## Basic Terms and Terminology Relating to Solving Equations with One Variable

• Algebra: A branch of mathematics that has at least one unknown number or variable.
• Variable: The unknown number in the algebraic equation
• Inverse mathematic calculations: Opposite calculations
• The inverse mathematic calculation for addition: Subtraction
• The inverse mathematic calculation for subtraction: Addition
• The inverse mathematic calculation for multiplication: Division
• The inverse mathematic calculation for division: Multiplication

## Algebraic Equations Algebraic expression notation:

1 – power (exponent)
2 – coefficient
3 – term
4 – operator
5 – constant term
x y c – variables/constants

Simply defined, algebra is a branch of mathematics that has at least one unknown number or variable.

A variable in algebra is the unknown number in the algebraic equation. Variables are expressed as any letter but x is the most commonly used variable letter that is used in algebra with one unknown variable, and any other letters are used when there is more than one unknown number or variable in an algebraic equation, such as the one shown in the picture above. The most commonly used letter for a second variable is y, however, as stated above, any letter can be used.

In this section, you will learn how to solve algebraic equations with only one unknown number or variable in an algebraic equation so, for this section, we will use only x as the unknown number or variable in an algebraic equation although any letter can be used, as stated above.

## Solving Algebraic Equations

The procedure for solving algebraic equations is as follows:

1. Put all of the unknown variables (x) on the left side of the equal sign (=)
2. Put all of the constants or known numbers on the right side of the equal sign (=)
3. Perform inverse mathematical calculations for all the unknowns or x until only 1 x or x is on the left side of the equal sign (=)

Inverse mathematic calculations are the opposite calculations.

For example:

• The inverse mathematic calculation for addition is subtraction.
• The inverse mathematic calculation for subtraction is addition.
• The inverse mathematic calculation for multiplication is division.
• The inverse mathematic calculation for division is multiplication.

Inverse mathematic calculations are important and necessary in algebra, but they are also useful in checking your answers to addition, subtraction, multiplication and division calculations.

Here are some examples of how addition calculations can be checked and verified by using the inverse mathematical calculation of subtraction:

24 + 56 = 80

Inverse Calculation of Subtraction:

80 – 24 = 56

80 – 56 = 24

2 + 560 = 562

Inverse Calculation of Subtraction:

562 – 2 = 560

562 - 560 = 2

Here are some examples of how subtraction calculations can be checked and verified by using the inverse mathematical calculation of addition:

Subtraction Calculation:

35 – 27 = 8

8 + 27 = 35

Subtraction Calculation:

55 – 20 = 35

35 + 25 = 55

Here are some examples of how multiplication calculations can be checked and verified by using the inverse mathematical calculation of division:

Multiplication Calculation:

35 x 50 = 1750

Inverse Calculation of Division:

1750 ÷ 50 = 35

1750 ÷ 35 = 50

Multiplication Calculation:

2.7 x 8.4 = 22.68

Inverse Calculation of Division:

22.68 ÷ 2.7 = 8.4

22.68 ÷ 8.4 = 2.7

Here are some examples of how division calculations can be checked and verified by using the inverse mathematical calculation of multiplication:

Division Calculation:

722 ÷ 4 = 180.5

Inverse Calculation of Multiplication:

180.5 x 4 = 722

Division Calculation:

82.6 ÷ 4 = 20.65

Inverse Calculation of Multiplication:

20.65 x 4 = 82.6

## Setting Up and Solving Algebraic Equations

Here are a few examples of setting up algebraic equations with x on the left side and the constant(s) on the right side of the algebraic equation:

Example 1: Setting Up and Arranging the Algebraic Equations

11 x + 2 = 46

• Set up with x on the left side and the constant(s) on the right side of the = sign:

11 x + 2 = 46

Example 2: Setting Up and Arranging the Algebraic Equations

x + 30 = 110

• Set up with x on the left side and the constant(s) on the right side of the = sign:

x + 30 = 110

Example 3: Setting Up and Arranging the Algebraic Equations

14 - 3 = 3 x + 2

• Set up with x on the left side and the constant(s) on the right side of the = sign:

3 x + 2 = 14 - 3

Example 4: Setting Up and Arranging the Algebraic Equations

46 -2 = 11 x

• Set up with x on the left side and the constant(s) on the right side of the = sign:

11 x = 46 - 2

Example 5: Setting Up and Arranging the Algebraic Equations

33.3 x + 7 = 276

• Set up with x on the left side and the constant(s) on the right side of the = sign:

33.3 x + 7 = 276

Now that the equations are set up, as above, it is now time to solve the equation and find out the precise value of the unknown x. The procedure for doing this is shown below for each of the five equations shown above.

