiply

    Multiplying is a way of adding a number many times
* 3 × 2 = 6

### Dividing

    Dividing is a way of subtracting a number many times.
* 6 ÷ 2 = 3

### Example Problems

* 2 + 1 = 3
* 8 + 2 = 10
* 8 – 4 = 4
* 5 – 2 = 3
* 6 × 2 = 12
* 2 × 3 = 6
* 12 ÷ 6 = 2
* 4 ÷ 2 = 2
Back to lesson

## Lesson 2. Exponents and Powers

Exponent is the number on the top that shows.
Base is the number to be multiplied by itself.

### Example Problems

* 6$^2$ = 36
* 2$^3$ = 8
* 4$^2$ = 16
* 5$^3$ = 125
* 2$^4$ = 16
Back to lesson

## Lesson 3. Order of Operations

Math problems are done in this order from top to bottom:
* Parenthesis ( )
* Exponent ^
* Multiply ×, Divide ÷ (Left to Right)
* Add +, Subtract – (Left to Right)

### Example Problem

| $2^{2}+(3\times 4)$ $2^{2}+(12)$ $2^{2}+12$ $4+12$ $16$ | Original problem. Do parenthesis first. Do exponent. Add. Answer |
Back to lesson

## Lesson 4. Working With Negative Numbers

A positive number is a number more than zero.
A negative number is a number less than zero.

### Example Problems

* 6 + (–3) = 3
* 3 + (–9) = –6
* –4 × 4 = -16
* 4 × (–9) = -36
* –2 × (–4) = 8
Back to lesson

## Lesson 5. Solving Equations Using Properties of Mathematics

It is very important to show math in the simplest way. For example, 5/10 is the same as 1/2, but 1/2 is better because it is easier to understand. The simplest answer is usually the best.

### Example Problems

    Find $x$ when $y=9$
* $x=8(y\div 3)$
* $x=24$
* $(x-4)=8+y$
* $x=21$
* $(14+x)\div y=3$
* $x=13$
Back to lesson

# Working with Numbers/Integers and the Number Line
## Vocabulary
### Integer

    A whole number (both positive and negative).

### Number Line

    a line marked periodically with numbers

### Origin

    Zero (0)

## Lesson

An integer is any number, positive or negative, that does not have a fraction or decimal point. Integers can be shown on a number line, like the one below.
This number line shows the numbers from -10 to (positive) 10. Notice the arrows on either side to show that the numbers continue in both directions even though they are not shown on the line.

## Practice Games

http://www.funbrain.com/linejump/index.html

## Practice Problems
# Working with Numbers/Absolute Value
## Vocabulary
### Absolute Value

    The absolute value of a number is its distance from zero (0) on a number line. This action ignores the "+" or "–" sign of a number because distance in mathematics is never negative. The symbol $|x|$ represents the absolute value of $x$. It is also called modulus $x$.

## Lesson

The absolute value of a number is its distance from zero (0) on a number line. This action ignores the “+” or “–“ sign of a number because distance in mathematics is never negative.
You identify an absolute value of a number by writing the number between two vertical bars referred to as absolute value brackets: |number|.
A helpful way of thinking about absolute value is relating it to a railroad track. If you were to stand on a railroad track, more specifically on any one of the railroad ties and mark that spot as zero, railroad ties to the left would represent negative numbers and railroad ties to the right would represent positive numbers.
The number –7 is 7 units away from zero on the negative side of the railroad track. So, the following is true, |-7| = 7. The number 16 is 16 units away from zero on the positive side of the railroad track. So, |16| = 16. The number 0 is 0 units from zero on the railroad track. So |0| = 0 Therefore, the absolute value of any number is a positive number or zero.
In summary… THE ABSOLUTE VALUE OF A NUMBER
If x is a positive number, then $|x|=x$. Example: $|5|=5$
If x is zero, then $|x|=0$. Example: $|0|=0$
If x is a negative number, $|x|=-x$. Example: $|-6|=-(-6)=6$
You can find the absolute value of expressions as well. When addressed with this you must treat the absolute value brackets as you would parentheses. You need to simplify everything inside the absolute value brackets by performing all the necessary operations by following the order of operations. Your last step once you have a single number inside the absolute value brackets is to take the absolute value. For example, $|-5+1\times 3|=|-5+3|=|-2|=2$.
You may use the absolute value to find the distance between two numbers on the number line. Let a and b be variables. Then $|a-b|$ is the distance between a and b. For example, if $a=3$ and $b=7$, then $|3-7|=|-4|=4$. Because you used the absolute value, the distance is the same if you switch the order of the two numbers; if $a=7$ and $b=3$, then $|7-3|=|4|=4$.
Two things to watch out for are an opposite sign and/or an operation outside the absolute value brackets. As stated above, simplify everything inside the absolute value brackets by performing all the necessary operations by following the order of operations. Your last step once you have a single number inside the absolute value brackets is to take the absolute value. Once you have taken the absolute value then perform the other necessary operations by following the order of operations from left to right in the expression. For example, $-|5|=-5$ and $7+|-5+1\times 3|=7+|-5+3|=7+|-2|=7+2=9$.

## Example Problems

* $|0|=0$
* $+|-21|=-21$
* $|7-1|+9=|6|+9=6+9=15$
* $|0.5|=0.5$
* $|-5\times 3|=|-15|=15$
* $\left|{\frac {-2}{3}}\right|={\frac {2}{3}}$
* $|25-16|=|9|=9$
* $|9|=9$
* $|3.5-5.7|=|-2.2|=2.2$
* $|-11.5|=11.5$
* $5-|3|=5-3=2$
* $-|6.7|=-6.7$
* $8+|-6\times 2|=8+|-12|=8+12=20$

## Practice Games

absolute value game

## Practice Problems
# Working with Numbers/Rational Numbers
## Lesson
### Rational numbers

A rational number is a fraction, written ${\frac {p}{q}}$ where $p$ and $q$ are integers. $p$ is called the numerator and $q$ the denominator. Applied to a cake, it means $p$ parts of a cake divided equally into $q$ parts. For example ${\frac {1}{2}}$ means a half. But note that $p$ and $q$ can be negative. $+{\frac {1}{2}}$ means gaining a half and $-{\frac {1}{2}}$ means losing a half.

### Fractions of negative numbers

If $p$ and $q$ are positive, then the fraction or rational number is positive. This is the way we commonly think of fractions ( ${\frac {1}{3}}$ of a cake...).
There is no difference whether $p$ is negative or $q$ is negative. The reason for this is simple : if you talk about losing parts of a cake ( $-{\frac {p}{q}}$ ), or about parts of a lost cake ( ${\frac {p}{-q}}$ ), in both cases, you talk about lost parts. In these cases, the fraction is said to be negative.
Finally, if $p$ and $q$ are negative, then their effect is canceled by each other and the fraction is positive. As rational numbers are on one axis, the second time you take the opposite you obtain the original fraction. Thus, the fraction ${\frac {-p}{-q}}$ is the fraction ${\frac {p}{q}}$.

## Example Problems

I have been given 1 piece of cake, my father who is very hungry has taken 2. My mother has taken 1 and my sister has taken 1 too. There were 10 pieces. What fraction of the cake has been eaten?
${\frac {1}{10}}+{\frac {2}{10}}+{\frac {1}{10}}+{\frac {1}{10}}={\frac {5}{10}}={\frac {1}{2}}$ of the cake, which is half of the cake.
Note that in this case, the addition is very simple because the denominator is always 10. We just have to add the numerators.

## Practice Games

put links here to games that reinforce these skills

## Practice Problems

(Note: put answer in parentheses after each problem you write)

# Working with Numbers/Adding Rational Numbers
## Lesson

It is easy to add fractions when the denominators are equal. For example, adding ${\frac {3}{10}}$ and ${\frac {2}{10}}$ is very simple, just add the numerators and you have the numerator of the resulting fraction:
* ${\frac {3}{10}}+{\frac {2}{10}}={\frac {5}{10}}={\frac {1}{2}}$
Notice the simplification: five parts out of ten is the half of the parts. Unfortunately, it is not always so simple. Sometimes we need to add fractions that have different denominators. Before we can add them, we must alter the fractions so that their denominators are the same. We can do this by multiplying each fraction by the number one which doesn't change the value of the fraction). However, the form of the number one will itself be represented as a fraction whose denominator and numerator are equal, and under our control. For example, all of these fractions are equal to one:
* ${\frac {2}{2}}={\frac {3}{3}}={\frac {107}{107}}=1$
Knowing this, we can change the denominators of the fractions so that the denominators of both are the same. For example:
* ${\frac {1}{3}}+{\frac {1}{2}}={\frac {1\times 2}{3\times 2}}+{\frac {1\times 3}{2\times 3}}={\frac {2}{6}}+{\frac {3}{6}}={\frac {5}{6}}$
In this case we changed both fractions so that they each had a denominator of 6.

## More complicated fractions

In these cases, we can guess which multiplication to do, but sometimes, it is not that easy. For example, adding ${\frac {123}{456}}$ and ${\frac {234}{120}}$.
The simplest general method is to multiply the numerator and denominator of the first fraction by the denominator of the second fraction and vice-versa. The resulting denominators will both be the product of the two original denominators. In this case :
* ${\frac {123}{456}}+{\frac {234}{120}}={\frac {123\mathbf {\times 120} }{456\mathbf {\times 120} }}+{\frac {234\mathbf {\times 456} }{120\mathbf {\times 456} }}={\frac {14760}{54720}}+{\frac {106704}{54720}}={\frac {14760+106704}{54720}}={\frac {121464}{54720}}$
We obtain generally big numbers which is not optimal because the fraction can most of the time be written with smaller numbers.
The second is more subtle. Instead of multiplying by the actual denominators, we multiply by the smallest possible number for each side so that we obtain the same denominator. For example:
* ${\frac {1}{6}}+{\frac {1}{4}}={\frac {1\times 2}{6\times 2}}+{\frac {1\times 3}{4\times 3}}={\frac {2}{12}}+{\frac {3}{12}}={\frac {5}{12}}$
We only multiplied by 2 in the first fraction and by 3 in the second fraction. The resulting fraction, ${\frac {5}{12}}$ is optimal, which we call irreducible.
Note that 2 is the half of $4=2\times 2$ and 3 the half of $6=3\times 2$. We did not multiply by the given denominators, we avoided to multiply by the factor 2. Let's take the previous example and find the factors composing the numbers...
    $123=\mathbf {3} \times 41$ and $456=2\times 228=2\times 2\times 114=2\times 2\times 2\times 57=2\times 2\times 2\times \mathbf {3} \times 19$
    $234=2\times 117=2\times 3\times 39=\mathbf {2\times 3} \times 3\times 13$ and $120=2\times 2\times 3\times 10=2\times \mathbf {2\times 3} \times 2\times 5$
We can see that we can simplify ${\frac {123}{456}}$ by 3 which gives ${\frac {41}{2\times 2\times 2\times 19}}$ and simplify ${\frac {234}{120}}$ by 2 × 3 which gives ${\frac {39}{2\times 2\times 5}}$. Remember that multiplying by the same number the numerator and the denominator does not change the value. The same is true when dividing by the same number.
Now comes a question : which is the smallest integer that contains the factors $2\times 2\times 2\times 19$ and the factors $2\times 2\times 5$. It is the number that has just all these factors in correct number: $2\times 2\times 2\times 5\times 19=760$.
To attain this number, we must multiply in the first fraction by 5 and in the second by 2 × 19. So, finally we have:
* ${\frac {123}{456}}+{\frac {234}{120}}={\frac {41}{2\times 2\times 2\times 19}}+{\frac {39}{2\times 2\times 5}}={\frac {41\mathbf {\times 5} }{760}}+{\frac {39\mathbf {\times 2\times 19} }{760}}={\frac {205}{760}}+{\frac {1482}{760}}={\frac {1687}{760}}$
This fraction is simpler as the first obtained ${\frac {121464}{54720}}$.
Both fractions are equal: ${\frac {1687}{760}}={\frac {121464}{54720}}$
But the factor between the two fractions is 72!

