# How to Do Algebra

Algebra can be a difficult subject to master. In addition to numbers, there are letters thrown into equations that represent unknown numbers. It may seem overwhelming at first, but by learning a few basic concepts and doing practice problems, you can be successful in algebra. Once you have learned the basics, you will begin to see how algebra is useful and applies to situations in everyday life!

### Part 1 Understanding Order of Operations

1. 1
Memorize PEMDAS. PEMDAS is an acronym to help you remember the order of operations in math. PEMDAS stands for Parentheses, Exponents, Multiplication and Division, Addition and Subtraction. Whenever you are solving a problem, start with the “parentheses” and work your way through the acronym, finishing with subtraction.
• For parentheses, perform all of the operations inside the parentheses using this same order.
• Multiplication and division are considered equal operations. You can solve them at the same time, so simply solve from left to right.
• Addition and subtraction are also equal operations, so solve from left-to-right.
• You can remember PEMDAS using the mnemonic, Please Excuse My Dear Aunt Sally.
2. 2
Use PEMDAS to solve problems. For any problem you solve in algebra, you will always solve it using this order. Many times, a problem has parentheses to denote all of the operations you will perform first. Multiplication and division rank equally, so simply solve either of these operations from left-to-right. The same goes for addition and subtraction.
• For example: Solve (3 + 6) x 7 – 4/2.
• Parentheses: 9 x 7 – 4/2
• Exponent: none
• Multiplication 1st: 63 – 4/2
• Then divide: 63 – 2
• Subtract: 61
• Final answer: (3 + 6) x 7 – 4/2 = 61.
3. 3
Practice with some examples. The more problems you practice with, the better you will be at solving them. Eventually, using this order of operations will become second nature and you won’t even think about it. Do as many problems as you need to feel confident in solving them.
• Example 1: 8 + (6 x 42 + 7) = 8 + (6 x 16 + 7) = 8 + (96 +7) = 8 + 103 = 111.
• Example 2: 30/2 + 52 - (6 x 3) = 30/2 + 52 - 18 = 30/2 + 25 – 18 = 15 + 25 – 18 = 40 – 18 = 22.
4. 4
Ask for help. When starting to learn algebra, the material can get overwhelming very quickly. Don’t be afraid to ask your teacher for help or seek out extra tutoring. Even asking a friend who may have a better understanding can be useful.

### Part 2 Solving Problems

1. 1
Recognize that algebra is just like solving a puzzle. Like any puzzle, there are pieces. Learning how to recognize the numbers and symbols for the placeholders that they are, make the solution much easier to grasp.
• Try to find the missing number in a problem where the final answer is given.
• For example: 1 + __ = 9
• The missing number is 8, because 1 plus 8 equals 9. Pretty simple, right? This is basic algebra.
2. 2
Perform operations on both sides of the equation. When solving an algebraic problem, you must remember that if you alter one side of the equation in any way, you must do the exact same thing to the other side of the equation. If you add, subtract, multiply, or divide, you must perform the same operation to the opposite side.
• For example: Solve x + 3 = 2x -1.
• Subtract x from both sides: x – x +3 = 2x – x – 1.
• Simplify: 3 = x – 1.
• Add one to both sides: 3 + 1 = x – 1 + 1.
• Simplify: x = 4.
3. 3
Isolate the variable on one side of the equation. When given an algebraic expression, you will notice that there are constants and variables. A constant is any number given, while a variable is generally a letter that represents an unknown number.
• To isolate the variable, add or subtract terms to get the variable on one side. If the variable has a coefficient, divide both sides by that coefficient to get the variable alone.
• For example: 6y + 6 = 48
• Subtract 6 from both sides: 6y – 6 = 48 – 6
• Simplify: 6y = 42
• Divide both sides by 6: 6y/6 = 42/6
• Simplify: y = 7
4. 4
Take the root of the number to cancel an exponent (and vice versa). If you are solving for a variable that is squared, you will need to take the square root of it to solve the problem. If the variable is a square root, then you will need to square it to solve the problem.
• Example 1: √x = 9
• Square both sides: √x2 = 92
• x = 81
• Example 2: x2 = 16
• Take the square root of both sides: √x2 = √16
• x = 4
5. 5
Combine like terms. Whenever you have terms that are identical (same variable), you can combine them together to simplify the problem. This helps to keep equations manageable and easier to solve. Remember, terms that have different exponents are not identical terms: x is not the same as x2.
• The following are like terms: 4x, -3x, 0.45x, -132x
• The following are not like terms: 5x, 8y2, -13y, 9z, 12xy
• For example: 4x + 3y – 7x has two like terms, 4x and -7x. Combining them gives you the simplified equation: -3x + 3y.
6. 6
Practice with more complex problems. The art of mastering any concept is practice. Try solving problems with increasing difficulty to truly check your comprehension. Use problems from your textbook or seek out extra problems online.
• Example 1: q + 18 = 9q – 6
• Add 6 to both sides: q + 18 + 6 = 9q – 6 + 6
• Simplify: q + 24 = 9q
• Subtract q from both sides: q – q + 24 = 9q – q
• Simplify: 24 = 8q
• Divide both sides by 8: 8q/8 = 24/8
• Solve: q = 3
7. 7
Check your answers. Make a habit of checking your answers when you have solved a problem. Once you have obtained the solution and discovered the value of the variable, check your work by inserting the number you have obtained into the original equation. If the expression is still true, then you have found the correct solution!
• Let’s try it: q = 3 for q + 18 = 9q – 6
• 3 + 18 = (9 x 3) - 6
• 21 = (27) - 6
• 21 = 21
• Right! q equals 3, and it is confirmed by a correctly completed equation.

