ORCCA Solving Multistep Linear Equations

Section 3.1 Solving Multistep Linear Equations

We have learned how to solve one-step equations in Section 2.5. In this section, we will learn how to solve multistep equations.

Subsection 3.1.1 Solving Two-Step Equations

Example 3.1.1.

A water tank can hold $140$ gallons of water, but it has only $5$ gallons of water. A tap was turned on, pouring $15$ gallons of water into the tank every minute. After how many minutes will the tank be full? Let's find a pattern first.

Table 3.1.2. Amount of Water in the Tank

Minutes since Tap	Amount of Water in
Was Turned on	the Tank (in Gallons)
$0$	$5$
$1$	$15\cdot1+5=20$
$2$	$15\cdot2+5=35$
$3$	$15\cdot3+5=50$
$4$	$15\cdot4+5=65$
$\vdots$	$\vdots$
$m$	$15m+5$

We can see that after $m$ minutes, the tank has $15m+5$ gallons of water. This makes sense since the tap pours $15m$ gallons of water into the tank in $m$ minutes and it had $5$ gallons to start with. To find when the tank will be full (with $140$ gallons of water), we can write the equation

\begin{equation*} 15m+5=140 \end{equation*}

First, we need to isolate the variable term, $15m\text{,}$ in the equation. In other words, we need to remove $5$ from the left side of the equals sign. We can do this by subtracting $5$ from both sides of the equation. Once the variable term is isolated, we can eliminate the coefficient and solve for $m\text{.}$ The full process is:

\begin{align*} 15m+5\amp=140\\ 15m+5\subtractright{5}\amp=140\subtractright{5}\\ 15m\amp=135\\ \divideunder{15m}{15}\amp=\divideunder{135}{15}\\ m\amp=9 \end{align*}

Next, we need to substitute $m$ with $9$ in the equation $15m+5=140$ to check the solution:

\begin{align*} 15m+5\amp=140\\ 15(\substitute{9})+5\amp\stackrel{?}{=}140\\ 135+5\amp\stackrel{\checkmark}{=}140 \end{align*}

The solution $9$ is checked. In summary, the tank will be full after $9$ minutes.

In solving the two-step equation in Example 3.1.1, we first isolated the variable expression $15m$ and then eliminated the coefficient of $15$ by dividing each side of the equation by $15\text{.}$ These two steps will be at the heart of our approach to solving linear equations. For more complicated equations, we may need to simplify some of the expressions first. Below is a general approach to solving linear equations that we will use as we solve more and more complicated equations.

List 3.1.3. Steps to Solve Linear Equations

Simplify: Simplify the expressions on each side of the equation by distributing and combining like terms.
Isolate: Use addition or subtraction to separate the variable terms and constant terms (numbers) so that they are on different sides of the equation.
Eliminate: Use multiplication or division to eliminate the variable term's coefficient.
Check: Check the solution. Substitute values into the original equation and use the order of operations to simplify both sides. It's important to use the order of operations alone rather than properties like the distributive law. Otherwise you might repeat the same arithmetic errors made while solving and fail to catch an incorrect solution.
Summarize: State the solution set or (in the case of an application problem) summarize the result in a complete sentence using appropriate units.

Let's look at some more examples.

Example 3.1.4.

Solve for $y$ in the equation $7-3y=-8\text{.}$

Explanation

To solve, we will first separate the variable terms and constant terms into different sides of the equation. Then, we will eliminate the variable term's coefficient.

\begin{align*} 7-3y\amp=-8\\ 7-3y\subtractright{7}\amp=-8\subtractright{7}\\ -3y\amp=-15\\ \divideunder{-3y}{-3}\amp=\divideunder{-15}{-3}\\ y\amp=5 \end{align*}

Checking the solution $y=5\text{:}$

\begin{align*} 7-3y\amp=-8\\ 7-3(\substitute{5})\amp\stackrel{?}{=}-8\\ 7-15\amp\stackrel{\checkmark}{=}-8 \end{align*}

Therefore, the solution to the equation $7-3y=-8$ is $5$ and the solution set is $\{5\}\text{.}$

Subsection 3.1.2 Solving Multistep Linear Equations

Example 3.1.5.

