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Inverse Property of Addition in Fraction Equations

Solve subtraction equations with fractions by using addition.

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Inverse Property of Addition in Fraction Equations
License: CC BY-NC 3.0

Cody is making a pie for his math class for pi day. His mom gives him a big bowl of blueberries and a package of blueberries to get started. Cody pours the blueberries from the package into a measuring cup and sees that it is \begin{align*}1 \frac{1}{2}\end{align*} cups of blueberries. He adds these blueberries to the big bowl. He is about to start adding the rest of the ingredients when he realizes he should know how many cups of blueberries he has! He measures all the blueberries in the bowl and finds that he has \begin{align*}6 \frac{1}{4}\end{align*} cups of blueberries. How can Cody figure out how many blueberries were in the bowl to start with?

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

Applying the Inverse Property of Addition in Fraction Equations

Recall that 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*}

You can use the inverse property of addition to help you to solve equations. Remember that when solving an equation you want to isolate the variable which means you want to get the variable by itself on one side of the equation. Sometimes, when you add or subtract the same amount from both sides of the equation you will be able to isolate the variable with the help of the inverse property of addition.

Here is an example.

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

 \begin{align*}x+1 \frac{1}{2}=3 \frac{3}{4}\end{align*}

First, notice that \begin{align*}1 \frac{1}{2}\end{align*} is added to \begin{align*}x\end{align*} on one side of the equation. You want to isolate the \begin{align*}x\end{align*}. Subtract \begin{align*}1 \frac{1}{2}\end{align*} from both sides of the equation.

 \begin{align*}x+1 \frac{1}{2}-1 \frac{1}{2}=3 \frac{3}{4}- 1 \frac{1}{2}\end{align*}

Next, simplify the left side of the equation by combining like terms. \begin{align*}1 \frac{1}{2}-1 \frac{1}{2}\end{align*} makes 0 which leaves the \begin{align*}x\end{align*} by itself.

 \begin{align*}x=3 \frac{3}{4}-1 \frac{1}{2}\end{align*}

Now, simplify the right side of the equation by combining like terms. Use what you have learned about mixed number subtraction. You will need to find a common denominator and then convert to improper fractions first.

\begin{align*}\begin{array}{rcl} x&=&3 \frac{3}{4}-1 \frac{2}{4}\\ \\ x&=&\frac{15}{4}-\frac{6}{4}\\ \\ x&=&\frac{9}{4}=2 \frac{1}{4} \end{array}\end{align*}

The answer is \begin{align*}x=2 \frac{1}{4}\end{align*}.

Examples

Example 1

Earlier, you were given a problem about Cody and his blueberry pie.

He has a big bowl filled with blueberries. He's not sure how many cups of blueberries were in the bowl to start with, but after adding \begin{align*}1 \frac{1}{2}\end{align*} cups of blueberries the bowl has \begin{align*}6 \frac{1}{4}\end{align*} cups of blueberries in it. Cody wants to know how many cups of blueberries were in the bowl originally.

Cody could set up an equation to represent this situation. He doesn't know how many cups of blueberries were in the bowl originally, so that can be his variable. Let \begin{align*}x\end{align*} be the original number of cups of blueberries in the bowl.

Cody knows that the original number of cups of blueberries plus the \begin{align*}1 \frac{1}{2}\end{align*} cups of blueberries he added will give him the \begin{align*}6 \frac{1}{4}\end{align*} cups of blueberries in the bowl now. He can turn this into an equation.

 \begin{align*}x+1 \frac{1}{2}=6 \frac{1}{4}\end{align*}

Now, he can solve this equation by isolating the \begin{align*}x\end{align*}. He can subtract \begin{align*}1 \frac{1}{2}\end{align*} from both sides of the equation.

 \begin{align*}x+1 \frac{1}{2}-1 \frac{1}{2}=6 \frac{1}{4}- 1 \frac{1}{2}\end{align*}

