4.7 Solve Equations with Fractions

Learning objectives

Learning objectives

• 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

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.

Example 4.95

Solution

1. Identify the test. A candidate value is a solution only if substitution makes both sides of $x-\frac{3}{10}=\frac{1}{2}$ equal.
2. Test $x=1$. Substitution gives $1-\frac{3}{10}=\frac{7}{10}$, and $\frac{7}{10}\ne\frac{1}{2}=\frac{5}{10}$, so $1$ is not a solution.
3. Test $x=\frac{4}{5}$. Substitution gives $\frac{4}{5}-\frac{3}{10}=\frac{8}{10}-\frac{3}{10}=\frac{5}{10}=\frac{1}{2}$, so $\frac{4}{5}$ is a solution.
4. Test $x=-\frac{4}{5}$. Substitution gives $-\frac{4}{5}-\frac{3}{10}=-\frac{8}{10}-\frac{3}{10}=-\frac{11}{10}$, which is not $\frac{5}{10}$, so $-\frac{4}{5}$ is not a solution.
5. Compare the results. Only $x=\frac{4}{5}$ makes the original equation a true statement.

Takeaway: Test each proposed value in the original equation; only $x=\frac{4}{5}$ makes $x-\frac{3}{10}=\frac{1}{2}$ true.

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.

Example 4.96

Solution

1. Isolate the variable. Subtract $\frac{9}{16}$ from both sides to get $y+\frac{9}{16}-\frac{9}{16}=\frac{5}{16}-\frac{9}{16}$, as the figures show.
2. Simplify each side. The left side becomes $y+0$, while the right side is $\frac{5-9}{16}=-\frac{4}{16}$.
3. Reduce the fraction. Since $4$ and $16$ share the factor $4$, $-\frac{4}{16}=-\frac{1}{4}$.
4. Check the solution. Substituting gives $-\frac{1}{4}+\frac{9}{16}=-\frac{4}{16}+\frac{9}{16}=\frac{5}{16}$, confirming the equality.
5. State the result. The solution is $y=-\frac{1}{4}$.

Takeaway: Apply the same fraction operation to both sides of an equation and verify the resulting solution by substitution.

We used the Subtraction Property of Equality in Example 4.96. Now we’ll use the Addition Property of Equality.

Example 4.97

Solution

1. Isolate the variable. Add $\frac{5}{9}$ to both sides: $a-\frac{5}{9}+\frac{5}{9}=-\frac{8}{9}+\frac{5}{9}$.
2. Simplify the left side. The opposite fractions cancel, leaving $a+0$.
3. Subtract on the right. The result is $\frac{-8+5}{9}=-\frac{3}{9}$.
4. Reduce and check. The fraction reduces to $-\frac{1}{3}$, and $-\frac{1}{3}-\frac{5}{9}=-\frac{3}{9}-\frac{5}{9}=-\frac{8}{9}$.
5. State the result. The solution is $a=-\frac{1}{3}$.

Takeaway: Adding the same fraction to both sides removes it from one side while preserving the equality.

The next example may not seem to have a fraction, but let’s see what happens when we solve it.

Example 4.98

Solution

1. Identify the coefficient. The variable $q$ is multiplied by $10$, so dividing both sides by $10$ will isolate it.
2. Divide both sides. This gives $\frac{10q}{10}=\frac{44}{10}$, and the factor $10$ cancels on the left.
3. Simplify the result. Reducing $\frac{44}{10}$ by the common factor $2$ gives $q=\frac{22}{5}$.
4. Check the solution. Substituting gives $10(\frac{22}{5})=2\cdot22=44$, which matches the original right-hand side.
5. State the result. The solution is $q=\frac{22}{5}$, left as an improper fraction.

Takeaway: Dividing both sides of $10q=44$ by $10$ isolates $q$ and gives $q=\frac{22}{5}$, which checks because $10(\frac{22}{5})=44$.

Solve Equations with Fractions Using the Multiplication Property of Equality

Consider the equation $\frac{x}{4}=3.$ 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.

Let’s use the Multiplication Property of Equality to solve the equation $\frac{x}{7}=-9.$

Example 4.99

Solution

1. Clear the denominator. Multiply both sides of $\frac{x}{7}=-9$ by $7$ to get $7\left(\frac{x}{7}\right)=7(-9)$.
2. Simplify the left side. Since $7\left(\frac{x}{7}\right)=x$, the equation becomes $\frac{7x}{7}=-63$.
3. Cancel and isolate. Dividing the left side by $7$ gives $x=-63$.
4. Check the solution. Substitution gives $\frac{-63}{7}=-9$, so the original equation is true.
5. State the result. The solution is $x=-63$.

