Solving Equations Containing Fractions Using the Addition, Subtraction, and Division Properties of Equality

Learning Outcomes

Determine whether a fraction is a solution of an equation
Solve equations with fractions using the Addition, Subtraction, and Division Properties of Equality

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.

Substitute the number for the variable in the equation.
Simplify the expressions on both sides of the equation.
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 [latex]x-\Large\frac{3}{10}=\Large\frac{1}{2}[/latex]

[latex]x=1[/latex]
[latex]x=\Large\frac{4}{5}[/latex]
[latex]x=-\Large\frac{4}{5}[/latex]

Solution:

1.
	[latex]x -\Large\frac{3}{10} =\Large\frac{1}{2}[/latex]
Substitute [latex]\color{red}{1}[/latex] for x.	[latex]\color{red}{1} -\Large\frac{3}{10} =\Large\frac{1}{2}[/latex]
Change to fractions with a LCD of [latex]10[/latex].	[latex]\color{red}{\Large\frac{10}{10}} -\Large\frac{3}{10} =\Large\frac{5}{10}[/latex]
Subtract.	[latex]\Large\frac{7}{10} \not=\Large\frac{5}{10}[/latex]

Since [latex]x=1[/latex] does not result in a true equation, [latex]1[/latex] is not a solution to the equation.

2.
	[latex]x -\Large\frac{3}{10} =\Large\frac{1}{2}[/latex]
Substitute [latex]\color{red}{\Large\frac{4}{5}}[/latex] for x.	[latex]\color{red}{\Large\frac{4}{5}} -\Large\frac{3}{10} =\Large\frac{1}{2}[/latex]
	[latex]\color{red}{\Large\frac{8}{10}} -\Large\frac{3}{10} =\Large\frac{5}{10}[/latex]
Subtract.	[latex]\Large\frac{5}{10} =\Large\frac{5}{10}\quad\checkmark[/latex]

Since [latex]x=\Large\frac{4}{5}[/latex] results in a true equation, [latex]\Large\frac{4}{5}[/latex] is a solution to the equation [latex]x-\Large\frac{3}{10}=\Large\frac{1}{2}[/latex].

3.
	[latex]x -\Large\frac{3}{10} =\Large\frac{1}{2}[/latex]
Substitute [latex]\color{red}{-\Large\frac{4}{5}}[/latex] for x.	[latex]\color{red}{-\Large\frac{4}{5}} -\Large\frac{3}{10} =\Large\frac{1}{2}[/latex]
	[latex]\color{red}{-\Large\frac{8}{10}} -\Large\frac{3}{10} =\Large\frac{5}{10}[/latex]
Subtract.	[latex]-\Large\frac{11}{10}\not=\Large\frac{5}{10}[/latex]

Since [latex]x=-\Large\frac{4}{5}[/latex] does not result in a true equation, [latex]-\Large\frac{4}{5}[/latex] is not a solution to the equation.

Try It

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

We also previously 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 [latex]a,b,\text{ and }c[/latex],

[latex]\text{if }a=b,\text{ then }a+c=b+c[/latex].	Addition Property of Equality
[latex]\text{if }a=b,\text{ then }a-c=b-c[/latex].	Subtraction Property of Equality
[latex]\text{if }a=b,\text{ then }\Large\frac{a}{c}=\Large\frac{b}{c}\normalsize ,c\ne 0[/latex].	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.

Solve Equations with Fractions Using the Subtraction Property of Equality

Let’s start by looking at equations that can be solved using the subtraction property of equality.

Example

Solve: [latex]y+\Large\frac{9}{16}=\Large\frac{5}{16}[/latex]

Show Solution

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: [latex]p+\Large\frac{1}{2}=\Large\frac{2}{3}[/latex]

Show Solution

Try It

Solve Equations with Fractions Using the Addition Property of Equality

We used the Subtraction Property of Equality in the example above. Now we’ll use the Addition Property of Equality.

Example

Solve: [latex]a-\Large\frac{1}{4}=-\Large\frac{2}{3}[/latex]

Show Solution

Try It

In the following video we show more examples of solving an equation with fractions where you are required to find a common denominator.

Solve Equations with Fractions Using the Division Property of Equality

Next, we will solve equations that require division to isolate the variable. First, let’s consider the division property of equality again.

The Division Property of Equality

For any numbers [latex]a,b[/latex], and [latex]c[/latex], [latex]c\ne 0[/latex]

[latex]\text{if }a=b,\text{ then }\Large\frac{a}{c}=\Large\frac{b}{c}[/latex].

If you divide both sides of an equation by the same quantity, you still have equality.