Ahmed has saved $\$2{,}500.00$ in his savings account and is going to start saving $\$550.00$ per month. Julia has saved $\$4{,}600.00$ in her savings account and is going to start saving $\$250.00$ per month. If this situation continues, how many months later would Ahmed catch up with Julia in savings?

Ahmed saves $\$550.00$ per month, so he can save $550m$ dollars in $m$ months. With the $\$2{,}500.00$ he started with, after $m$ months he has $550m+2500$ dollars. Similarly, after $m$ months, Julia has $250m+4600$ dollars. To find when those two accounts will have the same amount of money, we write the equation

\begin{equation*} 550m+2500=250m+4600\text{.} \end{equation*}

\begin{align*} 550m+2500\amp=250m+4600\\ 550m+2500\subtractright{2500}\amp=250m+4600\subtractright{2500}\\ 550m\amp=250m+2100\\ 550m\subtractright{250m}\amp=250m+2100\subtractright{250m}\\ 300m\amp=2100\\ \divideunder{300m}{300}\amp=\divideunder{2100}{300}\\ m\amp=7 \end{align*}

Checking the solution $7\text{:}$

\begin{align*} 550m+2500\amp=250m+4600\\ 550(\substitute{7})+2500\amp\stackrel{?}{=}250(\substitute{7})+4600\\ 3850+2500\amp\stackrel{?}{=}1750+4600\\ 6350\amp\stackrel{\checkmark}{=}6350 \end{align*}

In summary, Ahmed will catch up with Julia's savings in $7$ months.

Example 3.1.6.

Solve for $x$ in $5-2x=5x-9\text{.}$

Explanation

\begin{align*} 5-2x\amp=5x-9\\ 5-2x\subtractright{5}\amp=5m-9\subtractright{5}\\ -2x\amp=5x-14\\ -2x\subtractright{5x}\amp=5x-14\subtractright{5x}\\ -7x\amp=-14\\ \divideunder{-7x}{-7}\amp=\divideunder{-14}{-7}\\ x\amp=2 \end{align*}

Checking the solution $2\text{:}$

\begin{align*} 5-2x\amp=5x-9\\ 5-2(\substitute{2})\amp\stackrel{?}{=}5(\substitute{2})-9\\ 5-4\amp\stackrel{?}{=}10-9\\ 1\amp\stackrel{\checkmark}{=}1 \end{align*}

Therefore, the solution is $2$ and the solution set is $\{2\}\text{.}$

Remark 3.1.7.

In Example 3.1.6, we could have moved variable terms to the right side of the equals sign, and number terms to the left side. We chose not to. There's no reason we couldn't have moved variable terms to the right side though. Let's compare:

\begin{align*} 5-2x\amp=5x-9\\ 5-2x\addright{9}\amp=5x-9\addright{9}\\ 14-2x\amp=5x\\ 14-2x\addright{2x}\amp=5x\addright{2x}\\ 14\amp=7x\\ \divideunder{14}{7}\amp=\divideunder{7x}{7}\\ 2\amp=x \end{align*}

Lastly, we could save a step by moving variable terms and number terms in one step:

\begin{align*} 5-2x\amp=5x-9\\ 5-2x\addright{2x+9}\amp\phantom{=}5x-9\addright{2x+9}\\ 14\amp=7x\\ \divideunder{14}{7}\amp=\divideunder{7x}{7}\\ 2\amp=x \end{align*}

This textbook will move variable terms and number terms separately throughout this chapter. Check with your instructor for their expectations.

Checkpoint 3.1.8.

The next example requires combining like terms.

Example 3.1.9.

