4.12: Solve Equations with Fractions (Part 1)
- Page ID
- 6067
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Determine whether a fraction is a solution of an equation
- Solve equations with fractions using the Addition, Subtraction, and Division Properties of Equality
- Solve equations using the Multiplication Property of Equality
- Translate sentences to equations and solve
Before you get started, take this readiness quiz. If you miss a problem, go back to the section listed and review the material.
- Evaluate \(x + 4\) when \(x = −3\) If you missed this problem, review Example 3.2.10.
- Solve: \(2y − 3 = 9\). If you missed this problem, review Example 3.5.2.
- Solve: \(y − 3 = −9\) If you missed this problem, review Example 4.2.10.
Determine Whether a Fraction is a Solution of an Equation
As we saw in Solve Equations with the Subtraction and Addition Properties of Equality and Solve Equations Using Integers; The Division Property of Equality, 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.
Step 1. Substitute the number for the variable in the equation.
Step 2. Simplify the expressions on both sides of the equation.
Step 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.
Determine whether each of the following is a solution of \(x − \dfrac{3}{10} = \dfrac{1}{2}\).
- \(x = 1\)
- \(x = \dfrac{4}{5}\)
- \(x = − \dfrac{4}{5}\)
Solution
| Substitute \(\textcolor{red}{1}\) for x. | \(\textcolor{red}{1} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{1}{2}\) |
| Change to fractions with a LCD of 10. | \(\textcolor{red}{\dfrac{10}{10}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{5}{10}\) |
| Subtract. | \(\dfrac{7}{10} \stackrel{?}{=} \dfrac{5}{10}\) |
Since \(x = 1\) does not result in a true equation, \(1\) is not a solution to the equation.
| Substitute \(\textcolor{red}{\dfrac{4}{5}}\) for x. | \(\textcolor{red}{\dfrac{4}{5}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{1}{2}\) |
| \(\textcolor{red}{\dfrac{8}{10}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{5}{10}\) | |
| Subtract. | \(\dfrac{5}{10} = \dfrac{5}{10} \; \checkmark\) |
Since \(x = \dfrac{4}{5}\) results in a true equation, \(\dfrac{4}{5}\) is a solution to the equation \(x − \dfrac{3}{10} = \dfrac{1}{2}\).
| Substitute \(\textcolor{red}{- \dfrac{4}{5}}\) for x. | \(\textcolor{red}{- \dfrac{4}{5}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{1}{2}\) |
| \(\textcolor{red}{- \dfrac{8}{10}} - \dfrac{3}{10} \stackrel{?}{=} \dfrac{5}{10}\) | |
| Subtract. | \(\dfrac{11}{10} \neq \dfrac{5}{10}\) |
Since \(x = − \dfrac{4}{5}\) does not result in a true equation, \(− \dfrac{4}{5}\) is not a solution to the equation.
Determine whether each number is a solution of the given equation. \(x − \dfrac{2}{3} = \dfrac{1}{6}\)
- \(x = 1\)
- \(x = \dfrac{5}{6}\)
- \(x = − \dfrac{5}{6}\)
- Answer a
-
no
- Answer b
-
yes
- Answer c
-
no
Determine whether each number is a solution of the given equation. \(y − \dfrac{1}{4} = \dfrac{3}{8}\)
- \(y = 1\)
- \(y = - \dfrac{5}{8}\)
- \(y = \dfrac{5}{8}\)
- Answer a
-
no
- Answer b
-
no
- Answer c
-
yes
Solve Equations with Fractions using the Addition, Subtraction, and Division Properties of Equality
In Solve Equations with the Subtraction and Addition Properties of Equality and Solve Equations Using Integers; The Division Property of Equality, we solved equations using the Addition, Subtraction, and Division Properties of Equality. We will use these same properties to solve equations with fractions.
For any numbers \(a\), \(b\), and \(c\),
| if \(a = b\), then \(a + c = b + c\). | Addition Property of Equality |
| if \(a = b\), then \(a − c = b − c\). | Subtraction Property of Equality |
| if \(a = b\), then \(a c = b c \), \(c ≠ 0\). | 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.