Takeaway: Multiply both sides by the denominator of a fraction to isolate the variable in one step.

Example 4.100

Solution

1. Clear the denominator. Multiply both sides of $\frac{p}{-8}=-40$ by $-8$ to get $(-8)\left(\frac{p}{-8}\right)=(-8)(-40)$.
2. Simplify the left side. The factors $-8$ cancel, leaving $p$ in the numerator over $-8$.
3. Multiply on the right. The two negative factors make a positive product, so $\frac{-8p}{-8}=320$.
4. Isolate and check. Dividing by $-8$ gives $p=320$, and $\frac{320}{-8}=-40$ confirms it.
5. State the result. The solution is $p=320$.

Takeaway: Two negative factors multiply to a positive number, so a negative divisor can reverse the sign of the right-hand side.

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 4.101.

Example 4.101

Solution

1. Expose the coefficient. Rewrite $-y=15$ as $-1y=15$ so the negative coefficient is visible.
2. Multiply both sides. Multiplying by $-1$ gives $(-1)(-1y)=(-1)(15)$, and multiplying a number by negative one reverses its sign.
3. Simplify each side. The left side becomes $y$, while the right side becomes $-15$.
4. Check the solution. Substituting $y=-15$ gives $-(-15)=15$, so the original equation is true.
5. State the result. The solution is $y=-15$.

Takeaway: To solve $-y=15$, multiply both sides by $-1$ to isolate $y$, giving $y=-15$ because the opposite of $15$ is $-15$.

Solve Equations with a Fraction Coefficient

When we have an equation with a fraction coefficient we can use the Multiplication Property of Equality to make the coefficient equal to $1.$

For example, in the equation:

The coefficient of $x$ is $\frac{3}{4}.$ To solve for $x,$ we need its coefficient to be $1.$ Since the product of a number and its reciprocal is $1,$ our strategy here will be to isolate $x$ by multiplying by the reciprocal of $\frac{3}{4}.$ We will do this in Example 4.102.

Example 4.102

Solution

1. Find the reciprocal. The coefficient of $x$ is $\frac{3}{4}$, whose reciprocal is $\frac{4}{3}$.
2. Multiply both sides. Multiplying by that reciprocal gives $\left(\frac{4}{3}\right)\left(\frac{3}{4}x\right)=\left(\frac{4}{3}\right)(24)$.
3. Cancel on the left. The factors $\frac{4}{3}$ and $\frac{3}{4}$ cancel, leaving $1x$.
4. Simplify the right side. The product is $\frac{4\cdot24}{3}=32$, so the equation becomes $x=32$.
5. Check the solution. Substitution gives $\frac{3}{4}(32)=24$, confirming that the original equation is true.

Takeaway: Multiplying both sides of $\frac{3}{4}x=24$ by the reciprocal $\frac{4}{3}$ isolates $x$ and gives $x=32$.

Example 4.103

Solution

1. Find the reciprocal. The reciprocal of the coefficient $-\frac{3}{8}$ is $-\frac{8}{3}$.
2. Multiply both sides. This gives $\left(-\frac{8}{3}\right)\left(-\frac{3}{8}w\right)=\left(-\frac{8}{3}\right)(72)$.
3. Cancel on the left. The factors $-\frac{8}{3}$ and $-\frac{3}{8}$ cancel, leaving $1w$.
4. Multiply and simplify. The right side becomes $-\frac{8\cdot72}{3}=-192$, so $w=-192$.
5. Check the solution. Substituting gives $\left(-\frac{3}{8}\right)(-192)=72$, confirming the equation.
6. State the result. The solution is $w=-192$.

Takeaway: The reciprocal of a negative fraction is negative, and multiplying by that reciprocal preserves the equation.

Translate Sentences to Equations and Solve

Now we have covered all four properties of equality—subtraction, addition, division, and multiplication. We’ll list them all together here for easy reference.

When you add, subtract, multiply or divide the same quantity from both sides of an equation, you still have equality.

In the next few examples, we’ll translate sentences into equations and then solve the equations. It might be helpful to review the translation table in Evaluate, Simplify, and Translate Expressions.

Example 4.104

Solution

1. Translate the phrase. The statement that $n$ divided by $6$ is $-24$ becomes $\frac{n}{6}=-24$.
2. Clear the denominator. Multiply both sides by $6$ to get $6\left(\frac{n}{6}\right)=6(-24)$.
3. Simplify. The left side becomes $n$, while the right side is $-144$.
4. Check the solution. Substitution gives $\frac{-144}{6}=-24$, confirming the translation and solution.
5. State the result. The translated equation is $\frac{n}{6}=-24$, and its solution is $n=-144$.