Solve for $n$ in $n-9+3n=n-3n\text{.}$

Explanation

To start solving this equation, we'll need to combine like terms. After this, we can put all terms containing $n$ on one side of the equation and finish solving for $n\text{.}$

\begin{align*} n-9+3n\amp=n-3n\\ 4n-9\amp=-2n\\ 4n-9\subtractright{4n}\amp=-2n\subtractright{4n}\\ -9\amp=-6n\\ \divideunder{-9}{-6}\amp=\divideunder{-6n}{-6}\\ n\amp=\frac{3}{2} \end{align*}

Checking the solution $\frac{3}{2}\text{:}$

\begin{align*} n-9+3n\amp=n-3n\\ \substitute{\frac{3}{2}}-9+3\left(\substitute{\frac{3}{2}}\right)\amp\stackrel{?}{=}\substitute{\frac{3}{2}}-3\left(\substitute{\frac{3}{2}}\right)\\ \frac{3}{2}-9+\frac{9}{2}\amp\stackrel{?}{=}\frac{3}{2}-\frac{9}{2}\\ \frac{12}{2}-9\amp\stackrel{?}{=}-\frac{6}{2}\\ 6-9\amp\stackrel{?}{=}-3\\ -3\amp\stackrel{?}{=}-3 \end{align*}

The solution to the equation $n-9+3n=n-3n$ is $\frac{3}{2}$ and the solution set is $\left\{\frac{3}{2}\right\}\text{.}$

Checkpoint 3.1.10.

Example 3.1.11.

Azul is designing a rectangular garden and they have $40$ meters of wood for the border. Their garden's length must be $4$ meters less than three times the width, and its perimeter must be $40$ meters. Find the garden's length and width.

Explanation

Reminder: A rectangle's perimeter formula is $P=2(L+W)\text{,}$ where $P$ stands for perimeter, $L$ stands for length, and $W$ stands for width.

Let Azul's garden width be $W$ meters. We can then represent the length as $3W-4$ meters since we are told that it is $4$ meters less than three times the width. It's given that the perimeter is $40$ meters. Substituting those values into the formula, we have:

\begin{align*} P\amp=2(L+W)\\ 40\amp=2(3W-4+W)\\ 40\amp=2(4W-4)\amp\text{Like terms were combined.} \end{align*}

The next step to solve this equation is to remove the parentheses by distribution.

\begin{align*} 40\amp=2(4W-4)\\ 40\amp=8W-8\\ 40\addright{8}\amp=8W-8\addright{8}\\ 48\amp=8W\\ \divideunder{48}{8}\amp=\divideunder{8W}{8}\\ 6\amp=W\text{.} \end{align*}

Checking the solution $W=6\text{:}$

\begin{align*} 40\amp=2(4W-4)\\ 40\amp\stackrel{?}{=}2(4(\substitute{6})-4)\\ 40\amp\stackrel{\checkmark}{=}2(20)\text{.} \end{align*}

To determine the length, recall that this was represented by $3W-4\text{,}$ which is:

\begin{align*} 3W-4\amp=3(\substitute{6})-4\\ \amp=14\text{.} \end{align*}

Thus, the width of Azul's garden is $6$ meters and the length is $14$ meters.

Checkpoint 3.1.12.

We should be careful when we distribute a negative sign into the parentheses, like in the next example.

Example 3.1.13.

Solve for $a$ in $4-(3-a)=-2-2(2a+1)\text{.}$

Explanation

To solve this equation, we will simplify each side of the equation, manipulate it so that all variable terms are on one side and all constant terms are on the other, and then solve for $a\text{:}$

\begin{align*} 4-(3-a)\amp=-2-2(2a+1)\\ 4-3+a\amp=-2-4a-2\\ 1+a\amp=-4-4a\\ 1+a\addright{4a}\amp=-4-4a\addright{4a}\\ 1+5a\amp=-4\\ 1+5a\subtractright{1}\amp=-4\subtractright{1}\\ 5a\amp=-5\\ \divideunder{5a}{5}\amp=\divideunder{-5}{5}\\ a\amp=-1 \end{align*}

Checking the solution $-1\text{:}$

\begin{align*} 4-(3-a)\amp=-2-2(2a+1)\\ 4-(3-(\substitute{-1}))\amp\stackrel{?}{=}-2-2(2(\substitute{-1})+1)\\ 4-(4)\amp\stackrel{?}{=}-2-2(-1)\\ 0\amp\stackrel{\checkmark}{=}0 \end{align*}

Therefore, the solution to the equation is $-1$ and the solution set is $\{-1\}\text{.}$

Subsection 3.1.3 Differentiating between Simplifying Expressions, Evaluating Expressions, and Solving Equations

Let's look at the following similar, yet different, examples.

