# 2.2.2: Single Variable Equations with Addition and Subtraction

- Page ID
- 4353

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)## Single Variable Addition Equations

John’s soccer team is raising money to attend the two-day camp. The team’s goal is to raise the $450 camp registration fee by washing cars. At the end of the day, the team has made $380. The team decides to split the $450 registration cost equally, with each player responsible for $45. They also split the $380 equally, so each receives $38 toward their total cost. Now, John and the team need to know how much more each team member needs to add to the $38 to reach its goal.

In this concept, you will learn to solve single-variable addition equation.

**Solving Single-Variable Addition Equations**

To solve a single-variable equation means figuring out the value of the variable or the unknown number. There are a number of different ways to do this. The first method is to use mental math.

Let’s look at an example.

5+x=10

Using mental math you can ask yourself, “Five plus what number is equal to 10?”

The answer is 5.

You can check your answer too. To do this, simply substitute the value for x into the equation and see if it forms a true statement.

5+x=10

5+5=10

10=10

This is a true statement, so the answer is correct.

Sometimes, it may be more difficult to figure out the value of the variable using mental math. The second way of solving a single-variable equation involves using the ** inverse operation**. An

**inverse operation**is the opposite of the given operation.

Here is an equation.

x+27=43

In the equation, the given operation is addition, so you can use the opposite of addition (subtraction) to solve the problem. You need to get the variable by itself to figure out its value and you use subtraction to do this.

First, identify the number that is being added to the variable, in this case, 27.

Next, subtract 27 from both sides of the equation. That will leave the variable on one side of the equation with 0 and 16 on the other side. Remember, any number or variable plus 0 is the number or variable.

x+27−27=43-27

x+0=16

x=16

The answer is x=16.

Check your work by substituting the value for x back into the original equation. If it is true, then one side of the equation will equal the other side.

x+27=43

16+27=43

43=43

The answer is correct.

Here is another equation.

45+x=67

Even though x is in the middle of the equation and not at the start of it, you can still use an inverse to sort it out.

The goal is to get the x alone on one side of the equal sign.

First, identify the number being added to the variable, in this case, 45.

Next, using the inverse of a positive 45, subtract 45 from both sides of the equation.

45+x−45=67−45

x=22

Notice again that 45−45 is equal to 0. So, the variable is by itself on the left side of the equals. On the right side, you subtracted 45 and got an answer of 22.

Check your work by substituting 22 for x in the original equation. One side of the equation should equal the other side.

45+x=67

45+22=67

67=67

The answer checks out.

**Examples**

Example 2.2.2.1

Earlier, you were given a problem about John and the soccer’s team’s car wash fundraiser.

Each of the ten team members has $38, and needs to pay an equal percentage of the remaining balance. They figure that each team member will be responsible for is $45. How can John figure out how much more each team member needs to pay?

**Solution**

First, begin by using the information to write a phrase.

38 and an unknown number is 45

Next, write the phrase as an **expression**

38+x=45

Then, identify the number that is being added to the variable x, in this case, 38.

Finally, using the inverse of 38, subtract 38 from both sides of the equation.

38+x−38=45−38

0+x=7

x=7

Now, check the answer by plugging 7 in the equation for x.

38+x=45

38+7=45

45=45

The answer checks out.

Each team member will need to pay an additional $7 to cover the registration fee.

**Solve the following equations then check your answer. Write your answers in the form:**

Variable=____.

Example 2.2.2.2

x+15=32

**Solution**

First, identify the number that is being added to the variable, in this case, 15.

Next, using the inverse of 15, subtract 15 from both sides of the equation.

x+15−15=32-15

x+0=17

The answer is x=17.

Finally, check your answer by plugging 17 in the equation for x.

x+15=32

17+15=32

32=32

The answer checks out.

Example 2.2.2.3

x+16=22

**Solution**

First, identify the number that is being added to the variable, in this case, 16.

Next, using the inverse of 16, subtract 16 from both sides of the equation.

x+16−16=22-16

x+0=6

The answer is x=6.

Finally, check your answer by plugging 6 in the equation for x.

x+16=22

6+16=22

22=22

The answer checks out.

Example 2.2.2.4

y+15=30

**Solution**

First, identify the number that is being added to the variable, in this case, 15.

Next, using the inverse of 15, subtract 15 from both sides of the equation.

y+15−15=30-15

y+0=15

The answer is y=15.

Finally, check your answer by plugging 15 in the equation for y.

y+15=30

15+15=30

30=30

The answer checks out.

Example 2.2.2.5

12+x=18

**Solution**

First, identify the number that is being added to the variable, in this case, 12.

Next, using the inverse of 15, subtract 15 from both sides of the equation.

x+12−12=18-12

x+0=6

The answer is x=6.

Finally, check your answer by plugging 6 in the equation for x.

x+12=18

6+12=18

18=18

The answer checks out.

**Review**

Solve each single-variable addition equation. Write your answer in the form:

Variable=_____.

- x+4=11
- x+11=22
- x+3=8
- x+12=20
- x+9=11
- x+8=30
- 22+x=29
- 18+x=25
- 15+x=20
- 13+x=24
- x+18=24
- 23+x=33
- y+18=31
- 21+x=54

### Review (Answers)

To see the Review answers, open this PDF file and look for section 12.5.

### Vocabulary

Term | Definition |
---|---|

Difference |
The result of a subtraction operation is called a difference. |

Expression |
An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols. |

Inverse |
Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction. |

Simplify |
To simplify means to rewrite an expression to make it as "simple" as possible. You can simplify by removing parentheses, combining like terms, or reducing fractions. |

Sum |
The sum is the result after two or more amounts have been added together. |

### Additional Resources

Video:

PLIX: Play, Learn, Interact, eXplore: **Climbing Mountains**

Practice: **Single Variable Equations with Addition and Subtraction**