Solve: \(y + \dfrac{9}{16} = \dfrac{5}{16}\).
Solution
| Subtract \(\dfrac{9}{16}\) from each side to undo the addition. | \(y + \dfrac{9}{16} \textcolor{red}{- \dfrac{9}{16}} = \dfrac{5}{16} \textcolor{red}{- \dfrac{9}{16}}\) |
| Simplify on each side of the equation. | \(y + 0 = - \dfrac{4}{16}\) |
| Simplify the fraction. | \(y = - \dfrac{1}{4}\) |
Check:
| Substitute y = (− \dfrac{1}{4}\). | \(\textcolor{red}{- \dfrac{1}{4}} + \dfrac{9}{16} \stackrel{?}{=} \dfrac{5}{16}\) |
| Rewrite as fractions with the LCD. | \textcolor{red}{- \dfrac{4}{16}} + \dfrac{9}{16} \stackrel{?}{=} \dfrac{5}{16}\) |
| Add. | \(\dfrac{5}{16} \stackrel{?}{=} \dfrac{5}{16} \; \checkmark\) |
Since \(y = − \dfrac{1}{4}\) makes \(y + \dfrac{9}{16} = \dfrac{5}{16}\) a true statement, we know we have found the solution to this equation.
Solve: \(y + \dfrac{11}{12} = \dfrac{5}{12}\).
- Answer
-
\(-\dfrac{1}{2}\)
Solve: \(y + \dfrac{8}{15} = \dfrac{4}{15}\).
- Answer
-
\(-\dfrac{4}{15}\)
We used the Subtraction Property of Equality in Example \(\PageIndex{2}\). Now we’ll use the Addition Property of Equality.
Solve: a − \(\dfrac{5}{9}\) = \(− \dfrac{8}{9}\).
Solution
| Add \(\dfrac{5}{9}\) from each side to undo the addition. | \(a - \dfrac{5}{9} \textcolor{red}{+ \dfrac{5}{9}} = - \dfrac{8}{9} \textcolor{red}{+ \dfrac{5}{9}}\) |
| Simplify on each side of the equation. | \(a + 0 = - \dfrac{3}{9}\) |
| Simplify the fraction. | \(a = - \dfrac{1}{3}\) |
Check:
| Substitute a = \(− \dfrac{1}{3}\). | \(\textcolor{red}{- \dfrac{1}{3}} - \dfrac{5}{9} \stackrel{?}{=} - \dfrac{8}{9}\) |
| Change to common denominator. | \(\textcolor{red}{- \dfrac{3}{9}} - \dfrac{5}{9} \stackrel{?}{=} - \dfrac{8}{9}\) |
| Subtract. | \(- \dfrac{8}{9} = - \dfrac{8}{9} \; \checkmark\) |
Since \(a = − \dfrac{1}{3}\) makes the equation true, we know that \(a = − \dfrac{1}{3}\) is the solution to the equation.
Solve: \(a − \dfrac{3}{5} = − \dfrac{8}{5}\).
- Answer
-
\(-1\)
Solve: \(n − \dfrac{3}{7} = − \dfrac{9}{7}\).
- Answer
-
\(-\dfrac{6}{7}\)
The next example may not seem to have a fraction, but let’s see what happens when we solve it.
Solve: \(10q = 44\).
Solution
| Divide both sides by 10 to undo the multiplication. | \(\dfrac{10q}{10} = \dfrac{44}{10}\) |
| Simplify. | \(q =\dfrac{22}{5}\) |
Check:
| Substitute \(q = \dfrac{22}{5}\) into the original equation. | \(10 \left(\dfrac{22}{5}\right) \stackrel{?}{=} 44\) |
| Simplify. | \(\stackrel{2}{\cancel{10}} \left(\dfrac{22}{\cancel{5}}\right) \stackrel{?}{=} 44\) |
| Multiply. | \(44 = 44 \; \checkmark\) |
The solution to the equation was the fraction \(\dfrac{22}{5}\). We leave it as an improper fraction.
Solve: \(12u = −76\).
- Answer
-
\(-\dfrac{19}{3}\)
Solve: \(8m = 92\).