Example 3.1.14.

Simplify the expression $10-3(x+2)\text{.}$

Explanation

\begin{align*} 10-3(x+2)\amp=10-3x-6\\ \amp=-3x+4 \end{align*}

An equivalent result is $4-3x\text{.}$ Note that our final result is an expression.

Example 3.1.15.

Evaluate the expression $10-3(x+2)$ when $x=2$ and when $x=3\text{.}$

Explanation

We will substitute $x=2$ into the expression:

\begin{align*} 10-3(x+2)\amp=10-3(\substitute{2}+2)\\ \amp=10-3(4)\\ \amp=10-12\\ \amp=-2 \end{align*}

When $x=2\text{,}$ $10-3(x+2)=-2\text{.}$

Similarly, we will substitute $x=3$ into the expression:

\begin{align*} 10-3(x+2)\amp=10-3(\substitute{3}+2)\\ \amp=10-3(5)\\ \amp=10-15=-5 \end{align*}

When $x=3\text{,}$ $10-3(x+2)=-5\text{.}$

Note that the final results here are values of the original expression.

Example 3.1.16.

Solve the equation $10-3(x+2)=x-16\text{.}$

Explanation

\begin{align*} 10-3(x+2)\amp=x-16\\ 10-3x-6\amp=x-16\\ -3x+4\amp=x-16\\ -3x+4\subtractright{4}\amp=x-16\subtractright{4}\\ -3x\amp=x-20\\ -3x\subtractright{x}\amp=x-20\subtractright{x}\\ -4x\amp=-20\\ \divideunder{-4x}{-4}\amp=\divideunder{-20}{-4}\\ x\amp=5 \end{align*}

Checking the solution $x=5\text{:}$

\begin{align*} 10-3(x+2)\amp=x-16\\ 10-3(\substitute{5}+2)\amp\stackrel{?}{=}\substitute{5}-16\\ 10-3(7)\amp\stackrel{?}{=}-11\\ 10-21\amp\stackrel{\checkmark}{=}-11 \end{align*}

We have checked that $x=5$ is a solution of the equation $10-3(x+2)=x-16\text{.}$

Note that the final results here are solutions to the equations.

List 3.1.17. A summary the differences among simplifying expressions, evaluating expressions and solving equations:

An expression like $10-3(x+2)$ can be simplified to $-3x+4$ (as in Example 3.1.14), but we cannot solve for $x$ in an expression.
As $x$ takes different values, an expression has different values. In Example 3.1.15, when $x=2\text{,}$ $10-3(x+2)=-2\text{;}$ but when $x=3\text{,}$ $10-3(x+2)=-5\text{.}$
An equation connects two expressions with an equals sign. In Example 3.1.16, $10-3(x+2)=x-16$ has the expression $10-3(x+2)$ on the left side of equals sign, and the expression $x-16$ on the right side.
When we solve the equation $10-3(x+2)=x-16\text{,}$ we are looking for a number which makes those two expressions have the same value. In Example 3.1.16, we found the solution to be $x=5\text{,}$ which makes both $10-3(x+2)=-11$ and $x-16=-11\text{,}$ as shown in the checking part.

Exercises 3.1.4 Exercises

Warmup and Review

1.

Solve the equation.

$\displaystyle{ {t+1}={-7} }$

2.

Solve the equation.

$\displaystyle{ {t+7}={3} }$

3.

Solve the equation.

$\displaystyle{ {x-4}={-1} }$

4.

Solve the equation.

$\displaystyle{ {x-10}={-3} }$

5.

Solve the equation.

$\displaystyle{ {12}={-6y} }$

6.

Solve the equation.

$\displaystyle{ {64}={-8y} }$

7.

Solve the equation.

$\displaystyle{ {{\frac{9}{10}}A} = {6} }$

8.

Solve the equation.

$\displaystyle{ {{\frac{4}{7}}B} = {5} }$

Solving Two-Step Equations

9.