## Practice Games

put links here to games that reinforce these skills

## Practice Problems

Use / as the fraction line and put spaces between the wholes and fractions!

# Working with Numbers/Subtracting Rational Numbers
## Lesson

Subtracting fractions when the denominators are equal is also easy
    ${\frac {3}{7}}-{\frac {2}{7}}={\frac {1}{7}}$
Notice that when the denominators are the same for either adding or subtracting fractions we only add or subtract the numerators. With this in mind we can now use the techniques learned in Lesson 4 to subtract more complex fractions by finding the factors of the following fractions
    ${\frac {121}{456}}-{\frac {61}{570}}={\frac {11\times 11}{2\times 2\times 2\times 3\times 19}}-{\frac {61}{2\times 3\times 5\times 19}}$
We can now see that multiplying 456 by 5 and 570 by 4 gives the smallest possible denominator
    ${\frac {121\times 5}{5\times 456}}-{\frac {4\times 61}{4\times 570}}={\frac {605-244}{2280}}={\frac {361}{2280}}$
Now the question is, is this fraction irreducible? We know the factors in the denominator so checking if 361 is divisible by the numbers 2, 3, 5, 19 will show that 361/19 = 19 therefore
    ${\frac {361}{2280}}={\frac {19}{120}}$
Which is now an irreducible fraction.

## Example Problems

${\frac {20(8-3)-4(10-3)}{10(2-6)}}+2(7+4)$
$={\frac {20(5)-4(7)}{10(-4)}}+2(11)$
$={\frac {100-28}{-40}}+22$
$=-{\frac {72}{40}}+22$
$=-{\frac {9}{5}}+{\frac {5(22)}{5}}$
$=-{\frac {9}{5}}+{\frac {110}{5}}$
$={\frac {101}{5}}$
$=20{\frac {1}{5}}$

## Practice Games

put links here to games that reinforce these skills

## Practice Problems

(Note: put answer in parentheses after each problem you write)

# Working with Numbers/Multiplying
## Lesson

Multiplication of rational fractions is perhaps easier than addition and subtraction (lessons 4 and 5). This is because the denominators do not have to be equal, so you do not need to find a common denominator before carrying out a calculation. Consider the following problem:
    ${\frac {5}{9}}\times {\frac {7}{4}}$
This may look like a difficult calculation but in reality it's rather easy. We simply multiply the two numerators together, then multiply the denominators. So, the answer to the above problem would be:
    ${\frac {5}{9}}\times {\frac {7}{4}}={\frac {5\times 7}{9\times 4}}={\frac {35}{36}}$
This fraction is irreducible as 35 and 36 share no common factors.
Notice that in the problem above there was a top heavy fraction (${\frac {7}{4}}$). When multiplying two fractions, if one is top heavy then leave it as it is until you have your final answer. Attempting to multiply a mixed number with a fraction will result in an incorrect answer.
Let us now consider a more complex problem. Say we had three large fractions which we had to multiply together:
    ${\frac {100}{101}}\times {\frac {263}{325}}\times {\frac {150}{175}}$
The first thing you should notice is that ${\frac {150}{175}}$ can be simplified to ${\frac {6}{7}}$. This should make this calculation a little easier. As above, we simply multiply the numerators together then multiply the denominators together.
    ${\frac {100}{101}}\times {\frac {263}{325}}\times {\frac {6}{7}}={\frac {100\times 263\times 6}{101\times 325\times 7}}={\frac {157800}{229775}}$
Now this is a huge number so trying to find common factors in order to reduce it will be very difficult and time consuming. If you have a scientific calculator to hand, simply enter the above fraction and it should give you an irreducible fraction out. My calculator gives the following result:
    ${\frac {6312}{9191}}$

## Practice Problems

Use / as the fraction line!

# Working with Numbers/Distributive Property
## Vocabulary
### [[../../Introduction_to_Basic_Algebra_Ideas/Simple_Operations#Adding|Sum]]

    The resulting quantity obtained by the addition of two or more terms.

### Real Number

    An element of the set of all rational and irrational numbers. All of these numbers can be expressed as decimals.

### Term

    A term is a number or a [[../../Introduction_to_Basic_Algebra_Ideas/Variables_and_Expressions#Lesson|variable]] or the product of a number and a variable(s).

### Monomial

    An algebraic expression consisting of one term.

### Binomial

    An algebraic expression consisting of two terms.

### Trinomial

    An algebraic expression consisting of three terms.

### [[../../Polynomials/Adding_and_Subtracting_Polynomials|Polynomial]]

    An algebraic expression consisting of two or more terms.

### Like Terms

    Like terms are expressions that have the same variable(s) and the same exponent on the variable(s). Remember that constant terms are all like terms. This follows from the definition because all constant terms can be seen to have a variable with an [[../../Introduction_to_Basic_Algebra_Ideas/Exponents_and_Powers|exponent]] of zero.

## Lesson

The distributive property is the short name for "the distributive property of multiplication over addition", although you will be using it to distribute multiplication over subtraction as well. When you are simplifying or evaluating you follow the [[../../Introduction_to_Basic_Algebra_Ideas/Order_of_Operations|order of operations]]. Sometimes you are unable to simplify any further because you cannot combine like terms. This is when the distributive property comes in handy.
Natural Language
When you first learned about multiplication it was described as grouping. You used multiplication as a way to condense the multiple addition of the same quantity. If you wanted to add $3+3+3+3$ you could think about it as four groups of three items.
|ooo| + |ooo| + |ooo| + |ooo|
You have 12 items. This is where $4\cdot 3$ comes in. So as you moved on you took this idea to incorporate variables as well. $3x$ is three groups of x.
$\underbrace {x+x+x} _{\text{three groups of x}}=3x$
And $3(2x)$ is three groups of $2x$ and $2x$ is $x+x$
$\underbrace {(x+x)+(x+x)+(x+x)} _{\text{three groups of 2x}}=3(2x)$
This gives you six x's or 6x. Now we need to take this idea and extend it even further. If you have $3(x+1)$ you might try to simplify using the order of operations first. This would have you do the addition inside the parentheses first. However, x and 1 are not like terms so the addition is impossible. We need to look at this expression differently if we are going to simplify it. What you have is $3(x+1)$ or in other words you have three groups of $(x+1)$
$\underbrace {(x+1)+(x+1)+(x+1)} _{\text{three groups of (x+1)}}=3(x+1)$
Here you can collect like terms. You have three x's and three 1's.
$\underbrace {x+x+x} _{\text{3x}}+\underbrace {1+1+1} _{\text{3}}=3x+3$
So you started with $3(x+1)$ and ended with $3x+3$
$3(x+1)=3x+3$
The last equation might make it easier to see what the distributive property says to do.
$3(x+1)=3(x)+3(1)=3x+3$
You are taking the multiplication by 3 and distributing that operation across the terms being added in the parentheses. You multiply the x by 3 and you multiply the 1 by 3. Then you just have to simplify using the order of operations.
What Is Coming Next
After you learn about the distributive property you will know how to multiply a monomial by a polynomial. Next, you can use this information to understand how to multiply a polynomial by a polynomial. You will probably move on to multiplying a binomial times a binomial. This will show up in something like (x+2)(3x+5). You can think of a problem like this as x(3x+5) + 2(3x+5). Breaking up the first binomial like this allows you to use your knowledge of the distributive property. Once you understand this use of the distributive property you can extend this understanding even further to justify the multiplication of any polynomial with any polynomial.
Sometimes while you are attempting to isolate a variable in an equation or inequality you will need to use the distributive property. You already know that you use inverse operations to isolate your desired variable, but before you do that you need to combine like terms that are on the same side of the equation (or inequality). Now there might be a step even before that. You will need to see if the distributive property needs to be used before you can combine like terms then proceed to use inverse operations to isolate a variable.
Word to the Wise
Remember that you still have the order of operations. If you can evaluate operations in a straightforward manner it is usually in your best interest to do so. The distributive property is like a back door to the order of operations for when you get stuck because you do not have like terms. Of course when you are dealing with only constant terms everything you encounter is like terms. The trouble happens when you introduce variables. This means that some terms cannot be combined. Remember that variables take the place of real numbers (at least in Algebra 1) so the same rules that govern real numbers will also govern the variables that hold their place and vice versa. You can use the distributive property even when you do not need to.

## Example Problems

Example Problem #1:
Simplify $2(x+4)$
Solution to Example Problem #1:
Normally, to follow the order of operations you would add the two terms in the parenthesis first, then do the multiplication by. This does not work for this expression because x and 4 are unlike terms so you cannot combine them. We use the distributive property to help us find a way around the order of operations while still being sure that we keep the value of the express.
We distribute the multiplication by 2 across the addition. We will have 2 multiplied by x and 2 multiplied by 4.
$2(x)+2(4)$
Now we just need to finish the multiplication. $2(4)$ is equal to 8.
$2x+8$
We are done because we just have two terms being added and we cannot add them because they are not like terms.
Example Problem #2:
Simplify $3x(2x-4)$
Solution to Example Problem #2:
Since the terms inside the parentheses are not like terms we cannot combine them. We can use the distributive property to multiply by $3x$.
$3x(2x-4)=3x(2x)-3x(4)$
This is the first example with subtraction in it. You keep this operation between the two terms just like we kept the addition between the two terms in the previous example. The next step is to multiply
$3x(2x)-3x(4)=6x^{2}-12x$
In order to complete the previous step you will already need to know how to multiply [[../../Polynomials/Multiplying_Monomials|monomials]].
To summarize all the steps...
$3x(2x-4)=3x(2x)-3x(4)=6x^{2}-12x$
Example Problem #3:
Solve for $x$ in $2(x+10)=60$
Solution to Example Problem #3:
To solve for a variable you must isolate it on one side of the equation. We need to get the $x$ out of the parentheses. Since we cannot go through the order of operations and just add x plus 10 then multiply by 2, we will have to use the distributive property. First, distribute the multiplication by 2 across the addition inside the parentheses.
$2(x+10)=60$
$2(x)+2(10)=60$
Now you can multiply
$2(x)+2(10)=60$
$2x+20=60$
Now we can work on getting the $x$ on one side by itself. You need to do the order of operations backwards so we can "undo" what is "being done to" $x$. To get rid of adding 20 you need to subtract 20. And remember that an equation sets up a relationship that we need to preserve. If you subtract 20 from one side you need to subtract 20 from the other side as well to keep the balance.
$2(x)+20=60$
$2(x)+20-20=60-20$
$2(x)+0=40$
$2(x)=40$
Now we need to "undo" the multiplication by 2, so we divide by 2. Whatever you do to one side must be done to the other. So divide both sides by 2.
${\frac {2(x)}{2}}={\frac {40}{2}}$
$x=20$
This is it. You know you are done when the variable $x$ is by itself on one side, and it is.

