# 2.2.5: Properties of Equality with Decimals

- Page ID
- 4357

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

Sean is saving up to buy a new video game. He keeps his money in a jar in his room. He knows that he currently has $28.91 saved up. Sean’s little sister Maria has just gotten interested in money. Sean thinks Maria might have taken some of his money out of the jar. He dumps out all the money in the jar and counts it up- only $25.74! How can Sean figure out how much money Maria took from the jar?

In this concept, you will learn how to apply the **inverse property of addition** to solve decimal equations.

### Applying the **Inverse** Property of Addition

In mathematics, **inverse operations** are operations that reverse one another. Addition and subtraction are inverse operations. For example, if you take any number and add 5 to it and then subtract 5 from the total, you will be back to the original number. The subtraction reversed the addition.

The **Inverse Property of Addition** states that the sum of any number and its opposite is zero. In symbols, it says that for any number a:

a+(−a)=0

The ** additive inverse** of a number is another word for the opposite of a number. The additive inverse of a is −a.

You can use the inverse property of addition to help you to solve equations that would be difficult to solve using mental math. Remember that when you are solving an equation, your goal is to figure out the value of the variable that will make both sides of the equation equal.

Here is an example.

Solve the following equation for x.

x+39.517=50.281

First, notice that 39.517 is added to x on one side of the equation. You want to **isolate** the x which means you want to get the x by itself on one side of the equation. Subtract 39.517 from both sides of the equation.

x+39.517−39.517=50.281−39.517

Next, simplify the left side of the equation by combining like terms. 39.517−39.517 makes 0 which leaves the x by itself.

x=50.281−39.517

Now, simplify the right side of the equation by combining like terms. Use what you have learned about decimal subtraction.

x=10.764

The answer is x=10.764.

You can check your solution by substituting that value for x back into the original equation and verifying that it makes both sides equal.

x+39.517=50.281

10.764+39.517=50.281

50.281=50.281

Your answer is correct.

Remember that when solving equations using the inverse property of addition, you will always need to add or subtract the same amount from * both* sides of the equation in order to keep both sides equal to one another.

Here is another example.

Solve the following equation for x.

x−43.27=182.205

First, notice that 43.27 is subtracted from x on one side of the equation. To isolate x, add 43.27 to both sides of the equation.

x−43.27+43.27=182.205+43.27

Next, simplify the left side of the equation by combining like terms. −43.27+43.27 makes 0 which leaves the x by itself.

x=182.205+43.27

Now, simplify the right side of the equation by combining like terms. Use what you have learned about decimal addition.

x=225.475

The answer is x=225.475.

Next, check your solution.

x−43.27=182.205

225.475−43.27=182.205

182.205=182.205

Your answer is correct.

### Examples

Example 2.2.5.1

Earlier, you were given a problem about Sean and his sister Maria.

Sean had $28.91 saved up in a jar in his room, but now the jar only has $25.74 in it. Sean wants to figure out how much money Maria took from the jar.

**Solution**

First, define a variable. In this problem, you are trying to figure out the amount of money that Maria took from the jar.

Let the variable x be equal to the amount of money Maria took from the jar.

Now, write an equation. You know that if you take the number of amount of money currently in the jar and add the amount Maria took, you will get the amount of money Sean started with. You know there is currently $25.74 in the jar and Sean started with $28.91.

25.74+x=28.91

Finally, solve the equation. To isolate x, subtract 25.74 from both sides of the equation.

25.74+x−25.74=28.91−25.74

Next, simplify the left side of the equation by combining like terms. 25.74−25.74 makes 0, leaving the x by itself.

x=28.91−25.74

Now, simplify the right side of the equation by combining like terms.

x=3.17

The answer is Maria took $3.17 from the jar.

Example 2.2.5.2

At Saturday’s track meet, Liz ran 1.96 kilometers less than Sonya. Liz ran 1.258 kilometers. How many kilometers did Sonya run?

**Solution**

First, notice that the key words “less than” are a clue that this might be a subtraction equation.

Next, define a variable. In this problem, you are trying to figure out the number of kilometers Sonya ran.

Let the variable x be equal to the number of kilometers Sonya ran.

Now, write an equation. You know that if you take the number of kilometers Sonya ran and subtract 1.96, you will get the number of kilometers Liz ran. You also know that Liz ran 1.258 kilometers.

x−1.96=1.258

Finally, solve the equation. To isolate x, add 1.96 to both sides of the equation.

x−1.96+1.96=1.258+1.96

Next, simplify the left side of the equation by combining like terms.

x=1.258+1.96

Now, simplify the right side of the equation by combining like terms.

x=3.218

The answer is Sonya ran 3.218 kilometers.

Example 2.2.5.3

Solve the following equation for x.

x+5.678=12.765

**Solution**

First, notice that 5.678 is added to x on one side of the equation. To isolate x, subtract 5.678 from both sides of the equation.

x+5.678−5.678=12.765−5.678

Next, simplify both sides of the equation by combining like terms.

x=7.087

The answer is x=7.087.

Next, check your solution.

x+5.678=12.765

7.087+5.678=12.765

12.765=12.765

Your answer is correct.

Example 2.2.5.4

Solve the following equation for x.

x−4.32=19.87

**Solution**

First, notice that 4.32 is subtracted from x on one side of the equation. To isolate x, add 4.32 to both sides of the equation.

x−4.32+4.32=19.87+4.32

Next, simplify both sides of the equation by combining like terms.

x=24.19

The answer is x=24.19.

Next, check your solution.

x−4.32=19.87

24.19−4.32=19.87

19.87=19.87

Your answer is correct.

Example 2.2.5.5

Solve the following equation for x.

x+123.578=469.333

**Solution**

First, notice that 123.578 is added to x on one side of the equation. To isolate x, subtract 123.578 from both sides of the equation.

x+123.578−123.578=469.333−123.578

Next, simplify both sides of the equation by combining like terms.

x=345.755

The answer is x=345.755.

Next, check your solution.

x+123.578=469.333

345.755+123.578=469.333

469.333=469.333

Your answer is correct.

### Review

Solve each equation for x.

- x+2.39=7.01
- x+5.64=17.22
- x+8.07=18.12
- x+14.39=17.342
- x+21.3=87.12
- x+31.9=77.22
- x+18.77=97.12
- x+21.31=27.09
- x+18.11=87.22
- 818.703=614.208+x
- x+55.27=100.95

Use equations to solve each word problem. Each answer should have an equation and a value for the variable.

- Jamal’s leek and potato soup calls for 2.45 kg more potatoes than leeks. Jamal uses 4.05 kg of potatoes. How many kilograms of leeks does he use? Write an equation and solve.
- The distance east from Waterville to Longford is 67.729 kilometers. The distance west from Waterville to Transtown is 61.234 kilometers. What is the difference between the two distances? Write an equation and solve.
- Sabrina spent $25.62 at the book fair. When she left the fair, she had $6.87. How much did money did she take to the fair? Write an equation and solve.
- Mr. Bodin has 11.25 liters of a cleaning solution, which is a combination of soap and water. If there are 2.75 liters of soap in the solution, how many liters of water are in the solution? Write an equation and solve.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.9.

### Vocabulary

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

Additive inverse |
The additive inverse or opposite of a number is -1(x). A number and its additive inverse always sum to zero.x |

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

Inverse Property of Addition |
The inverse property of addition states that the sum of any real number and its additive inverse (opposite) is zero. If a is a real number, then a+(−a)=0. |

### Additional Resources

Video:

Practice: **Properties of Equality with Decimals**

Real World Application: **Cell Phone Plans**