Example 1: Solving the Algebraic Equation

1. 11 x + 2 = 46
2. 11 x = 46 – 2
3. 11 x = 44
4. x = 44 ÷ 11
5. x = 4

Example 2: Solving the Algebraic Equation

• x + 30 = 110
• x = 110 – 30
• x = 80

Example 3: Solving the Algebraic Equation

• 3 x + 2 = 14 - 3
• 3 x = 14 – 3 - 2
• 3 x = 9
• x = 9 ÷ 3
• x = 3

Example 4: Solving the Algebraic Equation

• 11 x = 46 - 2
• 11 x = 44
• x = 44 ÷ 11
• x = 4

Example 5: Solving the Algebraic Equation

• 33.3 x + 7 = 276
• 33.3 x = 276 -7
• 33.3 x = 269
• x = 269 ÷ 33.3
• x = 8.08 rounded off to the nearest hundredth

Here are a few examples of performing inverse, or opposite, calculations for the unknowns (x) in an algebraic equation:

Example 1

Algebraic equation: 2 x + 7 = 19

Solving the equation with inverse calculations:

• 2 x + 7 = 19

Perform the inverse calculation. Subtract 7 from both sides of the equal sign (=)

• 2 x + 7 - 7 = 19 – 7
• 2 x = 12
• Divide both sides of the equal sign (=) by 2.
• 2 x = 12 = x = 6
2 2

To check this answer, substitute 6 for each x in the original algebraic equation, as shown below:

• 2 (6) + 7 = 19
• 12 + 7 = 19
• 19 = 19

When the numbers (19) are the same on both sides of the equal sign, the calculation of x as 6 is correct. When the numbers are different for each side of the equal sign, the calculation is incorrect and wrong.

Example 2

Algebraic equation: 2 x - 7 = 19

Solving the equation with inverse calculations:

• 2 x - 7 = 19

Perform the inverse calculation. Add 7 to both sides of the equal sign (=)

• 2 x - 7 + 7 = 19 + 7
• 2 x = 26
• Divide both sides of the equal sign (=) by 2.
• 2 x = 26 = x = 13
2 2

To check this answer, substitute 13 for each x in the original algebraic equation, as shown below:

• 2 x - 7 = 19
• 2 (13) - 7 = 19
• 26 - 7 = 19
• 19 = 19

When the numbers (19) are the same on both sides of the equal sign, the calculation of x as 13 is correct. When the numbers are different for each side of the equal sign, the calculation is incorrect and wrong.

## Simplified Algebraic Equations

In addition to solving algebraic equations as you learned above, you may also see other types of algebra calculations on your TEAS. For example you may get simplified algebra calculations and solving for x with a word problem.

You may be asked what the solution, or answer, is for an algebra calculation such as the ones below.

Example 1

What is the answer to 2 x + 6 x – 2 when x is 4?

• 2 (4) + 6 (4) – 2
• 8 + 24 – 2 = 30

Example 2

What is the answer to 3 x + 2 x + 2 when x is 8?

• 3 (8) + 2 (8) + 2
• 24 + 16 + 2 = 42

Example 3

What is the answer to 32 x + 2 x + 2 when x is 5?

• 9 (5) + 2 (5) + 2
• 45 + 10 + 2 = 57

Example 4

What is the answer to 22 x - 2 x / 2 when x is 8?

• 4 (8) – 2 (8) / 2
• 32 -16 ÷ 2
• 16 ÷ 2 = 8

## Algebraic Word Problems

You may have some word problems on your TEAS examination that have to be solved and calculated with algebra.

An easy way to determine if a word problem has to be solved and calculated with algebra is to ask the question, "Are there any unknowns in this word problem?" If the answer to this question is yes, then this word problem has to be solved and calculated with algebra. If the answer to this question is no, then this word problem does NOT have to be solved and calculated with algebra.

Below are some algebraic word problems:

Algebraic Word Problem # 1:

How much would you have to put into your savings account in order to triple your money in this account when, at the current time, you have \$657.

• x = 3 x 657
• x = 1971

Algebraic Word Problem # 2:

What is 5 more than 2 dozen?

• x = 2 (12) + 5
• x = 24 + 5
• x = 29

Answer: 29 is 5 more than 2 dozen. There are 12 in one dozen.

Algebraic Word Problem # 3:

What is 2/7 (one seventh) of the weight of an object when 3/7 (three sevenths) is 68 pounds? (This calculation is a two step calculation. First, you have to calculate what 1/7 is and then you have to calculate what 2/7 of the weight of the object is.

Finding what 1/7 is:

• 3/7 : 68 = 2/7 : x
• 3/7 x = 68 x 2/7
• 3 x = 68
• x = 19.43 rounded off to the nearest hundredth
• 1/7 of the weight is 19.43 pounds when 3/7 of the weight is 68 pounds
• The next step is to determine 2/7 of the weight when 1/7 is 19.43.
• x = 2 (19.43)
• x = 38.86

Answer: There are 38.86 pounds in 2/7 of the total weight when the total weight for 3/7 of the total weight is 68 pounds.

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