## Practice Games

http://www.quia.com/ba/15357.html
http://www.studystack.com/matching-1870
http://www.slideshare.net/rfant/distributive-property-algebra-1/ (slide show)
http://www.phschool.com/atschool/academy123/html/bbapplet_wl-problem-430723.html ( video explanation)

## Practice Problems
# Working with Numbers/Combining Like Terms
## Lesson

Algebra is used to make many problems simpler, and that is why a lot of algebra is about finding simple expressions which mean the same thing as harder ones. Variables are given different letters and symbols in algebra so they can be kept apart, so every time $x$ is used in an expression it means the same thing, and every time $y$ is used it means the same thing, but a different thing to $x$ (of course this is only in the same expression, different expressions can use the same letters to mean different things). Since the different letters keep the variables apart this means that an expression with many variables in many places can be made simpler by bringing them together.

## Example Problems

Here is an example of variables keeping numbers apart even if we don't know them, and this lets us combine them without changing their value: Albert has some books in his bag, he does not know how many. Beth also has some books and she does not know how many. Chris does not know how many books he has, but he knows it is the same as Beth. Dora knows she has the same number of books as Albert. In this example there are 4 lots of books, so we could write the total number of books as:
$a+b+c+d$
Since we know that Albert and Dora have the same number of books, and Chris and Beth have the same number of books, we could also write:
$a+b+a+b$
This is the same as writing:
$2a+2b$
Here we have grouped both a terms and both b terms. We could also go further, since everything is being multiplied by 2, and write:
$2(a+b)$
This is the simplest way of writing how many books there are. Not only were the variables combined, but so were the constants (in this case the number 2). We can check if they are the same by seeing what happens when Albert has 2 books and Beth has 5.
$a+b+c+d=2+5+2+5=14$
$2(a+b)=2(2+5)=2\times 7=14$

## Practice Games

Put links here to games that reinforce these skills.
* http://www.math.com/school/subject2/practice/S2U2L4/S2U2L4Pract.html
* http://www.studystack.com/matching-47007

## Practice Problems
# Working with Numbers/Dividing Rational Numbers
## Lesson

Dividing rational numbers covers a general area of equations. For an equation that is such has only to have a numerator and denominator that are both rational numbers. In turn, one will come out with a quotient that fits the terms applied to a "rational number". Given the fact that you already understand rational numbers, you will understand this unit. If, on the other hand, you have no clue what a rational number is, then you should do some research concerning this subject so that you can understand the explanation of dividing such numbers that follows this text.
Anyway, dividing rational numbers, sometimes worded "quotients of rational expressions", is simply dividing a rational number by a rational number. For instance, look at the example problems, dividing rational numbers is very easy. If you have a fraction dividing another fraction then you simply flip the dividend and, by multiplying, one will come out with exactly the same number. The knowledge of expressing how this works is beyond the scope of this lesson. But, it works every time. You are still dividing, but you have switched your means of doing so. When you come to more complicated problems that have unknown variables the same method works. So if you have a fraction of 7 over 5 divided by 3 over 4, you will simply flip the 3 over 4 and multiply the fractions instead of dividing. This is a method that will be used again and again in math, so know it well. Look at the examples given and, although this is easy, make sure you know it.

## Example Problems

Example 1
    ${\frac {2}{7}}\div {\frac {14}{16}}$
    ${\frac {2}{7}}\times {\frac {16}{14}}$ (Change the division to multiplication and flip the fraction on the right.)
    ${\frac {1}{7}}\times {\frac {16}{7}}$ (Reduce fractions with any common factors on top and on bottom.)
    ${\frac {1\times 16}{7\times 7}}$ (Multiply the tops and bottoms together.)
    ${\frac {16}{49}}$ (Simplify)
    ${\frac {14}{2}}\div {\frac {14}{2}}$
    ${\frac {14\times 2}{2\times 14}}=$
    ${\frac {28}{28}}=1$
    ${\frac {31}{2}}=15.5$
    ${\frac {77}{9}}=8.{\overline {55}}$
    ${\frac {12}{6}}=2$
    ${\frac {36}{8}}=4.5$
    ${\frac {55}{3}}=18.{\overline {33}}$

## Practice Games

put links here to games that reinforce these skills

## Practice Problems

Note: Use parentheses for the repeating part, ex. ${\frac {1}{9}}=0.{\overline {1}}=$0.(1)

# Working with Numbers/Formulas
## Lesson

In this lesson we will be learning formulas. A formula is a standard procedure for solving a class of mathematical problems
There are many different kinds of formulas and like it or not most of us high school kids need to know some of them.
To give you an example of a formula we will show you the formula for finding area.
To find the area of a rectangle you multiply the length times the width. If the rectangle on the right is five inches wide and six inches long you would multiply five times six and get thirty, that would be the area of this rectangle. That is the formula for the area of a rectangle: Length * Width.
Of course there are other harder formulas you will have to work with but now you know the basics of what a formula is.
A formula is an established form of words or symbols for use in a ceremony or procedure.

## Example Problems

Example A:
Solve for 'y':
(a = 7)
a +5 + y = 20
Answer:
a + 5 + y = 20 (Put the 7 in place of the 'a')
7 + 5 + y = 20 (Add the integers, add 7+5=12)
12 + y = 20
12-12 + y = 20 - 12 (to get the 12 to the other side, add -12 to each side.)
y = 8 (20-12=8, so this leaves y=8)
Check:
a + 5 + y = 20 (to check it, put the 7 in place of the a, and the 8 in place of the y.)
7 + 5 + 8 = 20 (7+5+8=20, it's correct!)
Example B:
Solve for 'x':
(a = 7)
5 + 2 + x + a = 24
Answer:
5 + 2 + x + 7 = 24
14 + x = 24
14-14 + x = 24-14
x = 10
Check:
5 + 2 + x + a = 24
5 + 2 + 10 + 7 = 24

## Practice Games

put links here to games that reinforce these skills

## Practice Problems

(Note: put answer in parentheses after each problem you write)

# Lesson 9. Formulas
## Games and Activities for Review

* http://www.quia.com/cm/19994.html (30 second integer practice - add, subtract, multiply and divide)
* http://www.quia.com/cb/25168.html (add and subtract positive and negative numbers)

# Working with Numbers/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)
* Example 2x+x^2+x(=3x+x^2(form1 level))
* Example combine1:2,3:2(=6:4:2)(form2 level)

# Solving Equations/Solving Equations with Addition and Subtraction

Solving simple equations like
x+5=10
is a very important topic in beginning algebra. In this section, we will introduce the concept of a variable.
Don't be scared! It's just a number you don't know yet. Solving an equation can be looked at like you are 'filling in the blank'.
For the example above, what number plus 5 makes 10? The answer is probably pretty apparent to you when you are asked the question that way, but it might seem fuzzy when you see the notation like above, even though it means the exact same thing.

## Vocabulary

Variable - An unknown number which is represented by a letter. Common variables are usually the last letters of the alphabet. i.e. x, y, z. Variables do not necessarily have a set value, as in an expression, x+5 can mean any number plus 5. Only when you have an equals sign does it have a certain value. It's value is determined by the value on the other side of the equals sign.
Constant - Any number that is not a variable.
Expression - A 'mathematical sentence' that does NOT have an equals sign. (x+5 instead of x+5=10.) Think of this like a recipe to make ANY cake using the ingredients given to you.
Equation - A 'mathematical sentence' that DOES have an equals sign. (x+5=1.) Think of this like a recipe to make a SPECIFIC cake.
Notation - In mathematics, this describes the set of symbols and numbers and how they are arranged to make logical sense. (x+5 instead of 'a number added to five'. Notation also helps to make a lot of information much more readable.
Isolate the variable - You'll hear this a lot, but all it means is 'get the variable on one side and the constants on the other side.' x+5=10 is not solved. x=5 is solved.

## Lesson

Zoey and Pual put three eggs on each side of a scale. When Zoey took an egg away on one side the scale, it tipped. When Pual took an egg from the other side of the scale, it is balanced again. The same thing can be used in math too.
What is $y$ in this equation?
$2+y=7$
We want to know what $y$ is, so we want to make $y$ by itself on one side of the equal sign.
First subtract 2 from both sides.
$2-2+y=7-2$
After we subtracted, $y$ is by itself and we can see that $y$ equals
$y=5$

## Example Problems

Example 1
$3+x-4=4$
Add 4 to each side to get rid of the -4.
$3+x-4+4=4+4$
$3+x=8$
Subtract 3 from each side of the equation.
$3+x-3=8-3$
$x=5$

## Practice Games

put links here to games that reinforce these skills

## Practice Problems
# Solving Equations/Solving Equations with Multiplication and Division
## Vocabulary

Coefficient - A natural number (the numbers used for counting) next to a variable (the letter for which we solve the equation, also known as an unknown). For example, a variable with a coefficient of 3 would be 3x.

## Lesson

Like in the previous section, the goal is to isolate the Variable. In other words, get the variable on one side, and the constants on the other.
However, what happens when we are dealing with multiplication on a Variable?
if x is some number, 3x means three times that number.
In order to isolate the variable, we have to get rid of the coefficient. You can choose to do this before OR after you move the constants like you did in the previous section, although it's generally easier to do so afterwards.
Fractional coefficients are Basically just division. 1/2 of some number x can be written as (1/2)x or x/2. This can be helpful to remember, because all that you have to do in either case, all you have to do is multiply both sides by the bottom number, and presto chango, you have a solved equation.

## Practice Games

put links here to games that reinforce these skills

## Practice Problems

Use / as the fraction line and put spaces between the wholes and fractions!

# Solving Equations/Solving Multi-Step Equations
## Lesson

A multi-step equation is an equation that needs two or more steps to do. Our goal is to let the variable to be on its own on one side of the equal sign. You have to get rid of the numbers other than the variable on the side before the equal sign. If you do anything to a side you have to do the same thing on the other side. Solving multi-step equation is just like solving a one-step equation with more steps.

## Example Problems

Solve 4 x - 5 = 15
First we must get rid of the five by addition on both sides.
4 x - 5 = 15
4 x - 5 + 5 = 15 + 5
4 x = 20
Then divide four on both sides.
4 x = 20
4 x ÷ 4 = 20 ÷ 4
x = 5
We found our answer, x = 5
Solve 3 x + 6 = 27
First we must get rid of the 6 by subtraction on both sides.
3 x + 6 = 27
3 x + 6 - 6 = 27 - 6
3 x = 21
Then we have to divide by three on both sides.
3 x = 21
3 x ÷ 3 = 21 ÷ 3
x = 7
The Answer is, x = 7

## Practice Games

put links here to games that reinforce these skills

## Practice Problems
# Solving Equations/Solving Equations with Variables on Both Sides of the Equation
## Vocabulary
### Variable

    A letter ($a$-$z$) that takes the place of a number.

