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# Solving Equations That Contain Fractions Using the Addition and Subtraction Properties of Equality

### Learning Outcomes

• Determine whether a fraction is a solution to an equation
• Use the equality properties of addition and subtraction to solve equations that contain fractions

## Determine Whether a Fraction is a Solution of an Equation

As we saw in previous lessons, a solution of an equation is a value that makes a true statement when substituted for the variable in the equation. In those sections, we found whole number and integer solutions to equations. Now that we have worked with fractions, we are ready to find fraction solutions to equations. The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number, an integer, or a fraction.

### Determine whether a number is a solution to an equation.

1. Substitute the number for the variable in the equation.
2. Simplify the expressions on both sides of the equation.
3. Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.

### Example

Determine whether each of the following is a solution of $x-\Large\frac{3}{10}=\Large\frac{1}{2}$
1. $x=1$
2. $x=\Large\frac{4}{5}$
3. $x=-\Large\frac{4}{5}$
Solution:
 1. $x -\Large\frac{3}{10} =\Large\frac{1}{2}$ Substitute $\color{red}{1}$ for x. $\color{red}{1} -\Large\frac{3}{10} =\Large\frac{1}{2}$ Change to fractions with a LCD of $10$. $\color{red}{\Large\frac{10}{10}} -\Large\frac{3}{10} =\Large\frac{5}{10}$ Subtract. $\Large\frac{7}{10} \not=\Large\frac{5}{10}$
Since $x=1$ does not result in a true equation, $1$ is not a solution to the equation.
 2. $x -\Large\frac{3}{10} =\Large\frac{1}{2}$ Substitute $\color{red}{\Large\frac{4}{5}}$ for x. $\color{red}{\Large\frac{4}{5}} -\Large\frac{3}{10} =\Large\frac{1}{2}$ $\color{red}{\Large\frac{8}{10}} -\Large\frac{3}{10} =\Large\frac{5}{10}$ Subtract. $\Large\frac{5}{10} =\Large\frac{5}{10}\quad\checkmark$
Since $x=\Large\frac{4}{5}$ results in a true equation, $\Large\frac{4}{5}$ is a solution to the equation $x-\Large\frac{3}{10}=\Large\frac{1}{2}$.
 3. $x -\Large\frac{3}{10} =\Large\frac{1}{2}$ Substitute $\color{red}{-\Large\frac{4}{5}}$ for x. $\color{red}{-\Large\frac{4}{5}} -\Large\frac{3}{10} =\Large\frac{1}{2}$ $\color{red}{-\Large\frac{8}{10}} -\Large\frac{3}{10} =\Large\frac{5}{10}$ Subtract. $-\Large\frac{11}{10}\not=\Large\frac{5}{10}$
Since $x=-\Large\frac{4}{5}$ does not result in a true equation, $-\Large\frac{4}{5}$ is not a solution to the equation.

### Try It

[ohm_question]146128[/ohm_question]

## Solve Equations with Fractions using the Addition and Subtraction Properties of Equality

We also solved equations using the Addition, Subtraction, and Division Properties of Equality. We will use these same properties to solve equations with fractions.

### Addition, Subtraction, and Division Properties of Equality

For any numbers $a,b,\text{ and }c$,
 $\text{if }a=b,\text{ then }a+c=b+c$. Addition Property of Equality $\text{if }a=b,\text{ then }a-c=b-c$. Subtraction Property of Equality $\text{if }a=b,\text{ then }\Large\frac{a}{c}=\Large\frac{b}{c}\normalsize ,c\ne 0$. Division Property of Equality
In other words, when you add or subtract the same quantity from both sides of an equation, or divide both sides by the same quantity, you still have equality.