Takeaway: Translate a quotient phrase carefully before solving, because the order of the divisor and dividend determines the equation.

Example 4.105

Solution

1. Translate the phrase. The quotient of $q$ and $-5$ is $70$, so write $\frac{q}{-5}=70$.
2. Clear the denominator. Multiply both sides by $-5$ to get $(-5)\left(\frac{q}{-5}\right)=(-5)(70)$.
3. Simplify. The left side is $q$, and the right side is $-350$.
4. Check the solution. Substituting gives $\frac{-350}{-5}=70$, confirming the equation.
5. State the result. The translated equation is $\frac{q}{-5}=70$, and its solution is $q=-350$.

Takeaway: In a quotient statement, the first quantity is the dividend and the second is the divisor.

Example 4.106

Solution

1. Translate the phrase. Two-thirds of $f$ is $18$, so write $\frac{2}{3}f=18$.
2. Isolate the variable. Multiply both sides by the reciprocal $\frac{3}{2}$ to get $\left(\frac{3}{2}\right)\left(\frac{2}{3}f\right)=\left(\frac{3}{2}\right)(18)$.
3. Cancel and multiply. The left side becomes $1f$, while the right side is $\frac{3\cdot18}{2}=27$.
4. Check the solution. Substitution gives $\frac{2}{3}(27)=18$, confirming the translation.
5. State the result. The translated equation is $\frac{2}{3}f=18$, and its solution is $f=27$.

Takeaway: A fractional phrase of a variable translates to multiplication, so multiplying by the reciprocal isolates the variable.

Example 4.107

Solution

1. Translate the quotient. The quotient of $m$ and $\frac{5}{6}$ is $\frac{3}{4}$, so write $\frac{m}{5/6}=\frac{3}{4}$.
2. Multiply both sides. Multiplying by $\frac{5}{6}$ gives $\frac{5}{6}\left(\frac{m}{5/6}\right)=\frac{5}{6}\cdot\frac{3}{4}$.
3. Cancel and simplify. The left side becomes $m$, while the right side is $\frac{15}{24}=\frac{5}{8}$, so $m=\frac{5}{8}$.
4. Check the quotient. Dividing gives $\frac{5}{8}\div\frac{5}{6}=\frac{5}{8}\cdot\frac{6}{5}=\frac{3}{4}$, as required.
5. State the result. The translated equation has the solution $m=\frac{5}{8}$.

Takeaway: The quotient of $m$ and $\frac{5}{6}$ translates to $\frac{m}{5/6}=\frac{3}{4}$, and multiplying by $\frac{5}{6}$ gives $m=\frac{5}{8}$.

Example 4.108

Solution

1. Translate the sum. The sum of $\frac{3}{8}$ and $x$ is $3\frac{1}{2}$, so write $\frac{3}{8}+x=3\frac{1}{2}$.
2. Isolate the variable. Subtract $\frac{3}{8}$ from both sides to obtain $x=3\frac{1}{2}-\frac{3}{8}$.
3. Convert to eighths. The mixed number $3\frac{1}{2}$ equals $\frac{7}{2}=\frac{28}{8}$, so $x=\frac{28}{8}-\frac{3}{8}$.
4. Subtract the fractions. The common denominator stays $8$, giving $x=\frac{25}{8}$.
5. Rewrite as a mixed number. Since $25=3\cdot8+1$, the solution is $x=3\frac{1}{8}$.
6. Check the sum. Adding gives $\frac{3}{8}+3\frac{1}{8}=3\frac{4}{8}=3\frac{1}{2}$, so the solution checks.

Takeaway: The equation $\frac{3}{8}+x=3\frac{1}{2}$ gives $x=3\frac{1}{8}$, which checks because $\frac{3}{8}+3\frac{1}{8}=3\frac{1}{2}$.

Section 4.7 Exercises

Section 4.7 Exercises

Determine Whether a Fraction is a Solution of an Equation

In the following exercises, determine whether each number is a solution of the given equation.

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

In the following exercises, solve.

Solve Equations with Fractions Using the Multiplication Property of Equality

In the following exercises, solve.

Mixed Practice

In the following exercises, solve.

Translate Sentences to Equations and Solve

In the following exercises, translate to an algebraic equation and solve.