### Equation

    An example would be like $8y-3=1+10y$ (The answer is $y=-2$ )

## Lesson

NOTE: WHAT YOU DO TO ONE SIDE YOU MUST DO TO THE OTHER SIDE! NO EXCEPTIONS!
1) do the distributive property.
2) Combine like terms on both sides.
3) add/subtract numbers next to a variable on both sides.
4) divide by the number next to the variable on both sides.
5) The answer should look like: $x=20$ or $20=x$.
(Note: the variables and numbers may vary in your answer.)

## Example Problems

A simple problem:
| $2(x+5)=5(x-10)$ $2x+10=5x-50$ $2x+10-2x=5x-2x-50$ $10=3x-50$ $10+50=3x-50+50$ $60=3x$ $60/3=3x/3$ $20=x$ | Problem Distributive Prop. Subtract the variables with numbers next to them. This is what you're left with. get rid of the 50 by subtracting This is what you're left with. Get rid of the three by dividing by three This is your answer. |

## Practice Games

Put links here to games that reinforce these skills
Purplemath.com: http://www.purplemath.com/modules/index.htm

## Practice Problems

Note: Use / as the fraction line and put spaces between wholes and fractions!

# Solving Equations/Equations with More than One Variable

Sometimes, there will be more than one variable that needs to be solved for in an equation. There are systems used by mathematicians to solve for these unknown quantities, which are discussed below.

## Vocabulary
### Variable

    A letter (A-Z) that takes the place of a number.

### Expression

    A group of numbers and variables, added, multiplied, subtracted or divided by each other (examples include: $2,7x,8x-7,{\sqrt {x}}+4$)

### Equation

    An expression that is equal to another expression

### Coefficient

    The number multiplied by a variable (the 3 in $3x$)

## Lesson

There are three simple ways to solve an equation with two variables.
The number of variables in an equation corresponds to the number of equations needed to solve for the variables.

### Graphing

The most simple way to solve for the variables is by graphing both equations and finding the point where they intersect. This method is not incredibly accurate, unless we measure it with a ruler however, since we cannot be sure exactly where an intersection is unless we measure it.

### The substitution method

In these, the first step is isolating one variable on one side of an equation. After this is accomplished, the expression on the other side of the equal sign can be substituted for the variable we solved for in the first step. Now, this second equation has only one variable. We simplify the equation, solving for the remaining variable. Then we plug this value back into one of the original equations and solve for the first variable.

### Elimination method

To use the elimination method, we make the coefficients of one variable opposites of each other by multiplying or dividing both sides of the equations by the desired numbers. Once this is accomplished, we arrange the equations so that each variable is above the same variable in the other equation. After this, we add the coefficients of each variable to each other to produce a new equation, with only one variable (since the coefficients of one set of variables will cancel. Then follow the steps listed under substitution to solve for the remaining variable(s). (Refer to example 3 to see this method in action)

## Example Problems
### Example 1

| ${\begin{cases}3x+2=y\\4x-6=2y\end{cases}}$ |
| $4x-6=2(3x+2)$ | Substitute (3x + 2) for y in the second equation |
| $4x-6=6x+4$ | Distribute the 2 |
| $4x=6x+10$ | Add 6 to each side |
| $-2x=10$ | Subtract 6x from each side |
| $x=-5$ | Divide side by -2 |
| Now, plug -5 in for x in one of the original equations and solve for y: |
| $3(-5)+2=y$ |
| $-15+2=y$ | Simplify |
| $-13=y$ | Simplify to get the answer |
So the solutions are $x=-5$ and $y=-13$. (If you graphed these lines, the intersection would be at the point $(-5,-13)$)

### Example 2

| ${\begin{cases}5x+3y=12\\x-y=4\end{cases}}$ |
| ${\begin{cases}5x+3y=12\\3x-3y=12\end{cases}}$ | Multiply the second equation by 3 on each side so that the coefficients of the y's will be 3 and -3 |
| $8x+0y=24$ | "Add" the equations together using the addition method |
| $x=3$ | Divide each side of the equation by 8 |
| Plug this 3 back into the first equation for x and solve for y: |
| $(3)-y=4$ $-y=1$ $y=-1$ |
So, x=3 and y= -1

### Example 3

| ${\begin{cases}2y=3x+20\\4y=6x+24\end{cases}}$ |
| ${\begin{cases}y={\frac {3}{2}}x+10\\y={\frac {3}{2}}x+6\end{cases}}$ | Divide the equations by 2 and 4, respectively |
| ${\frac {3}{2}}x+10={\frac {3}{2}}x+6$ | Substitute |
| $10=6$ | Subtract ${\frac {3}{2}}x$ from each side |
This absurd equation is obviously false, therefore there are no solutions. This is because the coefficients of the x were the same while in slope-intercept form, so the lines were parallel (never touching) so there are no solutions. (For a further explanation of slope-intercept form, see Slope-Intercept Form of a Line)
NOTE: When you get and equation that is always true (3=3 or 5=5) there are infinitely many solutions.

## Practice Games

There is a 2-variable game on the following webpage:

## Practice Problems

Use / as the fraction line and put spaces between the wholes and fractions!

# Solving Equations/Formulas
## Basic Formulas

* $(a+b)^{2}=a^{2}+2ab+b^{2}$
* $(a-b)^{2}=a^{2}-2ab+b^{2}$
* $a^{2}-b^{2}=(a+b)(a-b)$
* $(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$
* $(a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}$
* $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$
* $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$
* $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$
* $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

### Derivation (1)

$(a+b)^{2}$ $=(a+b)(a+b)$ $=a(a+b)+b(a+b)$ $=a^{2}+ab+ab+b^{2}$ $=a^{2}+2ab+b^{2}$

### Derivation (2)

$(a-b)^{2}$ $=(a-b)(a-b)$ $=a(a-b)-b(a-b)$ $=a^{2}-ab-ab+b^{2}$ $=a^{2}-2ab+b^{2}$

### Derivation (3)

$a^{2}-b^{2}$ $=a^{2}+ab-ab-b^{2}$ $=a(a+b)-b(a+b)$ $=(a+b)(a-b)$

### Derivation (4)

$(a+b)^{3}$ $=(a+b)\times (a+b)^{2}$ $=(a+b)\left(a^{2}+2ab+b^{2}\right)$ $=a\left(a^{2}+2ab+b^{2}\right)+b\left(a^{2}+2ab+b^{2}\right)$ $=a^{3}+2a^{2}b+ab^{2}+a^{2}b+2ab^{2}+b^{3}$ $=a^{3}+3a^{2}b+3ab^{2}+b^{3}$

### Derivation (5)

$(a^{3}-b^{3})$ $=(a-b)\times (a-b)^{2}$ $=(a-b)\left(a^{2}-2ab+b^{2}\right)$ $=a\left(a^{2}-2ab+b^{2}\right)-b\left(a^{2}-2ab+b^{2}\right)$ $=a^{3}-2a^{2}b+ab^{2}-a^{2}b+2ab^{2}-b^{3}$ $=a^{3}-3a^{2}b+3ab^{2}-b^{3}$

### Derivation (9)
# Solving Equations/Chapter Review

Lesson 1. Solving Equations with Addition and Subtraction
Lesson 2. Solving Equations with Multiplication and Division
Lesson 3. Solving Multi-Step Equations
Lesson 4. Solving Equations with Variables on Both Sides of the Equation
Lesson 5. Equations with More than One Variable
Lesson 6. Formulas

# Solving Equations/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)
* Example 2x/2+4x/4=4(x=2,medium)

# Proportions and Proportional Reasoning/Proportions
## Vocabulary

Ratio: The comparison of two numbers.
Reciprocal: The multiplicative inverse of a number -- when the numerator and denominator of a fraction are switched around.
Equivalent Ratios: Two ratios that reduce to the same ratio.
Proportion: An equation stating two ratios are equal.
To Cross-multiply: see below.

## Lesson

There are several ways to express a "ratio". Let's compare the number of boys with the number of girls in a particular classroom. Let's say that our classroom has 25 students, 10 of whom are boys. That means there are 15 girls. So, the ratio of boys to girls is 10 to 15.
Writing a Ratio
There are three ways of expressing a ratio. You can simply use words like we did above, or you can separate the two numbers using a colon or a fraction (the mathematician's choice).
10:15 or 10/15 or ${\frac {10}{15}}$
In mathematics, we always use fractions to represent ratios. In this classroom example, many other ratios that can be made:
${\frac {10}{25}}$, the number of boys out of the total number of students
${\frac {15}{25}}$, the number of girls out of the total number of students
${\frac {15}{10}}$, the number of girls to the number of boys
Simplifying a Ratio
Since we're using a fraction to represent ratios, the ratios can sometimes be reduced. For example, the ratio of boys to girls in our hypothetical classroom is ${\frac {10}{15}}$, but can be reduced to ${\frac {2}{3}}$. So, if we would be correct we said that the ratio of boys to girls in our hypothetical classroom is 2:3, or two boys for every 3 girls.
Proportions
The following equation is a proportion:
${\frac {10}{15}}={\frac {2}{3}}$
Any proportion is simply an equation the states two ratios are equal to one another. Sometimes, a proportion may contain variables:
${\frac {2}{3}}={\frac {x}{30}}$
If we wish to solve such an equation, we can use a process called cross-multiplication.
Solving Proportions using Cross-Multiplication
If ${\frac {a}{b}}={\frac {c}{d}}$, then the products that are formed by diagonals across the equal sign are also equal: $ad=bc$. Note that this would be equivalent to saying $bc=ad$.

## Example Problems

1) Is the ratio ${\frac {4}{6}}$ equal to the ratio ${\frac {12}{16}}$?
We can test by cross-multiplying. If the cross-products are equal, then so are the original two ratios.
$4\cdot 16$ ?= $6\cdot 12$
$64$ ?= $72$
No, $64\not =72$, so ${\frac {4}{6}}\not ={\frac {12}{16}}$.
2) Solve the following proportion for x.
${\frac {3}{4}}={\frac {x}{10}}$
Cross-multiply.
$3\cdot 10=4x$
$30=4x$
Solve for x by dividing both sides of the equation by 4. Your result: $x={\frac {30}{4}}=7.5$.
3) Solve for x:
${\frac {x+1}{4}}={\frac {3}{8}}$
Cross multiply:
$8\cdot (x+1)=3\cdot 4$
$8x+8=12$
Now, solve for x:
Subtract 8 from both sides: $8x=4$
Divide both sides by 8: $x={\frac {1}{2}}$
4) A 9th-grade algebra classroom has a ratio of boys to girls of \frac{1} {2}. If there are 14 girls, how many boys are in the class?
We can create a proportion that compares the ratios of boys to girls, where x is the number of boys in the class:
${\frac {1}{2}}={\frac {x}{14}}$
Solve the proportion by cross-multiplying:
$2x=14$
$x=7$
There are 7 boys in the class.