### Example

Solve: $y+\Large\frac{9}{16}=\Large\frac{5}{16}$

 $y +\Large\frac{9}{16} =\Large\frac{5}{16}$ Subtract $\Large\frac{9}{16}$ from each side to undo the addition. $y +\Large\frac{9}{16}\color{red}{-}\color{red}{\Large\frac{9}{16}} =\Large\frac{5}{16}\color{red}{-}\color{red}{\Large\frac{9}{16}}$ Simplify on each side of the equation. $y + 0 = -\Large\frac{4}{16}$ Simplify the fraction. $y = -\Large\frac{1}{4}$ Check: $y +\Large\frac{9}{16} =\Large\frac{5}{16}$ Substitute $y=-\Large\frac{1}{4}$ . $\color{red}{-\Large\frac{1}{4}} +\Large\frac{9}{16} \stackrel{?}{=}\Large\frac{5}{16}$ Rewrite as fractions with the LCD. $\color{red}{-\Large\frac{4}{16}} +\Large\frac{9}{16}\stackrel{?}{=}\Large\frac{5}{16}$ Add. $\Large\frac{5}{16} =\Large\frac{5}{16}\quad\checkmark$
Since $y=-\Large\frac{1}{4}$ makes $y+\Large\frac{9}{16}=\Large\frac{5}{16}$ a true statement, we know we have found the solution to this equation.

In our next example, we will solve an equation with fractions whose denominators are different. We will need to make an additional step to find the common denominator.

### Example

Solve: $p+\Large\frac{1}{2}=\Large\frac{2}{3}$

 $p+\Large\frac{1}{2}=\Large\frac{2}{3}$ Subtract $\Large\frac{1}{2}$ from each side to undo the addition. $p+\Large\frac{1}{2}\color{red}{-}\color{red}{\Large\frac{1}{2}} =\Large\frac{2}{3}\color{red}{-}\color{red}{\Large\frac{1}{2}}$ Simplify on each side of the equation. $p + 0 =\Large\frac{2}{3}-\Large\frac{1}{2}$ Simplify the fraction by finding a common denominator. $p =\Large\frac{2}{3}\cdot\color{red}{\Large\frac{2}{2}}-\Large\frac{1}{2}\cdot\color{red}{\Large\frac{3}{3}}$ $p =\Large\frac{4}{6}-\Large\frac{3}{6}$ $p =\Large\frac{1}{6}$ Check: $p+\Large\frac{1}{2}=\Large\frac{2}{3}$ Substitute $p=\Large\frac{1}{6}$ .  Rewrite as fractions with the LCD. $\color{red}{\Large\frac{1}{6}} +\Large\frac{1}{2} \stackrel{?}{=}\Large\frac{2}{3}=\color{red}{\Large\frac{1}{6}}+\Large\frac{3}{6}\stackrel{?}{=}\Large\frac{4}{6}$ $\Large\frac{4}{6} =\Large\frac{4}{6}\quad\checkmark$
We know we have found the solution to this equation since $\Large\frac{4}{6} =\Large\frac{4}{6}$.

### Try It

[ohm_question]146492[/ohm_question]
We used the Subtraction Property of Equality in the example above. Now we’ll use the Addition Property of Equality.

### Example

Solve: $a-\Large\frac{1}{4}=-\Large\frac{2}{3}$

 $a-\Large\frac{1}{4}=-\Large\frac{2}{3}$ Add $\Large\frac{5}{9}$ from each side to undo the addition. $a-\Large\frac{1}{4}\color{red}{+}\color{red}{\Large\frac{1}{4}} =-\Large\frac{2}{3}\color{red}{+}\color{red}{\Large\frac{1}{4}}$ Find a common denominator. $a =-\Large\frac{2}{3}\cdot\color{red}{\Large\frac{4}{4}}+\Large\frac{1}{4}\color{red}{\Large\frac{3}{3}}$ Simplify on each side of the equation. $a = -\Large\frac{8}{12}+\Large\frac{3}{12}$ Simplify the fraction. $a = -\Large\frac{5}{12}$ Check: $a-\Large\frac{1}{4}=-\Large\frac{2}{3}$ Substitute $a=-\frac{5}{12}$ . $\color{red}{-\Large\frac{5}{12}} -\Large\frac{1}{4}\stackrel{?}{=}-\Large\frac{2}{3}$ Change to common denominator. $\color{red}{-\Large\frac{5}{12}} -\Large\frac{3}{12}\stackrel{?}{=}-\Large\frac{8}{12}$ Subtract. $-\Large\frac{8}{12} = -\Large\frac{8}{12}\quad\checkmark$
Since $a=-\Large\frac{5}{12}$ makes the equation true, we know that $a=-\Large\frac{5}{12}$ is the solution to the equation.

### Try It

[ohm_question]146494[/ohm_question]
In the following video we show more examples of solving an equation with fractions where you are required to find a common denominator. https://youtu.be/O7SPM7Cs8Ds