Let’s put this idea in practice with an example. We are looking for the number you multiply by [latex]10[/latex] to get [latex]44[/latex], and we can use division to find out.

Example

Solve: [latex]10q=44[/latex]

Solution:

		[latex]10q=44[/latex]
Divide both sides by [latex]10[/latex] to undo the multiplication.		[latex]\Large\frac{10q}{10}=\Large\frac{44}{10}[/latex]
Simplify.		[latex]q=\Large\frac{22}{5}[/latex]
Check:
Substitute [latex]q=\Large\frac{22}{5}[/latex] into the original equation.	[latex]10\left(\Large\frac{22}{5}\right)\stackrel{?}{=}44[/latex]
Simplify.	[latex]\stackrel{2}{\overline{)10}}\left(\Large\frac{22}{\overline{)5}}\right)\stackrel{?}{=}44[/latex]
Multiply.	[latex]44=44\quad\checkmark[/latex]

The solution to the equation was the fraction [latex]\Large\frac{22}{5}[/latex]. We leave it as an improper fraction.

Try It

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		[latex]y +\Large\frac{9}{16} =\Large\frac{5}{16}[/latex]
Subtract [latex]\Large\frac{9}{16}[/latex] from each side to undo the addition.		[latex]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}}[/latex]
Simplify on each side of the equation.		[latex]y + 0 = -\Large\frac{4}{16}[/latex]
Simplify the fraction.		[latex]y = -\Large\frac{1}{4}[/latex]
Check:	[latex]y +\Large\frac{9}{16} =\Large\frac{5}{16}[/latex]
Substitute [latex]y=-\Large\frac{1}{4}[/latex] .	[latex]\color{red}{-\Large\frac{1}{4}} +\Large\frac{9}{16} \stackrel{?}{=}\Large\frac{5}{16}[/latex]
Rewrite as fractions with the LCD.	[latex]\color{red}{-\Large\frac{4}{16}} +\Large\frac{9}{16}\stackrel{?}{=}\Large\frac{5}{16}[/latex]
Add.	[latex]\Large\frac{5}{16} =\Large\frac{5}{16}\quad\checkmark[/latex]

		[latex]p+\Large\frac{1}{2}=\Large\frac{2}{3}[/latex]
Subtract [latex]\Large\frac{1}{2}[/latex] from each side to undo the addition.		[latex]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}}[/latex]
Simplify on each side of the equation.		[latex]p + 0 =\Large\frac{2}{3}-\Large\frac{1}{2}[/latex]
Simplify the fraction by finding a common denominator.		[latex]p =\Large\frac{2}{3}\cdot\color{red}{\Large\frac{2}{2}}-\Large\frac{1}{2}\cdot\color{red}{\Large\frac{3}{3}}[/latex]
		[latex]p =\Large\frac{4}{6}-\Large\frac{3}{6}[/latex]
		[latex]p =\Large\frac{1}{6}[/latex]
Check:	[latex]p+\Large\frac{1}{2}=\Large\frac{2}{3}[/latex]
Substitute [latex]p=\Large\frac{1}{6}[/latex] . Rewrite as fractions with the LCD.	[latex]\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}[/latex]
	[latex]\Large\frac{4}{6} =\Large\frac{4}{6}\quad\checkmark[/latex]

		[latex]a-\Large\frac{1}{4}=-\Large\frac{2}{3}[/latex]
Add [latex]\Large\frac{5}{9}[/latex] from each side to undo the addition.		[latex]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}}[/latex]
Find a common denominator.		[latex]a =-\Large\frac{2}{3}\cdot\color{red}{\Large\frac{4}{4}}+\Large\frac{1}{4}\cdot\color{red}{\Large\frac{3}{3}}[/latex]
Simplify on each side of the equation.		[latex]a = -\Large\frac{8}{12}+\Large\frac{3}{12}[/latex]
Simplify the fraction.		[latex]a = -\Large\frac{5}{12}[/latex]
Check:	[latex]a-\Large\frac{1}{4}=-\Large\frac{2}{3}[/latex]
Substitute [latex]a=-\frac{5}{12}[/latex] .	[latex]\color{red}{-\Large\frac{5}{12}} -\Large\frac{1}{4}\stackrel{?}{=}-\Large\frac{2}{3}[/latex]
Change to common denominator.	[latex]\color{red}{-\Large\frac{5}{12}} -\Large\frac{3}{12}\stackrel{?}{=}-\Large\frac{8}{12}[/latex]
Subtract.	[latex]-\Large\frac{8}{12} = -\Large\frac{8}{12}\quad\checkmark[/latex]

UNIT 4