### Part 3 Multiplying with the FOIL Method

1. 1
Define FOIL. The FOIL method stands for First Outside Inside Last. It is a method used to multiply binomials together. A binomial is an algebraic expression with two terms, like (3x + 5)(2x – 4). You start by multiplying the First two terms together, then the two Outside terms, then the Inside terms, then the Last terms of each binomial.
• The First terms are the two terms that start each binomial: 3x and 2x.
• The Outside terms are the two terms that are the outermost of both binomials: 3x and 4.
• The Inside terms are the two middle terms: 5 and 2x.
• The Last terms are the two terms that end each binomial: 5 and 4.
• Remember to simplify all of the terms before writing your final answer.
2. 2
Multiply the two first terms together. Identify the first two terms of each binomial. This is the first term in each set of parentheses. In the example, (3x + 5)(2x – 4), the first two terms are 3x and 2x. You will then multiply these two numbers together.
• For example: (3x + 5)(2x – 4)
• First two terms: 3x and 2x
• Multiply: 3x * 2x = 6x2
3. 3
Multiply the two outside terms together. Identify the two outside terms. These are the outermost terms of each set of parentheses. In the example, (3x + 5)(2x – 4),the two outside terms are 3x and 4. Now, multiply them together.
• For example: (3x + 5)(2x – 4)
• Two outside terms: 3x and 4
• Multiply: 3x * 4 = 12x
4. 4
Find the product of the two inside terms. The two inside terms are the two terms found in the middle of the expression; one is the second term of the first binomial and the other is the first term of the second binomial. In the example, (3x + 5)(2x – 4),the two inside terms are 5 and 2x. To find the product, multiply them together.
• For example: (3x + 5)(2x – 4)
• Two inside terms: 5 and 2x
• Multiply: 5 * 2x = 10x
5. 5
Multiply the last two terms together. The final multiplication step is with the last two terms. These are the second terms of both binomials. In the example, (3x + 5)(2x – 4), the two last terms are 5 and -4. Multiply them together.
• For example: (3x + 5)(2x – 4)
• Last two terms: 5 and -4
• Multiply: 5 * -4 = -20
6. 6
Combine all terms and simplify. Once you have finished multiplying all of the terms together, combine them into one expression and simplify. After putting the expression together, you can combine like terms to simplify the expression fully.
• For example: (3x + 5)(2x – 4)
• Write the expression: 6x2 + 12x + 10x – 4
• Combine like terms: 6x2 + 22x – 4

### Part 4 Working with Exponents

1. 1
Simplify exponents of numbers. When a number has an exponent that means you multiply that number by itself as many times as the exponent says. To simplify any number that has an exponent simply multiply it the appropriate number of times.
• For example: 43 = 4 * 4 * 4 = 64.
• If there is a negative sign and no parentheses, the exponent is simplified and then the negative sign gets added: -22 = -(2 * 2) = -4.
• If there is a negative sign, but the number is in parenthesis, the negative number is part of the exponent: (-2)2 = -2 * -2 = 4.
2. 2
Combine like terms with the same exponents. It may be confusing at first to see a variable with an exponent. Just remember, any variable with the same exponent number can be added or subtracted. If the letters are the same, but the exponents are different, they cannot be combined.
• Example 1: 6x2 + 5x2 = 11x2
• Example 2: 4xy3 - 8xy3 = -4xy3
• Example 3: 5z + 5z2; terms cannot be combined.
3. 3
Add the exponents together when multiplying variables. If two variables are being multiplied together and they both have exponents, you can add the exponents together to get the resulting exponent. This only applies to variables of the same letter.
• Example 1: x2x3 = x2+3 = x5
• Example 2: (a3b5c2)(ab2) = a3+1b5+2c2 = a4b7c2
4. 4
Subtract the exponents when dividing variables. If an exponent is negative, that means you are dividing: x-1 = 1/x. If you want to divide two variables that have exponents, simply subtract the bottom exponent from the top exponent.
• Example 1: a6/a3 = a6-3 = a3
• Example 2: (x4y2)/(x6y2) = x4-6y2-2 = x-2 = 1/x2

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Categories: Algebra