Solve the equation.

$\displaystyle{ {4m+6}={34} }$

10.

Solve the equation.

$\displaystyle{ {9n+5}={23} }$

11.

Solve the equation.

$\displaystyle{ {6q-3}={-9} }$

12.

Solve the equation.

$\displaystyle{ {3x-1}={23} }$

13.

Solve the equation.

$\displaystyle{ {-66} = {9r+6} }$

14.

Solve the equation.

$\displaystyle{ {-20} = {6t+4} }$

15.

Solve the equation.

$\displaystyle{ {-26} = {3b-2} }$

16.

Solve the equation.

$\displaystyle{ {8} = {9A-1} }$

17.

Solve the equation.

$\displaystyle{ {-6B+8}={20} }$

18.

Solve the equation.

$\displaystyle{ {-10m+5}={75} }$

19.

Solve the equation.

$\displaystyle{ {-4n-3}={-27} }$

20.

Solve the equation.

$\displaystyle{ {-7q-10}={32} }$

21.

Solve the equation.

$\displaystyle{ {17} = {-x+7} }$

22.

Solve the equation.

$\displaystyle{ {5} = {-r+1} }$

23.

Solve the equation.

$\displaystyle{ {5t+30} = {0} }$

24.

Solve the equation.

$\displaystyle{ {2b+4} = {0} }$

Application Problems for Solving Two-Step Equations

25.

A gym charges members ${\$30}$ for a registration fee, and then ${\$34}$ per month. You became a member some time ago, and now you have paid a total of ${\$438}$ to the gym. How many months have passed since you joined the gym?

months have passed since you joined the gym.

26.

Your cell phone company charges a ${\$22}$ monthly fee, plus ${\$0.13}$ per minute of talk time. One month your cell phone bill was ${\$80.50}\text{.}$ How many minutes did you spend talking on the phone that month?

You spent talking on the phone that month.

27.

A school purchased a batch of T-shirts from a company. The company charged ${\$8}$ per T-shirt, and gave the school a ${\$100}$ rebate. If the school had a net expense of ${\$2{,}540}$ from the purchase, how many T-shirts did the school buy?

The school purchased T-shirts.

28.

Barbara hired a face-painter for a birthday party. The painter charged a flat fee of ${\$95}\text{,}$ and then charged ${\$4.50}$ per person. In the end, Barbara paid a total of ${\$189.50}\text{.}$ How many people used the face-painter’s service?

people used the face-painter’s service.

29.

A certain country has $741.81$ million acres of forest. Every year, the country loses $9.39$ million acres of forest mainly due to deforestation for farming purposes. If this situation continues at this pace, how many years later will the country have only $450.72$ million acres of forest left? (Use an equation to solve this problem.)

After years, this country would have $450.72$ million acres of forest left.

30.

Dave has ${\$71}$ in his piggy bank. He plans to purchase some Pokemon cards, which costs ${\$2.35}$ each. He plans to save ${\$54.55}$ to purchase another toy. At most how many Pokemon cards can he purchase?

Write an equation to solve this problem.

Dave can purchase at most Pokemon cards.

Solving Equations with Variable Terms on Both Sides

31.

Solve the equation.

${8r+9} = {r+79}$

32.

Solve the equation.

${7t+4} = {t+46}$

33.

Solve the equation.

$\displaystyle{ {-10b+7} = {-b-29} }$

34.

Solve the equation.

$\displaystyle{ {-6c+2} = {-c-3} }$

35.

Solve the equation.

$\displaystyle{ {3-6B} = {10B+51} }$

36.

Solve the equation.

$\displaystyle{ {9-10m} = {10m+109} }$

37.

Solve the equation.

$\displaystyle{ {8n+10}={3n+4} }$

38.

Solve the equation.

$\displaystyle{ {3q+8}={10q+10} }$

39.

Solve the equation.

$\displaystyle{ {6x+9} = {3x+36} }$
$\displaystyle{ {3n+9} = {6n-21} }$

40.

Solve the equation.

$\displaystyle{ {8r+5} = {5r+14} }$
$\displaystyle{ {5c+5} = {8c-4} }$