## Practice Games

Online games/resources about RATIOS: All About Ratios
A list of games/resources about RATIOS and PROPORTIONS: Ratios and Proportions from eThemes

## Practice Problems

Use / as the fraction line and put spaces between wholes and fractions!

# Proportions and Proportional Reasoning/Percents
## Vocabulary

Percent: Parts per one hundred

## Lesson

Suppose we had a grid of 100 squares, of which 17 are shaded in. There are several ways to express this as a ratio, as you learned in the previous lesson, such as ${\frac {17}{100}}$, the number of shaded squares compared to the number of squares in total. However, we can also write this as a percent. Since there are 17 shaded squares and 100 total, we say that $17\%$ of the squares are shaded. The $\%$ symbol stands for "percent".
Now suppose that instead of 100 squares, we have 50 squares, with 9 of them shaded. The percent shaded would not be $9\%$, because percent means "per 100", and we have 9 shaded squares out of 50 total squares. To find the percent, we need some number over 100, so we can set up a proportion.
          ${\frac {9}{50}}={\frac {x}{100}}$
Cross-multiplying gives us $9\times 100=50x$, and $x=18$.
We also could have noticed that $50\times 2=100$, so we just need to multiply the numerator, 9, by 2 to get our answer, 18.

## Practice Problems

Use / as the fraction line and put spaces between the wholes and fractions!

# Proportions and Proportional Reasoning/Percent of Change
## Vocabulary

percent - a ratio that compares a number to 100
percent of change - the percent an amount changes from its original amount.
percent of change = (amount of change) / (original amount), where the amount of change is the new value minus the original value.
percent of increase - when a value increases from its original amount, you call the percent of change the percent of increase
percent of decrease - when a value decreases from its original amount, you call the percent of change the percent of decrease

## Lesson

Finding the percent of change is using the ratio of the amount of change to the original amount. If the amount increases then the percent of change is called the percent of increase and will result in a positive value. If the amount decreases then the percent of change is called the percent of decrease and will result in a negative value.
The first question to ask yourself when finding the percent of change is: Is it an increase or a decrease? Once you have determined what type of change you are dealing with then you can calculate what that percent of change is.
The percent of change is calculated by dividing the amount of change by the original amount. The amount of change is the new value minus the original value. An easy way to remember how to find percent of change is to consider the no over o method. In the no over o method, n stands for the new amount and the o stands for the original amount.
Here is how to use the no over o method.
(the new amount - the original amount) over the original amount.
(n-o)/o

## Example Problems

A. Solve and describe the percent of change as a percent of increase or decrease. Round to the nearest percent.
1) $12 to $9
new amount = $9
original amount = $12
(n-o)/o
(9-12)/12 = -3/12 = -1/4 = -0.25
no over o is negative which means it decreased so there is a 25% decrease
2) 19 in to 25 in
new amount = 25 in
original amount = 19 in
(n-o)/o
(25-19)/19 = 6/19 = 0.3158
no over o is positive which means it increased so there is approximately a 32% increase
B. Solve the percent of change word problem
1) Anna's grade in Algebra changed from 88 to 94. What is the percent of change in her grade?
new amount = 94
original amount = 88
(n-o)/o
(94-88)/88 = 6/88 = 3/44 = .068 = 6.8%
therefore it was a 6.8% increase
2) Between 1940 and 1980 the Federal budget went from $725.3 billion to $9.5 billion. What was the percent of change?
new amount = $9.5 billion
original amount = $725.3 billion
(n-o)/o
(9.5-725.3)/725.3 = -715.8/725.3 = -0.986 = -98.6%
therefore it was a 98.6% decrease
C. Solve
1) If 46 is decreased by 20%, what is the result?
new amount = x
original amount = 46
(n-o)/o
(x-46)/46 = -0.20
x-46 = -9.2
x = 36.8
2) If 16 is increased by 30%, what is the result?
new amount = x
original amount = 16
(n-o)/o
(x-16)/16 = 0.30
x-16 = 4.8
x = 20.8

## Practice Games

(good to check your answers)

## Practice Problems
# Proportions and Proportional Reasoning/Weighted Averages

What Is a Weighted Average? Weighted average is a calculation that takes into account the varying degrees of importance of the numbers in a data set. In calculating a weighted average, each number in the data set is multiplied by a predetermined weight before the final calculation is made.
Key takeaways:
* Weighted average is the average of a set of numbers, each with different associated “weights” or values.
* To find a weighted average, multiply each number by its weight, then add the results.
* If the weights don’t add up to one, find the sum of all the variables multiplied by their weight, then divide by the sum of the weights.
The weighted average method is a tool used in classrooms, statistical analysis and accounting offices, among others. A weighted average helps the user gather a more accurate look at a set of data than the normal average alone. The accuracy of the numbers you arrive at with this method is determined by the weight you give specific variables in the data set.
In this article, we explore how to calculate weighted average using two methods.

## What is weighted average?

A weighted average is the average of a data set that recognizes certain numbers as more important than others. Weighted averages are commonly used in statistical analysis, stock portfolios and teacher grading averages. It is an important tool in accounting for stock fluctuations, uneven or misrepresented data and ensuring similar data points are equal in the proportion represented.

## Weighted average example

Weighted average is one means by which accountants calculate the costs of items. In some industries where quantities are mixed or too numerous to count, the weighted average method is useful. This number goes into the calculation for the cost of goods sold. Other costing methods include last in, first out and first in, first out, or LIFO and FIFO respectively.
Example:
A manufacturer purchases 20,000 units of a product at $1 each, 15,000 at $1.15 each and 5,000 at $2 each. Using the units as the weight and the total number of units as the sum of all weights, we arrive at this calculation:
$1(20,000) + $1.15 (15,000) + $2 (5,000) / (20,000 + 15,000 + 5,000) = ($20,000 + $17,250 + $10,000) / ($20,000 + 15,000 + 5,000) = $47,250 / 40,000 = $1.18
This equals a weighted average cost of $1.18 per unit.

## How to calculate weighted average

Weighted average differs from finding the normal average of a data set because the total reflects that some pieces of the data hold more “weight,” or more significance, than others or occur more frequently. You can calculate the weighted average of a set of numbers by multiplying each value in the set by its weight, then adding up the products.
For a more in-depth explanation of the weighted average formula above, follow these steps:
* Determine the weight of each data point
* Multiply the weight by each value
* Add the results of step two together

### 1. Determine the weight of each data point

You determine the weight of your data points by factoring which numbers are most important. Teachers often weigh tests and papers more heavily than quizzes and homework, for example. In large statistical data sets, such as consumer behavior data mining or a population census, randomized data trees are used to determine the importance of a variable in a data set. This helps ensure the distribution of importance is unbiased. This process is typically performed with the aid of a computer program. For accounting and finance purposes, the number of units of a product is used as the weighting factor.
Example:
* You score a 76 on a test that is 20% of your final grade. The percentage of your grade is the weight it carries.
* An investor purchases 50 stocks at $100 each. The stocks purchased serve as the weight.
Related: 21 Job Interview Tips: How to Make a Great Impression

### 2. Multiply the weight by each value

Once you know the weight of each value, multiply the weight by each data point.
Example:
In a data set of four test scores where the final test is more heavily weighted than the others:
* 50(.15) = 7.5
* 76(.20) = 15.2
* 80(.20) = 16
* 98(.45) = 44.1

### 3. Add the results of step two together

Calculate the sum of all the weighted values to arrive at your weighted average.
Example:
7.5 + 15.2 + 16 + 44.1 = 82.8
The weighted average is 82.8%. Using the normal average where we calculate the sum and divide it by the number of variables, the average score would be 76%. The weighted average method stresses the importance of the final exam over the others.

## How to calculate weighted average when the weights don't add up to one

Sometimes you may want to calculate the average of a data set that doesn't add up perfectly to 1 or 100%. This occurs in a random collection of data from populations or occurrences in research. You can calculate the weighted average of this set of numbers by multiplying each value in the set by its weight, then adding up the products and dividing the products' sum by the sum of all weights.
For a more in-depth explanation of the weighted average formula above when the weights don’t add up to one, follow these steps:
* Determine the weight of each number
* Find the sum of all weights
* Calculate the sum of each number multiplied by its weight
* Divide the results of step three by the sum of all weights
Related: Learn About Being a Data Scientist

### 1. Determine the weight of each number

To determine the weight of each number, consider its importance to you or the frequency of occurrence. If you are trying to calculate the average number of business leads you pursue, you may want leads that turn into sales to weigh more heavily than cold calls. To find the weighted average without added bias, calculate the frequency a number occurs as the variable's weight. This reflects its influence over the entire data set.
Example: Calculate the average time you spend exercising four days a week over the period of a month or four weeks. The time you spent exercising on any given day is the data set. The number of days you exercised for an average time is the weight you'll use.
* 7 days you exercised for 20 minutes
* 3 days you exercised for 45 minutes
* 4 days you exercised for 15 minutes
* 2 days you were supposed to exercise and did not

### 2. Find the sum of all weights

The next step to finding the weighted average of a data set that doesn't equal 1 is to add the sum of the total weight. From our previous example, you should have a total of 16 days spent exercising:
* 7+3+4+2 = 16

### 3. Calculate the sum of each number multiplied by its weight

Using the frequency numbers, multiply each by the time you spent exercising. The combined total gives you the sum of the variables multiplied by their respective weights.
Example:
* 20(7) = 140
* 45(3) = 135
* 15(4) = 60
* 0(2) = 0
* 140 + 135 + 60 + 0 = 335

### 4. Divide the results of step three by the sum of all weights

The formula for finding the weighted average is the sum of all the variables multiplied by their weight, then divided by the sum of the weights.
Example:
Sum of variables (weight) / sum of all weights = weighted average
335/16 = 20.9
The weighted average of the time you spent working out for the month is 20.9 minutes.

# Proportions and Proportional Reasoning/Solving Similar Triangles using Proportions
## Vocabulary

Proportions - Proportions are equations resulting from two or more ratios. (Ex. 2/3 = 4/x)
Similar triangles - have equal corresponding angles and proportional corresponding sides (Ex. If triangle ABC has side lengths 1, square root of 3, and 2, and angles of 30$^0$, 60$^0$, and 90$^0$, and triangle DEF has side lengths 2, 2*square root of 3, and 4, and angles of 30$^0$, 60$^0$, and 90$^0$, then triangles ABC and DEF are similar, because their corresponding angles are equal, and their corresponding sides are proportional to each other.)