- Answer
-
\(\dfrac{23}{2}\)
Solve Equations with Fractions Using the Multiplication Property of Equality
Consider the equation \(\dfrac{x}{4} = 3\). We want to know what number divided by \(4\) gives \(3\). So to “undo” the division, we will need to multiply by \(4\). The Multiplication Property of Equality will allow us to do this. This property says that if we start with two equal quantities and multiply both by the same number, the results are equal.
For any numbers \(a\), \(b\), and \(c\), if \(a = b\), then \(ac = bc\). If you multiply both sides of an equation by the same quantity, you still have equality.
Let’s use the Multiplication Property of Equality to solve the equation \(\dfrac{x}{7} = −9\).
Solve: \(\dfrac{x}{7} = −9\).
Solution
| Use the Multiplication Property of Equality to multiply both sides by 7 . This will isolate the variable. | \(\textcolor{red}{7} \cdot \dfrac{x}{7} = \textcolor{red}{7} (-9)\) |
| Multiply. | \(\dfrac{7x}{7} = -63\) |
| Simplify. | \(x = -63\) |
| Check. Substitute \(\textcolor{red}{-63}\) for x in the original equation. | \(\dfrac{\textcolor{red}{-63}}{7} \stackrel{?}{=} -9\) |
| The equation is true. | \(-9 = -9 ; \checkmark\) |
Solve: \(\dfrac{f}{5} = −25\).
- Answer
-
\(-125\)
Solve: \(\dfrac{h}{9} = −27\).
- Answer
-
\(-243\)
Solve: \(\dfrac{p}{−8} = −40\).
Solution
Here, \(p\) is divided by \(−8\). We must multiply by \(−8\) to isolate \(p\).
| Multiply both sides by −8. | \(\textcolor{red}{-8} \left(\dfrac{p}{-8}\right) = \textcolor{red}{-8} (-40)\) |
| Multiply. | \(\dfrac{-8p}{-8} = 320\) |
| Simplify. | \(p = 320\) |
Check:
| Substitute p = 320. | \(\dfrac{\textcolor{red}{320}}{-8} \stackrel{?}{=} -40\) |
| The equation is true. | \(-40 = -40 \; \checkmark\) |
Solve: \(\dfrac{c}{−7} = −35\).
- Answer
-
\(245\)
Solve: \(\dfrac{x}{−11} = −12\).
- Answer
-
\(132\)
Solve Equations with a Coefficient of \(−1\)
Look at the equation \(−y = 15\). Does it look as if y is already isolated? But there is a negative sign in front of \(y\), so it is not isolated.
There are three different ways to isolate the variable in this type of equation. We will show all three ways in Example \(\PageIndex{7}\).
Solve: \(−y = 15\).
Solution
One way to solve the equation is to rewrite \(−y\) as \(−1y\), and then use the Division Property of Equality to isolate \(y\).
| Rewrite −y as −1y. | \(-1y = 15\) |
| Divide both sides by −1. | \(\dfrac{-1y}{\textcolor{red}{-1}} = \dfrac{15}{\textcolor{red}{-1}}\) |
| Simplify each side. | \(y = -15\) |
Another way to solve this equation is to multiply both sides of the equation by −1.
| Multiply both sides by −1. | \(\textcolor{red}{-1} (-y) = \textcolor{red}{-1} (15)\) |
| Simplify each side. | \(y = -15\) |
The third way to solve the equation is to read \(−y\) as “the opposite of \(y\).” What number has \(15\) as its opposite? The opposite of \(15\) is \(−15\). So \(y = −15\).
For all three methods, we isolated \(y\) is isolated and solved the equation.
Check:
| Substitute y = −15. | \(-(\textcolor{red}{15}) \stackrel{?}{=} (15)\) |
| Simplify. The equation is true. | \(15 = 15 \; \checkmark\) |
Solve: \(−y = 48\).
- Answer
-
\(-48\)
Solve: \(−c = −23\).
- Answer
-
\(23\)
Contributors and Attributions
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at http://cnx.org/contents/fd53eae1-fa2...49835c3c@5.191."