5.4 Solve Equations with Decimals

Learning objectives

Learning objectives

• Determine whether a decimal is a solution of an equation

• Solve equations with decimals

• Translate to an equation and solve

Determine Whether a Decimal is a Solution of an Equation

Solving equations with decimals is important in our everyday lives because money is usually written with decimals. When applications involve money, such as shopping for yourself, making your family’s budget, or planning for the future of your business, you’ll be solving equations with decimals.

Now that we’ve worked with decimals, we are ready to find solutions to equations involving decimals. 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, a fraction, or a decimal. We’ll list these steps here again for easy reference.

Example 5.40

Solution

1. State the solution test. A candidate value is a solution only if substituting it for $x$ makes both sides of $x-0.7=1.5$ equal.
2. Test the value $x=1$. Substitution gives $1-0.7=0.3$, and $0.3\neq 1.5$, so $1$ is not a solution.
3. Test the value $x=-0.8$. Substitution gives $-0.8-0.7=-1.5$, and $-1.5\neq 1.5$, so $-0.8$ is not a solution.
4. Test the value $x=2.2$. Substitution gives $2.2-0.7=1.5$, so the two sides are equal and $2.2$ is a solution.

Takeaway: Substitution confirms that only $x=2.2$ satisfies $x-0.7=1.5$, because it is the only proposed value that makes both sides equal.

Solve Equations with Decimals

In previous chapters, we solved equations using the Properties of Equality. We will use these same properties to solve equations with decimals.

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

Example 5.41

Solution

1. Identify the operation to undo. The variable $y$ is added to $2.3$, so subtract $2.3$ from both sides of $y+2.3=-4.7$.
2. Subtract on both sides. The equation becomes $y+2.3-2.3=-4.7-2.3$, and the Subtraction Property of Equality preserves the equality.
3. Simplify the result. Removing the matching $2.3$ terms and evaluating $-4.7-2.3$ gives $y=-7$.
4. Check the solution. Substituting $y=-7$ gives $-7+2.3=-4.7$, so the original equation is true.

Takeaway: Subtracting $2.3$ from both sides isolates $y$ in $y+2.3=-4.7$ and gives $y=-7$, which satisfies the original equation.

Example 5.42

Solution

1. Identify the operation on $a$. The equation $a-4.75=-1.39$ contains the subtraction $a-4.75$, so add $4.75$ to both sides to cancel it: $a-4.75+4.75=-1.39+4.75$.
2. Add the decimals with aligned place values. The calculation is $\begin{array}{r}4.75\\-1.39\\\hline3.36\end{array}$, giving $a=-1.39+4.75=3.36$.
3. Substitute the candidate. Replace $a$ with $3.36$ to obtain $3.36-4.75=-1.39$. Aligning the decimals, $\begin{array}{r}3.36\\-4.75\\\hline-1.39\end{array}$, confirms that the two sides are equal, as the check in the diagram shows.
4. State the solution. The value that satisfies the original equation is $a=3.36$.

Takeaway: Adding the same quantity to both sides isolates the variable while preserving the equality.

Example 5.43

Solution

1. Identify the decimal coefficient. The equation $-4.8=0.8n$ has $n$ multiplied by $0.8$, so divide both sides by the nonzero number $0.8$.
2. Divide both sides by $0.8$. The equation becomes $\frac{-4.8}{0.8}=\frac{0.8n}{0.8}$, and the Division Property of Equality preserves the equality.
3. Simplify to solve for $n$. Since $-4.8\div 0.8=-6$ and $0.8n\div 0.8=n$, the result is $n=-6$.
4. Check the solution. Substituting $n=-6$ gives $0.8(-6)=-4.8$, so the original equation is true.

Takeaway: Dividing both sides of $-4.8=0.8n$ by $0.8$ gives $n=-6$, and substitution confirms that this value satisfies the equation.