## Example Problems

Proportions:
2/3 = 4/x Cross-multiply: 2x = 12 Divide both sides by 2: x = 6
Similar Triangles: If Triangles ABC and DEF are similar:
Find DE if AB=3, BC=4, EF=8. Since AB/DE = BC/EF, 3/DE = 4/8 Solving the proportion: DE=6 Find angle C when angle F=60$^0$. Since corresponding angles are congruent in similar triangles, angle C=angle F=60$^0$

## Practice Problems
# Proportions and Proportional Reasoning/Chapter Review

Lesson 1. Proportions
Lesson 2. Percents
Lesson 3. Percent of Change
Lesson 4. Weighted Averages
Lesson 5. Direct and Inverse Variation
Lesson 6. Solving Similar Triangles using Proportions

# Proportions and Proportional Reasoning/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)
* 2+2=x (x=4,easy)

# Lines (Linear Functions)/Graphing Points

Graphing points is as simple as counting. First, a grid is needed which we will call a graph. A set of given coordinates are needed to plot a point which can be called ordered pairs. Ordered pairs follow this format: (x, y). The first number represents where at the point lies in the x-direction. The second number represents where at the point lies in the y-direction, or how many units the point lies from the origin. For example, the coordinates (2, 3) tell you to move 2 units to the right on the x-axis, and 3 units up on the y-axis, then draw a point. For instance, if the coordinates given were (-5, 0) you would draw a point that is 5 units left of the origin (which is (0, 0)) and the point wouldn't go up or down because the second coordinate is 0. These points are often labeled as letters such as A, B, or C.

# Lines (Linear Functions)/Slope of a Line
## Lesson

To understand the Cartesian Coordinate graph and how it works, you must know how number lines work. A number line is a line of numbers that streches infinitely on both sides. The Negative numbers extend infinitely to the left and the Positive numbers go infinitely to the right. Zero sits right smack in the dead center of the line however long it might be. A cartesian coordinate graph is just two infinite number lines one vertical(the X-axis) and one horizontal(the Y-axis) on top of each other forming a cross with 0 as the center point or Origin as it is called. The graph is also divided into 4 quadrants. these are numbered with Roman numerals.
http://commons.wikimedia.org/wiki/Image:2D_Cartesian_Coordinates.svg
However, for understandability's sake, this graph is usually shortened to -6 and + 6 on both the X and Y axis. Also, however, this graph is also ugly and not very user friendly. (Gridboard)
Now, how to determine the slope of a line: Let's pretend that we are baking giant chocolate cookies for a bake sale. Right before you start your Bake-O-Rama, You make three of them. You figure you can make two an hour. You also take inventory every hour like this:

## Example Problems

slope (m) = ${\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}$
Using the points in the diagram above, x$_1$ and y$_1$ are both 2 and x$_2$ is 6 and y$_2$ is 4. So we substitute each value into the slope equation.

## Practice Games

Sorry, none at this time!

## Practice Problems
# Lines (Linear Functions)/Slope-Intercept Form of a Line
## Vocabulary
### Domain

    The x position of a point

### Slope-intercept form

    y=mx+b, where x and y are constants and m=slope and b= y-intercept

## Example Problems

Graph:
* y=2x+3
* y=3/2x-5

## Practice Games

Linear Equations Review Project

## Practice Problems
# Lines (Linear Functions)/Find the Equation of the Line Using Two Points
## Vocabulary
## Lesson

You can find the y intercept(b) of a line by using "point slope" with a pair of cordinates.
-Find the slope (y$_2$ - y$_1$) / (x$_2$ - x$_1$) -Use one of the coordinates (points) and use this formula: y-y$_1$=m(x-x$_1$) -Then you end up with y=mx+b

## Example Problems

Find the y-intercept of the following coordinates:
(2,1) (3,-7)
(1,3) (3,4)
(0,2) Example 1:
First find the slope m of between the two points:
m = (y-y1) / (x-x1)
m = (6-2) / (3-5)
m = -2
We know that the equation has the form y = mx + b, and we also know that this function passes both of the points, so let's use point #1 to find b:
point #1 (2,1)
2 = m(5) + b
2 = (-2)(5) + b
b = 12
Alternatively, we can use point #2 and get to the same result.
The line equation is:
y = -2x + 12

## Practice Problems
# Lines (Linear Functions)/Functions and Relations
## Lesson

y = x^2
Is this a function?
versus, y^2 = x
Is this a function?
The first equation is a function because x has only one y-value, whereas the second equation has many values.
In a function, each domain value or x-value should only have one y-value, otherwise it is not a function.
Definition: In two sets, A and B, a function is a mapping from A to B.
This mapping should satisfy the following conditions in order for it to be a function:
* All the elements of set A should be paired with any element of set B.
* An element of set A should not be paired with more than one element of set B. (i.e. the first coordinates of an ordered pairs should not be repeated)

## Practice Problems
# Lines (Linear Functions)/Chapter Review

Lesson 1. Graphing Points
Lesson 2. Equations in Two Variables
Lesson 3. Graphing Lines(Linear Functions)
Lesson 4. Slope of a Line
Lesson 5. Slope-Intercept Form of a Line
Lesson 6. Find Parallel and Perpendicular Lines
Lesson 7. Find the Equation of the Line Using Two Points
Lesson 8. Functions and Relations

# Lines (Linear Functions)/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)

# Linear Inequalities/Solving Inequalities
## Vocabulary

* Inequality
* Inverse Operation
* Graphing
* Number Line
* Filled Circle
* Empty Circle

## Practice Problems
# Linear Inequalities/Chapter Review

Lesson 1. Solving Inequalities
Lesson 2. Solving "AND" Compound Inequalities
Lesson 3. Solving "OR" Compound Inequalities
Lesson 4. Solving Absolute Value Inequalities
Lesson 5. Graphing Linear Inequalities

# Linear Inequalities/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)

# Systems of Linear Equations/Solving Linear Systems by Graphing

In Algebra, the linear equation is defined as an algebraic equation, in which each term has an exponent value of 1. If the linear equation is graphed, it should always result in a straight line. The system of linear equations contains two or more linear equations. While solving the system of linear equations graphically, we have to graph the equations in the same coordinate system. The point at which the straight lines intersects is called the solution of the linear equation. The solution to the system of linear equations is an ordered pair, which satisfies the given equations.
Now, let us solve the system of linear equations by the graphical method.
Consider two linear equations,
y = x-1
y= 2x+2
Now, graph the equations in the coordinate plane.
Now, the two linear equations intersect at the common point (-3, -4), which is the solution for the given system of linear equations.

# Systems of Linear Equations/Solving Linear Systems by Substitution
## Vocabulary

* Variable
* System
* Solution set

## Lesson

To solve a system of linear equations without graphing, you can use the substitution method. This method works by solving one of the linear equations for one of the variables, then substituting this value for the same variable in the other linear equation and solving for the other variable. It does not matter which equation you choose first, or which variable you solve for first; the values for both variables will be the same.

## Example Problems

For example, given the system of linear equations:
$y=2x-8$
$y=-8x-3$
The first step would be to choose one of the equations and solve it for either x or y. In the second equation y is not multiplied by a constant so it can be isolated in fewer steps.
$4x+y=24$
$y=24-4x$
Now you have a value, $24-4x$, for y. Substitute this value of y into the first equation.
$2x-3y=-2$
$2x-3(24-4x)=-2$
Solve this equation for the variable x.
$2x-3(24-4x)=-2$
$2x-72+12x=-2$
$14x-72=-2$
$14x=70$
$x=5$
We now have a value for x that can be substituted into either equation to solve for y.
$4x+y=24$
$4(5)+y=24$
$20+y=24$
$y=4$
The solution to this system of linear equations is $x=5$, $y=4$. This can also be written as $(x,y)=(5,4)$.
NOTE: If we substitute the value of x into the other equation, the value of y will remain the same.
$2x-3y=-2$
$2(5)-3y=-2$
$10-3y=-2$
$-3y=-12$
$y=4$

## Practice Games

Practice Problems
Math Drills

## Practice Problems

Note: Use / as the fraction line and put spaces between wholes and fractions!

# Systems of Linear Equations/Solving Linear Systems by Linear Combinations
## Practice Problems

Use / as the fraction line!

# Systems of Linear Equations/Chapter Review

Lesson 1. Solving Linear Systems by Graphing
Lesson 2. Solving Linear Systems by Substitution
Lesson 3. Solving Linear Systems by Linear Combinations
Lesson 4. Using Matricies to Solve Linear Systems

# Systems of Linear Equations/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example
$x+11=36$:
    ${\begin{cases}x+11=y+111\\y=-3x\end{cases}}$
$x+11=-3x+111$
$x=25$ ,($25+11=36$)
($y=36-111=-75$, medium)

# Polynomials/Exponents
## Vocabulary

Base: The number directly preceding an exponent
EX: a$^2$ -> a is the base
Exponent: The number (written in superscript) used to express how many times a base is multiplied by itself
EX: a$^4$ = a * a * a * a -> 4 is the exponent
EX: 4$^3$ = 4 * 4 * 4 = 64 -> 3 is the exponent

## Lesson

Exponents are a simple way to represent repeated multiplication. For example a x a = a$^2$. There are a few simple rules for exponents that help reduce very large problems to simple little ones. The rules are as follows:
1) The exponent of any number is always a one (1): a = a$^1$
2) When we multiply the same base we add our exponenents: a$^3$ x a$^2$ = a$^{3+2}$ = a$^5$
3) When we divide the same base we subtract our exponents: a$^6$ / a$^4$ = a$^{6-4}$ = a$^2$
4) When we raise a power to a power we multiply our exponents: (a$^2$)$^3$ = a$^{2*3}$ = a$^6$
5) When we raise a PRODUCT to a power we raise both parts of the product to the power: (ab)$^3$ = a$^3$b$^3$ [NOTE: This ONLY works with multiplication and NOT addition: (a + b)$^3$ ≠ a$^3$ + b$^3$]
6) When we raise a QUOTIENT to a power we raise both parts of the quotient to the power: (a/b)$^2$ = a$^2$ / b$^2$ [NOTE: This ONLY works with division and NOT subtraction: (a - b)$^2$ ≠ a$^2$ - b$^2$]

## Practice Problems

Use ^ for exponentiation and remember Order of Operations

# Polynomials/Zero and Negative Exponents
## Lesson

There are two very important things you need to know when working with Zero Power or Negative Exponents.
First, any number to the Zero Power always equals one. For example (-50)$^0$ = 1
There is one number that CANNOT be raised to the Zero Power, 0$^0$ does not exist!
When dealing with Negative Exponents there is a simple trick. Whatever part of a fraction the negative exponent is in, switch it and the exponent becomes positive.
a$^{-2}$ = 1/a$^2$
1/a$^{-3}$ = a$^3$
If we have something a little more complicated, we only move things with Negative Exponents. These processes only work with multiplication. If there is addition/subtraction involved, then we are in something a little more complicated than Algebra 1...
(a$^{-2}$c$^3$)/b$^{-1}$ = (bc$^3$)/a$^2$
Something like this wouldn't follow the aforementioned rules
(a$^{-2}$ + b$^5$)/(c$^6$)
This problem would require a little more work: splitting up the fraction and working with both parts individually and having an answer with two fractions instead of one nice one. It's possible but it doesn't flow like the other examples or the practice problems.

## Example Problems

(-2)$^2$ = 4. -2$^2$ = -4.