Next, he can simplify the left side of the equation by combining like terms. \begin{align*}1 \frac{1}{2}-1 \frac{1}{2}\end{align*} makes 0 which leaves the \begin{align*}x\end{align*} by itself.

 \begin{align*}x=6 \frac{1}{4}- 1 \frac{1}{2}\end{align*}

Now, he can simplify the right side of the equation by combining like terms. He will need to find a common denominator and then convert to improper fractions first.

\begin{align*}\begin{array}{rcl} x&=&6 \frac{1}{4}-1 \frac{2}{4}\\ \\ x&=&\frac{25}{4}-\frac{6}{4}\\ \\ x&=&\frac{19}{4}=4 \frac{3}{4} \end{array}\end{align*}

The answer is the bowl had \begin{align*}4 \frac{3}{4}\end{align*} cups of blueberries in it originally.

Example 2

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

\begin{align*}x+\frac{2}{3}=\frac{5}{6}\end{align*}

First, notice that \begin{align*}\frac{2}{3}\end{align*} is added to \begin{align*}x\end{align*} on one side of the equation. To isolate the \begin{align*}x\end{align*}, subtract \begin{align*}\frac{2}{3}\end{align*} from both sides of the equation.

\begin{align*}x+\frac{2}{3}-\frac{2}{3}=\frac{5}{6}-\frac{2}{3}\end{align*}

Next, simplify the left side of the equation by combining like terms. \begin{align*}\frac{2}{3}-\frac{2}{3}\end{align*} makes 0 which leaves the \begin{align*}x\end{align*} by itself.

\begin{align*}x=\frac{5}{6}-\frac{2}{3}\end{align*}

Now, simplify the right side of the equation by combining like terms. Use what you have learned about fraction subtraction. You will need to find a common denominator first.

\begin{align*}\begin{array}{rcl} x &=& \frac{5}{6}- \frac{4}{6}\\ \\ x &=& \frac{1}{6} \end{array}\end{align*}

The answer is \begin{align*}x=\frac{1}{6}\end{align*}.

Example 3

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

\begin{align*}\frac{1}{7}+x=\frac{5}{7}\end{align*}

First, isolate the \begin{align*}x\end{align*} by subtracting \begin{align*}\frac{1}{7}\end{align*} from both sides of the equation.

\begin{align*}\frac{1}{7}+x-\frac{1}{7}=\frac{5}{7}-\frac{1}{7}\end{align*}

Next, simplify the left side of the equation by combining like terms. \begin{align*}\frac{1}{7}-\frac{1}{7}\end{align*} makes 0 which leaves the \begin{align*}x\end{align*} by itself.

\begin{align*}x=\frac{5}{7}-\frac{1}{7}\end{align*}

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

\begin{align*}x=\frac{4}{7}\end{align*}

The answer is \begin{align*}x=\frac{4}{7}\end{align*}.

Example 4

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

\begin{align*}\frac{1}{3}+y=\frac{5}{6}\end{align*}

First, isolate the \begin{align*}y\end{align*} by subtracting \begin{align*}\frac{1}{3}\end{align*} from both sides of the equation.

\begin{align*}\frac{1}{3}+y-\frac{1}{3}=\frac{5}{6}-\frac{1}{3}\end{align*}

Next, simplify the left side of the equation by combining like terms. \begin{align*}\frac{1}{3}-\frac{1}{3}\end{align*} makes 0 which leaves the \begin{align*}y\end{align*} by itself.

 \begin{align*}y=\frac{5}{6}-\frac{1}{3}\end{align*}

Now, simplify the right side of the equation by combining like terms. Use what you have learned about fraction subtraction. You will need to find a common denominator first. Don't forget to simplify your answer.

\begin{align*}\begin{array}{rcl} y &=& \frac{5}{6}-\frac{2}{6}\\ \\ x &=& \frac{3}{6}=\frac{1}{2} \end{array}\end{align*}

The answer is \begin{align*}x=\frac{1}{2}\end{align*}.