Example 5.44

Solution

1. Choose the inverse operation. Since $p$ is divided by $-1.8$, multiply both sides of $\frac{p}{-1.8}=-6.5$ by $-1.8$.
2. Multiply both sides. The equation becomes $-1.8\left(\frac{p}{-1.8}\right)=-1.8(-6.5)$, and the Multiplication Property of Equality preserves the equality.
3. Simplify to solve for $p$. The fraction on the left becomes $p$, while the right side equals $11.7$, so $p=11.7$.
4. Check the solution. Substituting $p=11.7$ gives $\frac{11.7}{-1.8}=-6.5$, so the original equation is true.

Takeaway: Multiplying both sides of $\frac{p}{-1.8}=-6.5$ by $-1.8$ isolates $p$ and gives $p=11.7$, which checks in the original equation.

Translate to an Equation and Solve

Now that we have solved equations with decimals, we are ready to translate word sentences to equations and solve. Remember to look for words and phrases that indicate the operations to use.

Example 5.45

Solution

1. Translate the phrase into an equation. The difference of $n$ and $4.3$ means $n-4.3$, and that difference equals $2.1$, so write $n-4.3=2.1$.
2. Add $4.3$ to both sides. The equation becomes $n-4.3+4.3=2.1+4.3$, which cancels the subtraction on the left and leaves $n=2.1+4.3$.
3. Add with aligned decimal places. Using $\begin{array}{r}2.10\\+4.30\\\hline6.40\end{array}$ shows that $2.1+4.3=6.4$, so $n=6.4$.
4. Check the result. Substituting gives $6.4-4.3=2.1$, and the matching final equality in the diagram confirms that the original equation is true. Therefore, the solution is $n=6.4$.

Takeaway: Translate the difference phrase as subtraction, then isolate the variable and verify the result.

Example 5.46

Solution

1. Write the product as multiplication. The product of $-3.1$ and $x$ is $5.27$, so the translated equation is $-3.1x=5.27$.
2. Divide both sides by $-3.1$. This gives $\frac{-3.1x}{-3.1}=\frac{5.27}{-3.1}$, and the left side simplifies to $x$.
3. Divide using aligned decimal places. The calculation $\begin{array}{r}5.27\\\div(-3.1)\\\hline-1.7\end{array}$ shows that $5.27\div(-3.1)=-1.7$, so $x=-1.7$.
4. Verify the candidate. Substitution gives $-3.1(-1.7)=5.27$, since the product of two negative numbers is positive, and the displayed check confirms that both sides agree. Thus, the solution is $x=-1.7$.

Takeaway: Preserve the negative factor when translating, then divide both sides by the same nonzero coefficient.

Example 5.47

Solution

1. Translate the quotient into an equation. The quotient of $p$ and $-2.4$ is $6.5$, so write $\frac{p}{-2.4}=6.5$.
2. Multiply both sides by $-2.4$. The equation becomes $-2.4\left(\frac{p}{-2.4}\right)=-2.4(6.5)$, and the factors $-2.4$ cancel on the left.
3. Multiply the right side. With decimal place values aligned, $\begin{array}{r}6.5\\\times(-2.4)\\\hline-15.6\end{array}$, so $p=-2.4(6.5)=-15.6$.
4. Check the candidate. Substituting gives $\frac{-15.6}{-2.4}=6.5$, since the quotient of two negative numbers is positive, and the diagram's final equality confirms the result. Therefore, the solution is $p=-15.6$.

Takeaway: A quotient becomes an equation whose solution is found by multiplying both sides by the divisor.

Example 5.48

Solution

1. Translate the sum into an equation. The sum of $n$ and $2.9$ is $1.7$, so write $n+2.9=1.7$.
2. Subtract $2.9$ from both sides. The equation becomes $n+2.9-2.9=1.7-2.9$, and the quantities $2.9$ cancel on the left.
3. Subtract with aligned decimal places. Using $\begin{array}{r}1.7\\-2.9\\\hline-1.2\end{array}$ shows that $1.7-2.9=-1.2$, so $n=-1.2$.
4. Verify the candidate. Substitution gives $-1.2+2.9=1.7$, because $2.9-1.2=1.7$, and the final equality shown in the diagram confirms the check. Therefore, the solution is $n=-1.2$.

