# 2.9: Inverse Property of Addition in Decimal Equations

**At Grade**Created by: CK-12

**License**: CC BY-NC 3.0

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.

### 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 \begin{align*}a\end{align*}:

\begin{align*}a+(-a)=0\end{align*}

The **additive inverse** of a number is another word for the opposite of a number. The additive inverse of \begin{align*}a\end{align*} is \begin{align*}-a\end{align*}.

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 \begin{align*}x\end{align*}.

\begin{align*}x+39.517=50.281\end{align*}

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

\begin{align*}x+39.517-39.517=50.281-39.517\end{align*}

Next, simplify the left side of the equation by combining like terms. \begin{align*}39.517 - 39.517\end{align*} makes 0 which leaves the \begin{align*}x\end{align*} by itself.

\begin{align*}x=50.281-39.517\end{align*}

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

\begin{align*}x=10.764\end{align*}

The answer is \begin{align*}x=10.764\end{align*}.

You can check your solution by substituting that value for \begin{align*}x\end{align*} back into the original equation and verifying that it makes both sides equal.

\begin{align*}\begin{array}{rcl} x+39.517 &=&50.281 \\ 10.764+39.517&=&50.281 \\ 50.281&=&50.281 \end{array}\end{align*}

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 \begin{align*}x\end{align*}.

\begin{align*}x-43.27=182.205\end{align*}

First, notice that 43.27 is subtracted from \begin{align*}x\end{align*} on one side of the equation. To isolate \begin{align*}x\end{align*}, add 43.27 to both sides of the equation.

\begin{align*}x-43.27+43.27=182.205+43.27\end{align*}

Next, simplify the left side of the equation by combining like terms. \begin{align*}-43.27 + 43.27\end{align*} makes 0 which leaves the \begin{align*}x\end{align*} by itself.

\begin{align*}x=182.205+43.27\end{align*}

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

\begin{align*}x=225.475\end{align*}

The answer is \begin{align*}x=225.475\end{align*}.

Next, check your solution.

\begin{align*}\begin{array}{rcl} x-43.27&=&182.205 \\ 225.475-43.27&=&182.205 \\ 182.205&=&182.205 \end{array}\end{align*}

Your answer is correct.

### Examples

#### Example 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.

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 \begin{align*}x\end{align*} 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.

\begin{align*}25.74+x=28.91\end{align*}

Finally, solve the equation. To isolate \begin{align*}x\end{align*}, subtract 25.74 from both sides of the equation.

\begin{align*}25.74+x-25.74=28.91-25.74\end{align*}

Next, simplify the left side of the equation by combining like terms. \begin{align*}25.74 - 25.74\end{align*} makes 0, leaving the \begin{align*}x\end{align*} by itself.

\begin{align*}x=28.91-25.74\end{align*}

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

\begin{align*}x=3.17\end{align*}

The answer is Maria took $3.17 from the jar.

#### Example 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?

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 \begin{align*}x\end{align*} 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.

\begin{align*}x-1.96=1.258\end{align*}

Finally, solve the equation. To isolate \begin{align*}x\end{align*}, add 1.96 to both sides of the equation.

\begin{align*}x-1.96+1.96=1.258+1.96\end{align*}

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

\begin{align*}x=1.258+1.96\end{align*}

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

\begin{align*}x=3.218\end{align*}

The answer is Sonya ran 3.218 kilometers.

#### Example 3

Solve the following equation for \begin{align*}x\end{align*}.

\begin{align*}x+5.678=12.765\end{align*}

First, notice that 5.678 is added to \begin{align*}x\end{align*} on one side of the equation. To isolate \begin{align*}x\end{align*}, subtract 5.678 from both sides of the equation.

\begin{align*}x+5.678-5.678=12.765-5.678\end{align*}

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

\begin{align*}x=7.087\end{align*}

The answer is \begin{align*}x=7.087\end{align*}.

Next, check your solution.

\begin{align*} \begin{array}{rcl} x+5.678 &=& 12.765 \\ 7.087+5.678 &=& 12.765 \\ 12.765 &=& 12.765 \end{array}\end{align*}

Your answer is correct.

#### Example 4

Solve the following equation for \begin{align*}x\end{align*}.

\begin{align*}x-4.32=19.87\end{align*}

First, notice that 4.32 is subtracted from \begin{align*}x\end{align*} on one side of the equation. To isolate \begin{align*}x\end{align*}, add 4.32 to both sides of the equation.

\begin{align*}x-4.32+4.32=19.87+4.32\end{align*}

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

\begin{align*}x=24.19\end{align*}

The answer is \begin{align*}x=24.19\end{align*}.

Next, check your solution.

\begin{align*}\begin{array}{rcl} x-4.32 &=&19.87 \\ 24.19-4.32&=&19.87 \\ 19.87&=&19.87 \end{array}\end{align*}

Your answer is correct.

#### Example 5

Solve the following equation for \begin{align*}x\end{align*}.

\begin{align*}x+123.578=469.333\end{align*}

First, notice that 123.578 is added to \begin{align*}x\end{align*} on one side of the equation. To isolate \begin{align*}x\end{align*}, subtract 123.578 from both sides of the equation.

\begin{align*}x+123.578-123.578=469.333-123.578\end{align*}

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

\begin{align*}x=345.755\end{align*}

The answer is \begin{align*}x=345.755\end{align*}.

Next, check your solution.

\begin{align*}\begin{array}{rcl} x+123.578 &=& 469.333 \\ 345.755+123.578 &=& 469.333 \\ 469.333 &=& 469.333 \end{array}\end{align*}

Your answer is correct.

### Review

Solve each equation for \begin{align*}x\end{align*}.

1. \begin{align*}x + 2.39 = 7.01\end{align*}

2. \begin{align*}x + 5.64 = 17.22\end{align*}

3. \begin{align*}x + 8.07 = 18.12\end{align*}

4. \begin{align*}x + 14.39 = 17.342\end{align*}

5. \begin{align*}x + 21.3 = 87.12\end{align*}

6. \begin{align*}x + 31.9 = 77.22\end{align*}

7. \begin{align*}x + 18.77 = 97.12\end{align*}

8. \begin{align*}x + 21.31 = 27.09\end{align*}

9. \begin{align*}x + 18.11 = 87.22\end{align*}

10. \begin{align*}818.703 = 614.208 + x\end{align*}

11. \begin{align*}x + 55.27 = 100.95\end{align*}

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

12. 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.

13. 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.

14. 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.

15. 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.

### Answers for Review Problems

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

### Resources

### Image Attributions

**[1]****^**License: CC BY-NC 3.0

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## Learning Objectives

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

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