## Practice Problems

Use ^ for exponentiation

# Polynomials/Adding and Subtracting Polynomials
## Vocabulary

Polynomial: Mathematical sentence with "many terms" (literal English translation of polynomial). Terms are separated by either a plus (+) or a minus (-) sign. There will always be one more term than there are plus (+) or minus (-) signs. Also, the number of terms will (generally speaking) be one higher than the lead exponent.
EX: A Quadratic function has a lead exponent of 2, but generally has three terms (ax$^2$ + bx + c; lead exponent = 2, # of + (or -) signs = 2, # of terms = 3)
Like Terms: Terms in a polynomial that have the same power of the variable
EX: 3x$^2$ and 2x$^2$ are like terms, but 3x$^2$ and 4x$^3$ are not!
PROPERTIES TO REMEMBER:
If we see a variable standing alone (it has no coefficient, no number next to it) then we assume that there is an invisible one (1) standing there:
x$^2$ = 1x$^2$
We are extremely lazy in math and do not like to write numbers that we feel are unnecessary. This is one of the cases in which we do not write what is actually there but we always remember it is. Another case is with fractions and whole numbers. The DEMONIMATOR of every whole number is a one (1) but we do not write this one (1) because we do not feel the need. We always remember it is there though.
4 = 4/1 [READ: 4 over 1]

## Lesson

There are many, many types of polynomials in the world of mathematics and are classified by the power (or exponent) of their leading term.
Some Common Functions
1) f(x) = ax + b (more commonly seen as y = mx + b). The leading term has an exponent of one (1) and is called a LINEAR FUNCTION because it creates a line when graphed. Because there are two terms, this function is called a binomial, or two-termed, function.
2) f(x) = ax$^2$ + bx + c The leading term has an exponent of two (2) and is called a QUADRATIC FUNCTION because the first x is squared and squares are QUADrilaterals. This function generally has three terms and is therefore called a trinomial. A QUADRATIC FUNCTION has amazing properties that span years of mathematical studies. Since this is the first polynomial to have more than two terms, it is the first polynomial able to be factored. However, there are special cases in which ax$^2$ + bx + c cannot be factored.
3) f(x) = ax$^3$ + bx$^2$ + cx + d The leading term has an exponent of three (3) and is called a CUBIC FUNCTION because the first x is cubed (raised to the third power). This function generally has four terms and will always be able to factor out at least one term of the form (x - h) [where h is any number].
There are an infinite number of polynomials and each one has amazing features unique to that function. However, there are a few universal traits to all functions. Every function with an even lead exponent (ax$^2$, ax$^4$, etc. . .) have a chance of not being factorable. Every function with an odd lead exponent (ax, ax$^3$, etc. . .) will be able to factor AT LEAST ONE term of the form (x - h) [where h is any number].
_______________________________________________________________________________________________________________________________
As with regular numbers, we can add and subtract polynomials. However, instead of only worrying about which numbers have an x and which numbers do not, we also have to keep in mind that the exponents have to be the same in order for us to add and subtract terms.
EX: Add (4x$^3$ + 3x + 1) + (-3x$^3$ + 2x$^2$ + 4)
Step 1: We have to match up our terms: (4x$^3$ + -3x$^3$) + (2x$^2$) + (3x) + (1 + 4)
Step 2: We combine the coefficients of the like terms: x$^3$ + 2x$^2$ + 3x + 5 <- We've solved the problem
(4x$^3$ + 3x + 1) + (-3x$^3$ + 2x$^2$ + 4) = x$^3$ + 2x$^2$ + 3x + 5
Subtracting polynomials is the same thing, except we add an extra step. When we subtract polynomials we use the distributive property first and multiply the second polynomial by a negative one (-1). This changes all the signs of the second polynomial to the OPPOSITE of what they are. [NOTE: When we add a negative number we actually subtract!!!]
EX: Subtract (3x$^4$ + 2x$^2$ + 2) - (x$^4$ + 6x$^2$ + 12x - 1)
Step 1: We distribute the negative one (-1) across the second polynomial and our new polynomial reads:
(-x$^4$ - 6x$^2$ - 12x + 1) <- Notice how the signs are all opposite of what we were given.
Step 2: We match up our terms: (3x$^4$ + -x$^4$) + (2x$^2$ + -6x$^2$) + (-12x) + (2 + 1)
Step 3: We combine our coefficients of the like terms: 2x$^4$ - 4x$^2$ - 12x + 3
(3x$^4$ + 2x$^2$ + 2) - (x$^4$ + 6x$^2$ + 12x - 1) = 2x$^4$ - 4x$^2$ - 12x + 3
Now we've successfully subtracted two polynomials.

## Practice Problems

Use ^ for exponentiation

# Polynomials/Multiplying Monomials
## Practice Problems

Use ^ for exponentiation

# Polynomials/Multiplying Monomials and Polynomials

Basic Algebra/Polynomials/Multiplying Monomials and Polynomials. 1 term=monomial,2 terms=binomial,3 terms=trinomial and more than 3 terms=polynomial.You can classify an algebraic expression according to the number of terms it has.A polynomial is an expression with any number of terms which are greater than three terms.A monomial is an expression with only one term,a binomial is an expression with two terms and a trinomial is an expression with three terms. WORKED EXAMPLES:1.-2(x)=-2x. 2.4(x+2)=4(x)+4(2)=4x+8 3.-3xy(x²y+2x² -x-1). Remember when calculating, don't forget the signs.When the signs are the same it becomes a addition sign.But when the signs are different it becomes subtraction sign.These signs are the ones which separate the terms e.g.addition(+)*addition (+)=addition (+). subtraction (-)*subtraction (-)=addition and addition (+)*subtraction (-)=subtraction (-). subtraction (-)*addition (+)=subtraction (-).An exercise you can do is :1.a)-2 and 2x-3y

# Polynomials/Chapter Review

Lesson 1. Exponents
Lesson 2. Zero and Negative Exponents
Lesson 3. Adding and Subtracting Polynomials
Lesson 4. Multiplying Monomials
Lesson 5. Powers of Monomials
Lesson 6. Multiplying Monomials and Polynomials
Lesson 7. Multiplying Polynomials

# Polynomials/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)

# Factoring/Factors of Integers
## Vocabulary

* Factor: A factor of an integer n is any number that divides n without remainder.
* Prime Number: A positive integer n is called prime if its only factors are 1 and n (and -1 and -n).
* Composite Number: An integer n is called composite if it is not prime. This means that n is composite if it has a factor that is not 1 and not n.

## Lesson

Sometimes numbers can be written as the product of other numbers. When we write a number n as a product (n = a x b) then say that a and b are factors of n. The equation n = a x b is called a factorization of n.
For example, 6 is a factor of 12 because 12 = 2 x 6. Also, 3 is a factor of 6 because 6 = 2 x 3. If we put these factorizations together we get 12 = 2 x 6 = 2 x (2x 3) = (2x2)x3 = 4 x 3 and so 3 and 4 are factors of 12.
But, for example, 4 is not a factor of 13 because 4 cannot multiply any other whole number and come out with an even 13.
If we have a number n we can always factor it as n = 1 x n. So 1 and n are always factors of n.
If we have a factorization n = a x b then n = (-a) x (-b). This means that if a is a factor of n then -a is also a factor of n. A positive integer can always be factored into positive integers.
We call a positive integer p a prime if it can only be factored into positive numbers as p = 1 x p or p = p x 1. The number 1 is a special number which we do not call prime.
There are many prime numbers: 2,3,5,7,11,13,17 and more.
When a positive integer is not prime we call it composite. Since we can write 12 = 2 x 6, we know that 12 is not prime. That means we call 12 composite.
There are many composite numbers: 4,6,8,12,14,15,16,18 and more.
Every positive number can be factored into a product of positive primes in only one way. For example, 30 = 2 x 15 = 2 x 3 x 5 where 2, 3 and 5 are prime. A factorization of a number into a product of primes is called a prime factorization. There is only one prime factorization of a number.
If we write a factorization of 30 that then it must either contain a composite number of be 2x3x5. The factorizations of 30 are listed below:
* 30 = 1 x 30,
* 30 = 6 x 5,
* 30 = 2 x 15,
* 30 = 10 x 3, and
* 30 = 2 x 3 x 5. This is the prime factorization of 30.
What, then, is the prime factorization of 100?
100 = 1 x 100, 100 = 2 x 50 100 = 4 x 25 100 = 2 x 2 x 5 x 5

## Example Problems

* Is 4 a factor of 20? Answer: Yes. The number 4 is a factor of 20 because 20 = 4 x 5.
* Is 6 a factor of 20? Answer: No. The number 6 is not a factor of 20 because 6 does not divide 20 without remainder; 20 = (6 x 3) + 2.
* Find all factorizations of 20. Which factorization of 20 is the prime factorization. Answer:
** 20 = 1x20,
** 20 = 2x10,
** 20 = 4x5, and
** 20 = 2x2x5 are the factorizations of 20. This last factorization is the prime factorization of 20.

## Practice Games

One way to think of factor games is as all the different ways you can package groups of given numbers. For instance 12 items can be packaged as a string of 12, like a giant skinny chocolate bar, or as 2 groups of six like a carton of eggs, or as 4 sets of 3 like two six packs of soft drinks. Experience with factors and numbers can make other people respect your ideas when you use your number sense to find more attractive or efficient ways to create or display something.
There are many online games you can play involving factorization. One of my favorites is the factor game. In order to win this game you must pick numbers with small prime factors while leaving numbers with special large composite factors for the end game.
A factor game you can play without a computer is to pick a range of prime numbers. You can give one player a time limit to construct a composite number with these primes. If there are just two players you can allow player number 2 to factor the number and then compare the factorizations. If player 1 made a mistake constructing the number then player 2 gets a point, if player 2 does not factor the number correctly then player 1 gets a point. This game can be extended for multiple players by allowing the players factoring the numbers to race for the factorization, and then checking the answers in the order which they are turned in. Something that makes this game interesting is that multiplications with larger prime numbers become harder to perform correctly and harder to decompose. You can easily handicap this game by giving some players more times to pick their number then other players. For instance how big a number can you build with the primes 2, 3, 5, 7, 11, and 13 in 30 seconds? It takes less time to double numbers than multiply them by 11, but these numbers are also easier to factor.

## Practice Problems

Easy
Medium
Hard

# Factoring/Squares of Binomials
## Vocabulary

Binomial - An algebraic expression with exactly two terms.
Square - Multiply a number by itself.
FOIL - The product of two binomials is the sum of the products dio of the First terms, the Outer terms, the Inner terms, and the Last terms.
Quantity -Total amount or number.
Please note that the ‘FOIL’ method as well as the shortcut shown below is ONLY for binomial(s).

## Technical Equation

(a+b)$^2$ = a$^2$ + 2ab + b$^2$
(a-b)$^2$ = a$^2$ - 2ab + b$^2$
NOTE: These equations will work if you substitute the first term for a and the second term for b. You may find these equations easier to understand after the concept is learned using the three steps described below.

## Lesson

Lesson 4 has shown you how to multiply binomials. In Lesson 5 we are going to learn how to square binomials. Squaring a binomial can be done using two different methods. The first method uses FOIL (refer to lesson 4). The second method is a shorter alternative to FOIL. The way we use the shortcut is to follow three simple steps.
Step 1: Square the first term of the binomial.
Step 2: Multiply the first term and last term of the binomial together and then double that quantity (in other words multiply by 2).
Step 3: Square the last term of the binomial.

