Section 2.2 Solving Multistep Linear Inequalities
We solved inequalities in Section 1.5 where only one step was needed to isolate the variable. Now we will work with inequalities that need more than one step.
Subsection 2.2.1 Solving Multistep Inequalities
When solving a linear inequality, we almost follow the exact same steps as we do in Process 2.1.5. One difference is that when we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol must switch. The other difference is that checking a solution set takes more effort.
Process 2.2.2. Steps to Solve Linear Inequalities.
- Simplify
- Simplify the expressions on each side of the inequality by distributing and combining like terms.
- Separate
- Use addition or subtraction to separate the terms so that the variable terms are on one side of the inequality and the constant terms are on the other side of the inequality.
- Clear the Coefficient
- Use multiplication or division to eliminate the variable termâs coefficient. If you multiply or divide each side by a negative number, switch the direction of the inequality symbol.
- Check
- A solution to a linear inequality has a âboundary numberâ. Using the original inequality, check (1) a number less than the boundary number, (2) the boundary number itself, and (3) a number greater than the boundary number to confirm what should and shouldnât be solutions are all working as expected. (This can take time, so use your judgment about when you might get away with skipping this checking.)
- Summarize
- State the solution set. Or in the case of an application problem, summarize the result in a complete sentence using appropriate units.
Example 2.2.3.
Solve for \(t\) in the inequality \(-3t+5\geq11\text{.}\) Write the solution set in both set-builder notation and interval notation.
Explanation.
Weâll handle this much like we would handle an equation, at least for the first step.
\begin{align*}
-3t+5\amp\geq11\\
-3t+5\subtractright{5}\amp\geq11\subtractright{5}\\
-3t\amp\geq6\\
\divideunder{-3t}{-3}\amp\mathbin{\secondhighlight{\le}}\divideunder{6}{-3}\\
t\amp\leq-2
\end{align*}
Note that when we divided both sides of the inequality by \(-3\text{,}\) we had to switch the direction of the inequality symbol. At this point we think that the solution set in set-builder notation is \(\{t\mid t\leq-2\}\text{,}\) and the solution set in interval notation is \((-\infty,-2]\text{.}\)
Since there are infinitely many solutions, itâs impossible to literally check them all. We believe that all values of \(t\) for which \(t\leq-2\) are solutions. We check that one number less than \(-2\) (any number, your choice) satisfies the inequality. And that \(-2\) satisfies the inequality. And that one number greater than \(-2\) (any number, your choice) does not satisfy the inequality. We choose to check the values \(-10\text{,}\) \(-2\text{,}\) and \(0\text{.}\)
\begin{align*}
\amp\amp -3t+5\amp\ge11\amp \amp\\
-3(\substitute{-10})+5\amp\wonder{\geq}11\amp -3(\substitute{-2})+5\amp\wonder{\geq}11\amp -3(\substitute{0})+5\amp\wonder{\geq}11\\
30+5\amp\wonder{\geq}11\amp 6+5\amp\wonder{\geq}11\amp 0+5\amp\wonder{\geq}11\\
35\amp\confirm{\geq}11\amp 11\amp\confirm{\geq}11\amp 5\amp\reject{\geq}11
\end{align*}
So both \(-10\) and \(-2\) are solutions as expected, while \(0\) is not. This is evidence that our solution set is correct. Making these checks would help us catch an error if we had made one. While it certainly does take time and space to make three checks like this, it has its value.
Example 2.2.4.
Solve for \(z\) in the inequality \((6z+5)-(2z-3)\gt-12\text{.}\) Write the solution set in both set-builder notation and interval notation.
Explanation.
Here, our first step will be simplifying the left side.