Everyday Math

Writing Exercises

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

1. Determine whether each number is a solution of the given equation.

$x-\frac{2}{3}=\frac{1}{6}$:

1. ⓐ $x=1$

2. ⓑ $x=\frac{5}{6}$

3. ⓒ $x=-\frac{5}{6}$

2. Determine whether each number is a solution of the given equation.

$y-\frac{1}{4}=\frac{3}{8}$:

1. ⓐ $y=1$

2. ⓑ $y=-\frac{5}{8}$

3. ⓒ $y=\frac{5}{8}$

3. Solve: $y+\frac{11}{12}=\frac{5}{12}.$

4. Solve: $y+\frac{8}{15}=\frac{4}{15}.$

1. Solve: $a-\frac{3}{5}=-\frac{8}{5}.$

2. Solve: $n-\frac{3}{7}=-\frac{9}{7}.$

3. Solve: $12u=-76.$

4. Solve: $8m=92.$

1. Solve: $\frac{f}{5}=-25.$

2. Solve: $\frac{h}{9}=-27.$

3. Solve: $-y=48.$

4. Solve: $\frac{5}{6}y=15.$

1. Solve: $-\frac{4}{7}a=52.$

2. Solve: $-\frac{7}{9}w=84.$

3. Translate and solve: $n$ divided by $7$ is equal to $-21.$

4. Translate and solve: $n$ divided by $8$ is equal to $-56.$

1. Translate and solve: The quotient of $q$ and $-8$ is $72.$

2. Translate and solve: The quotient of $p$ and $-9$ is $81.$

3. Translate and solve: Two-fifths of $f$ is $16.$

4. Translate and solve: Three-fourths of $f$ is $21.$

1. Translate and solve. The quotient of $n$ and $\frac{2}{3}$ is $\frac{5}{12}.$

2. Translate and solve The quotient of $c$ and $\frac{3}{8}$ is $\frac{4}{9}.$

3. Translate and solve: The sum of five-eighths and $x$ is one-fourth.

4. Translate and solve: The difference of one-and-three-fourths and $x$ is five-sixths.

1. $y-\frac{1}{3}=\frac{5}{12}$:

1. ⓐ $y=1$

2. ⓑ $y=\frac{3}{4}$

3. ⓒ $y=-\frac{3}{4}$

2. $k+\frac{2}{5}=\frac{5}{6}$:

1. ⓐ $k=1$

2. ⓑ $k=\frac{13}{30}$

3. ⓒ $k=-\frac{13}{30}$

3. $m+\frac{3}{8}=\frac{7}{8}$

4. $h+\frac{5}{6}=\frac{1}{6}$

1. $c-\frac{1}{4}=-\frac{5}{4}$

2. $z-(-\frac{5}{12})=-\frac{7}{12}$

3. $p-\frac{3}{10}=\frac{5}{8}$

4. $k+(-\frac{1}{3})=-\frac{4}{5}$

1. $7k=18$

2. $-9v=33$

3. $\frac{b}{3}=-9$

4. $\frac{x}{8}=-32$

1. $\frac{q}{-4}=-40$

2. $\frac{s}{-15}=-3$

3. $-y=42$

4. $-k=-\frac{17}{20}$

1. $\frac{3}{10}p=30$

2. $\frac{5}{2}m=-40$

3. $-\frac{3}{7}b=9$

4. $-\frac{5}{12}v=-15$

1. $8y=0$

2. $7g=\frac{7}{9}$

3. $q+\frac{5}{6}=\frac{1}{12}$

4. $\frac{1}{4}n=\frac{7}{10}$

1. $-\frac{2}{3}=y+\frac{3}{8}$

2. $\frac{8}{15}=-d$

3. $n$ divided by six is $-24.$

4. $m$ divided by $-7$ is $-8.$

1. The quotient of $f$ and $-4$ is $-20.$

2. The quotient of $g$ and nine is $14.$

3. Two-fifths of $q$ is $20.$

4. Four-ninths of $p$ is $-28.$

1. The quotient of $h$ and $2$ is $43.$

2. The quotient of $a$ and $\frac{2}{3}$ is $\frac{3}{4}.$

3. The sum of three-fourths and $x$ is $\frac{1}{8}.$

4. The difference of $y$ and one-third is $-\frac{1}{6}.$

1. Playhouse The table in a child’s playhouse is $\frac{3}{5}$ of an adult-size table. The playhouse table is $18$ inches high. Find the height of an adult-size table by solving the equation $\frac{3}{5}h=18.$

2. Richard thinks the solution to the equation $\frac{3}{4}x=24$ is $16.$ Explain why Richard is wrong.