## Example Problems

Here is an example to follow.
Using the binomial x$^2$+12x+36 we will square it creating the problem (x+6)$^2$
Step 1 Square the first term of the binomial.
(x)$^2$=x$^2$
Step 2 Multiply the first term and last term of the binomial together and then double that quantity (in other words multiply by 2).
[(x)*(6)]*2 = (6x)*(2) = 12x
Step 3 Square the last term of the binomial.
(6)$^2$ = 36
Finally we put the three term we have acquired together and get the answer
(x+6)$^2$ = x$^2$ + 12x + 36
Let’s try another problem that may be a bit more difficult.
Let’s square the binomial (x$^2$-4x) giving us (x$^2$-4x)$^2$
Step 1 Square the first term of the binomial.
(x$^2$)$^2$ = x$^4$
Step 2 Multiply the first term and last term of the binomial together and then double that quantity (in other words multiply by 2).
*Notice we keep the negative sign with the second term
[(x$^2$)(-4x)]*2 = (-4x$^3$)*(2) = -8x$^3$
Step 3 Square the last term of the binomial.
(-4x)$^2$ = (-4)$^2$(x)$^2$ = 16x$^2$
Our final answer will be the answers from the three steps combined
(x$^2$-4x)$^2$ = x$^4$ -8x$^3$ + 16x$^2$
Problem 3 (2x-6y)$^2$ = 4x$^2$ - 24xy + 36y$^2$

## Online Lesson

You can use this link to view this lesson illustrated by a teacher.
http://www.phschool.com/atschool/academy123/html/bbapplet_wl-problem-431067.html

## Practice Problems

Use ^ for exponentiation.

# Factoring/Factoring a^2-b^2 Binomials
## Lesson

Difference of Squares
Any binomial of the form $a^{2}-b^{2}$ may be written as $(a+b)\cdot (a-b)$. That is
$a^{2}-b^{2}=(a-b)\cdot (a+b)$.
Example 1: Factor $x^{2}-9$.
This is clearly seen just take $a^{2}=x^{2}$ and $b^{2}=9$ so that $b=3$. So $x^{2}-9=(x-3)(x+3)$
Example 2:: $32w^{4}-162$.
Here it is unclear where we can use the difference of squares as 32 is NOT a perfect square. However if we look we see that we can factor out a common factor of 2.
$32w^{4}-162=2(16w^{4}-81)$
Now we see we can use the difference of two squares to simplify matters take $a^{2}=16w^{4}$ and $b^{2}=81$:
$2(16w^{4}-81)=2(4w^{2}-9)(4w^{2}+9)$
Now we notice that we can use the difference of squares again in the first factor to get:
$2(4w^{2}-9)(4w^{2}+9)=2(2w+3)(2w-3)(4w^{2}+9)$
This is now completely factored.
This is brings us to our next point that is that $a^{2}+b^{2}$ is NOT FACTORABLE (at least for the purposes of this class).

## Example Problems

Let a = b
Therefore: a^2 = ab
Therefore: a^2 - b^2 = ab - b^2
Therefore: (a + b)(a - b) = b(a - b)
Now divide both sides by (a - b)
Therefore: a + b = b
But since a = b and substituting b for a
Therefore: b + b = b
Therefore: 2b = b
Now divide both sides by b
Therefore: 2 = 1
QED
This doesn't work because in line 5 both sides of the equation are divided by (a-b). Now, we know a = b and therefore, (a - b) = 0. This means that in line 5, both sides of the equation are divided by 0, which is not allowed. There are many ways that unrealities can be shown, if division by zero is permitted which is why this is not possible.

## Practice Problems

Use ^ for exponentiation

# Factoring/Chapter Review

Lesson 1. Factors of Integers
Lesson 2. Dividing Monomials
Lesson 3. Monomial Factors of Polynomials
Lesson 4. Multiplying Binomials
Lesson 5. Squares of Binomials
Lesson 6. Factoring x^2 + bx + c
Lesson 7. Factoring x^2 - bx + c
Lesson 8. Factoring x^2 - bx - c and x^2 + bx - c
Lesson 9. Factoring by Grouping
Lesson 10. Factoring Cubic Polynomials

# Factoring/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)

# Solving Quadratic Equations/The Zero Product Property
## Lesson

The zero property is based on the elementary concept that when you multiply two numbers and the answer is zero, then one of the numbers multiplied must be zero.
Rule: if a × b = 0, then a = 0 or b = 0.
Meaning that the zero property establishes that a or b has to equal zero in order for the result to be zero, a or b could be any random number while the other variable is zero so that result would still be zero.

## Example Problems

(x+5)(x-3)=0
x+5=0 x-3=0
x=-5 x=3

## Practice Problems

Easy
Medium
Hard

# Solving Quadratic Equations/Solving Quadratic Equations by Factoring
## Lesson

The easiest way to solve a quadratic equation is to factor the problem. Steps:
* Make "Y" a zero.
* list all the factors of "a" in y=ax$^2$+bx+c.
* List all the factors of "c" in y=ax$^2$+bx+c.
* See which factors can be used in such a way that when added or subtracted together they equal "b"

## Example Problems

y=x$^2$+2x+1
* a=1 1*1=1
* c=1 1*1=1
* b=2 1+1=2
0=(x+1)(x+1) or 0=(x+1)$^2$ x=-1

## Practice Problems

Easy
Medium
Hard

# Solving Quadratic Equations/Solving Quadratic Equations By Taking The Square Root of Both Sides
## Lesson

One way to solve a quadratic is to use square root. Steps:
* Factor a quadratic so that it is in the form $y=(x+z)^{2}$ ($z$ can be determined by completing the square)
* Take the square root of both sides, so $x+z=\pm {\sqrt {y}}$
* If y is 0, the square root is zero. In this case, there will only be one solution instead of two.
* Solve for x

## Example problems
### Problem 1

${\begin{aligned}200&=2x^{2}+38\\162&=2x^{2}\\81&=x^{2}\\x&=\pm 9\end{aligned}}$

### Problem 2

${\begin{aligned}64&=x^{2}+6x+9\\64&=(x+3)^{2}\\x+3&=\pm 8\\x&=-3\pm 8{\text{, i.e., }}x=\{-11,5\}\end{aligned}}$

# Solving Quadratic Equations/Chapter Review

Lesson 1. The Zero Product Property
Lesson 2. Solving Quadratic Equations by Factoring
Lesson 3. Solving Quadratic Equations By Taking The Square Root of Both Sides
Lesson 4. Solving Quadratic Equations By Completing The Square
Lesson 5. Solving Quadratic Equations By Using The Quadratic Formula
Lesson 6. The Discriminant
Lesson 7. Quadratic Equation Word Problems

# Solving Quadratic Equations/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)

# Quadratic Functions/Chapter Review

Lesson 1. Graphing a Parabola
Lesson 2. Using the Discrimnant to Help Graph a Parabola
Lesson 3. Finding the Axis of Symmetry of a Parabola
Lesson 4. Finding the Vertex of a Parabola
Lesson 5. Graphing Quadratic Inequalities

# Quadratic Functions/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)

# Exponential Functions/Multiplication of Exponents
## Example Problems

Ex.1
(c+3)(c-2)-(c+2)(c-3)=
(c²-2c+3c-6)-(c²-3c+2c+6)=
c²-2c+3c-6-c²+3c-2c-6=
Answer: 2c-12
Ex.2
3(2s+4)(s-3)-2(s+3)(2s-4)=
(6s+12)(s-3)-(2s+6)(2s-4)=
(6s²-18s+12s-36)-(4s²-8s+12s-24)=
(6s²-6s-36)-(4s²+4s-24)=
6s²-6s-36-4s²-4s+24=
Answer: 2s²-10s-12

## Practice Problems

Use ^ for exponentiation

# Exponential Functions/Zero and Negative Exponents
## Lesson

There are two very important things you need to know when working with zero or negative exponents. The rules of them.
First, any number the zero power always equals one. For example -500$^0$ = 1
Second, any number to the negative power is the opposite fraction of the exponent. For example 2$^{-2}$ = 1/2$^2$ = 1/4 because 2$^2$ = 4 and when you have 2$^{-2}$ it becomes its reciprocal to get 1/4.
This is a review of section 2 of the polynomials chapter.

## Practice Problems

Easy
Medium
Hard

# Exponential Functions/Graphs of Exponential Functions
## Lesson

An exponential function is a function that includes exponents, such as the function y=e$^x$. A Graph of an exponential function becomes a curved line that steadily gets steeper, like the one at the right.

## Practice Problems

Easy
Medium
Hard

# Exponential Functions/Chapter Review

Lesson 1. Multiplication of Exponential Expressions
Lesson 2. Zero and Negative Exponents
Lesson 3. Graphs of Exponential Functions
Lesson 4. Division of Exponential Expressions
Lesson 5. Scientific Notation
Lesson 6. Growth and Decay Functions

# Exponential Functions/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)

# Rational Expressions and Equations/Simplifying Rational Expressions
## Lesson

To simplify rational expressions, first we need to know what they are. They are fractions that have polynomials on both their numerators and denominators.
    ${\frac {2x^{2}+8x+8}{6x-2}}$
You just can simplify factors, not terms. That means you have to factorize the nominator and the denominator:
    ${\frac {2x^{2}+8x+8}{6x-2}}$
    $={\frac {2(x+2)^{2}}{2(3x-1)}}$
Now, both are divisible by 2. So just divide them by 2, the result would be:
    $={\frac {(x+2)^{2}}{3x-1}}$

## Practice Problems

Easy
Medium
Hard

# Rational Expressions and Equations/Adding and Subtracting with Like Denominators
## Vocabulary
## Lesson

"Write" the original expression
5/2x + x-5/2x
"Add" the numerators
5x + (x-5)/2x
"Combine" like terms
x/2x
"Simplify" the expression
1/2

## Practice Problems

Easy
Medium
Hard

# Rational Expressions and Equations/Adding and Subtracting When the Denominators are the Same
## Lesson

You have to leave the same denominator and add the nominators:

## Practice Problems

Easy
Medium
Hard

# Rational Expressions and Equations/Adding and Subtracting When the Denominators are Different
## Lesson

If you have to add two rational fractions with different denominators, as the first step, you have to find the LCM:
Now divide the LCM by both denominators and multiply by their respectives numerators:
The sum of the two results would be the new nominator:
This is another example:
We factorize both denominators and find the LCM
Now we divide and multiply:
We add the results to obtain the nominator; the denominator is the LCM:
We can factorize the nominator to simplify the result:

## Practice Problems

Easy
Medium
Hard

# Rational Expressions and Equations/Chapter Review

Lesson 1. Direct and Inverse Variation
Lesson 2. Simplifying Rational Expressions
Lesson 3. Multiplying and Dividing Rational Expressions
Lesson 4. Adding and Subtracting with Like Denominators
Lesson 5. Adding and Subtracting When the Denominators are the Same
Lesson 6. Adding and Subtracting When the Denominators are Different
Lesson 7. Solving Rational Equations

# Rational Expressions and Equations/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)

# Radical Expressions and Equations/Chapter Review

Lesson 1. Square Root Functions
Lesson 2. Equations with Radicals
Lesson 3. Rational Exponents
Lesson 4. The Pythagorean Theorem
Lesson 5. The Distance Formula

# Radical Expressions and Equations/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)