Example 5

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

 \begin{align*}y+\frac{1}{2}=\frac{5}{6}\end{align*}

First, isolate the \begin{align*}y\end{align*} by subtracting \begin{align*}\frac{1}{2}\end{align*} from both sides of the equation.

\begin{align*}y+\frac{1}{2}-\frac{1}{2}=\frac{5}{6}-\frac{1}{2}\end{align*}

Next, simplify the left side of the equation by combining like terms. \begin{align*}\frac{1}{2}-\frac{1}{2}\end{align*} makes 0 which leaves the \begin{align*}y\end{align*} by itself.

\begin{align*}y=\frac{5}{6}-\frac{1}{2}\end{align*}

Now, simplify the right side of the equation by combining like terms. Use what you have learned about fraction subtraction. You will need to find a common denominator first. Don't forget to simplify your answer.

\begin{align*}\begin{array}{rcl} y &=& \frac{5}{6}-\frac{3}{6}\\ \\ y &=& \frac{2}{6}=\frac{1}{3} \end{array}\end{align*}

The answer is \begin{align*}x=\frac{1}{3}\end{align*}.

Review

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

  1. \begin{align*}x+\frac{2}{5}=\frac{5}{5}\end{align*}
  2. \begin{align*}x+\frac{2}{8}=\frac{6}{8}\end{align*}
  3. \begin{align*}x+\frac{3}{9}=\frac{4}{9}\end{align*}
  4. \begin{align*}x+\frac{5}{7}=\frac{6}{7}\end{align*}
  5. \begin{align*}x+\frac{10}{12}=\frac{11}{12}\end{align*}
  6. \begin{align*}x+\frac{3}{15}=\frac{10}{15}\end{align*}
  7. \begin{align*}x+\frac{9}{13}=\frac{12}{13}\end{align*}
  8. \begin{align*}x+\frac{6}{14}=\frac{12}{14}\end{align*}
  9. \begin{align*}x+\frac{1}{5}=\frac{6}{10}\end{align*}
  10. \begin{align*}x+\frac{1}{2}=\frac{11}{12}\end{align*}
  11. \begin{align*}x-1 \frac{1}{2}=4\end{align*}
  12. \begin{align*}2 \frac{1}{4}+x=3 \frac{3}{4}\end{align*}
  13. \begin{align*}x-\frac{7}{8}=2 \frac{3}{4}\end{align*}
  14. Ludmilla, Brent, and Rudy have \begin{align*}8 \frac{5}{6}\end{align*} feet of taffy that they have to sell to raise money for the school drama club. Brent has already sold \begin{align*}3 \frac{2}{3}\end{align*} feet of taffy and Rudy plans to sell exactly \begin{align*}2 \frac{3}{4}\end{align*} feet. How much taffy does Ludmilla have to sell, if they sell all of the taffy?
  15. Ron, Jung-Ho, and Sarah have a lawn mowing business. Today they are cutting an enormous lawn. Sarah agrees to start and will mow \begin{align*}\frac{3}{8}\end{align*} of the lawn. Ron will only mow \begin{align*}\frac{1}{7}\end{align*} of the lawn, but he's willing to work during the hottest time of the day. How much of the lawn is Jung-Ho responsible for completing?

Review (Answers)

To see the Review answers, open this PDF file and look for section 3.9.

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Vocabulary

TermDefinition
Equivalent Fractions Equivalent fractions are fractions that can each be simplified to the same fraction. An equivalent fraction is created by multiplying both the numerator and denominator of the original fraction by the same number.
Estimation Estimation is the process of finding an approximate answer to a problem.
improper fraction An improper fraction is a fraction in which the absolute value of the numerator is greater than the absolute value of the denominator.
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.
Lowest Common Denominator The lowest common denominator of multiple fractions is the least common multiple of all of the related denominators.

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

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