Takeaway: Translate the sum phrase as addition, then use an equality property to isolate the variable.

Section 5.4 Exercises

Section 5.4 Exercises

Determine Whether a Decimal is a Solution of an Equation

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

Solve Equations with Decimals

In the following exercises, solve the equation.

Mixed Practice

In the following exercises, solve the equation. Then check your solution.

Translate to an Equation and Solve

In the following exercises, translate 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.

ⓑ On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

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

$x-0.6=1.3:\,$ⓐ $\,x=0.7$ⓑ $\,x=1.9$ⓒ $\,x=-0.7$

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

$y-0.4=1.7:\,$ⓐ $\,y=2.1$ⓑ $\,y=1.3$ⓒ $\,-1.3$

3. Solve: $y+2.7=-5.3.$

4. Solve: $y+3.6=-4.8.$

1. Solve: $a-3.93=-2.86.$

2. Solve: $n-3.47=-2.64.$

3. Solve: $-8.4=0.7b.$

4. Solve: $-5.6=0.7c.$

1. Solve: $\frac{c}{-2.6}=-4.5.$

2. Solve: $\frac{b}{-1.2}=-5.4.$

3. Translate and solve: The difference of $y$ and $4.9$ is $2.8.$

4. Translate and solve: The difference of $z$ and $5.7$ is $3.4.$

1. Translate and solve: The product of $-4.3$ and $x$ is $12.04.$

2. Translate and solve: The product of $-3.1$ and $m$ is $26.66.$

3. Translate and solve: The quotient of $q$ and $-3.4$ is $4.5.$

4. Translate and solve: The quotient of $r$ and $-2.6$ is $2.5.$

1. Translate and solve: The sum of $j$ and $3.8$ is $2.6.$

2. Translate and solve: The sum of $k$ and $4.7$ is $0.3.$

3. $x-0.8=2.3$ⓐ $\,x=2$ⓑ $\,x=-1.5$ⓒ $\,x=3.1$

4. $\frac{h}{1.5}=-4.3$ ⓐ $\,h=6.45$ⓑ $\,h=-6.45$ⓒ $\,h=-2.1$

1. $y+2.9=5.7$

2. $f+3.45=2.6$

3. $a+6.2=-1.7$

4. $c+1.15=-3.5$

1. $n-2.6=1.8$

2. $x-0.4=-3.9$

3. $j-1.82=-6.5$

4. $m-0.25=-1.67$

1. $0.5x=3.5$

2. $-1.7c=8.5$

3. $-1.4p=-4.2$

4. $-120=1.5q$

1. $0.24x=4.8$

2. $-3.4z=-9.18$

3. $\frac{a}{0.4}=-20$

4. $\frac{x}{0.7}=-0.4$

1. $\frac{p}{-5}=-1.65$

2. $\frac{r}{-1.2}=-6$

3. $x-5=-11$

4. $p+8=-2$

1. $-4.2m=-33.6$

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

3. $\frac{7}{8}m=\frac{1}{10}$

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

1. $\frac{11}{20}=-f$

2. $-4.2a=3.36$

3. $r-1.25=-2.7$

4. $\frac{h}{-3}=-8$

1. The difference of $n$ and $1.9$ is $3.4.$

2. The product of $-6.2$ and $x$ is $-4.96.$

3. The quotient of $y$ and $-1.7$ is $-5.$

4. The sum of $n$ and $-7.3$ is $2.4.$

1. Shawn bought a pair of shoes on sale for $\$78$. Solve the equation $0.75p=78$ to find the original price of the shoes, $p.$

2. Think about solving the equation $1.2y=60,$ but do not actually solve it. Do you think the solution should be greater than $60$ or less than $60?$ Explain your reasoning. Then solve the equation to see if your thinking was correct.