\begin{align*}
(6z+5)-(2z-3)\amp\gt-12\\
6z+5-2z+3\amp\gt-12\\
4z+8\amp\gt-12\\
4z+8\subtractright{8}\amp\gt-12\subtractright{8}\\
4z\amp\gt-20\\
\divideunder{4z}{4}\amp\gt\divideunder{-20}{4}\\
z\amp\gt -5
\end{align*}
Note that we divided both sides of the inequality by \(4\) and since this is a positive number we did not need to switch the direction of the inequality symbol. At this point we think that the solution set in set-builder notation is \(\{z\mid z\gt-5\}\text{,}\) and the solution set in interval notation is \((-5,\infty)\text{.}\)
Since there are infinitely many solutions, itâs impossible to literally check them all. We believe that all values of \(z\) for which \(z\gt-5\) are solutions. We check that one number less than \(-5\) (any number, your choice) does not satisfy the inequality. And that \(-5\) does not satisfy the inequality. And that one number greater than \(-5\) (any number, your choice) does satisfy the inequality. We choose to check the values \(-10\text{,}\) \(-5\text{,}\) and \(0\text{.}\)
\begin{align*}
(6(\substitute{-10})+5)-(2(\substitute{-10})-3)\amp\wonder{\gt}-12\\
(-60+5)-(-20-3)\amp\wonder{\gt}-12\\
-55-(-23)\amp\wonder{\gt}-12\\
-32\amp\reject{\gt}-12
\end{align*}
\begin{align*}
(6(\substitute{-5})+5)-(2(\substitute{-5})-3)\amp\wonder{\gt}-12\\
(-30+5)-(-10-3)\amp\wonder{\gt}-12\\
-25-(-13)\amp\wonder{\gt}-12\\
-12\amp\reject{\gt}-12
\end{align*}
\begin{align*}
(6(\substitute{0})+5)-(2(\substitute{0})-3)\amp\wonder{\gt}-12\\
(0+5)-(0-3)\amp\wonder{\gt}-12\\
5-(-3)\amp\wonder{\gt}-12\\
8\amp\confirm{\gt}-12
\end{align*}
So both \(-10\) and \(-5\) are not solutions as expected, while \(0\) is a solution. This is evidence that our solution set is correct. The solution set in set-builder notation is \(\{z\mid z\gt-5\}\text{.}\) The solution set in interval notation is \((-5,\infty)\text{.}\)
Checkpoint 2.2.5.
Solve the inequality \(-2-2(2x+1)\gt4-(3-x)\text{.}\) Graph the solution set on a number line. State the solution set using both interval notation and set-builder notation.
Explanation.
We start by simplifying the two sides.
\begin{equation*}
\begin{aligned}
-2-2(2x+1)\amp\gt4-(3-x)\\
-2-4x-2\amp\gt4-3+x\\
-4x-4\amp\gt1+x
\end{aligned}
\end{equation*}
And now add/subtract terms to separate the variable terms and the constant terms.
\begin{equation*}
\begin{aligned}
-4x-4\addright{4}\subtractright{x}\amp\gt1+x\addright{4}\subtractright{x}\\
-5x\amp\gt5
\end{aligned}
\end{equation*}
And now divide each side by \(-5\text{.}\) Since we are dividing by a negative number, the inequality sign will change direction.
\begin{equation*}
x\lt-1
\end{equation*}
Graphically, we represent this solution set as:
Using interval notation, we write the solution set as \((-\infty,-1)\text{.}\) Using set-builder notation, we write it as \(\{x \mid x \lt -1\}\text{.}\)
We should check that some number less than \(-11\) is a solution, that \(-1\) itself is not a solution, and that some number greater than \(-1\) is not a solution.
\begin{equation*}
\begin{aligned}
-2-2(2(\substitute{-2})+1)\amp\wonder{\gt}4-(3-(\substitute{-2}))\\
-2-2(-4+1)\amp\wonder{\gt}4-5\\
-2-2(-3)\amp\wonder{\gt}-1\\
-2+6\amp\wonder{\gt}-1\\
4\amp\confirm{\gt}-1
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
-2-2(2(\substitute{-1})+1)\amp\wonder{\gt}4-(3-(\substitute{-1})) \amp -2-2(2(\substitute{0})+1)\amp\wonder{\gt}4-(3-(\substitute{0}))\\
-2-2(-2+1)\amp\wonder{\gt}4-4 \amp -2-2(0+1)\amp\wonder{\gt}4-3\\
-2-2(-1)\amp\wonder{\gt}0 \amp -2-2(1)\amp\wonder{\gt}1\\
-2+2\amp\wonder{\gt}0 \amp -2-2\amp\wonder{\gt}1\\
0\amp\reject{\gt}0 \amp -4\amp\reject{\gt}1
\end{aligned}
\end{equation*}
Everything worked out as expected, so our solution is reasonably checked.
Subsection 2.2.2 Applications
Example 2.2.6. Rate Problem.
When an experiment started, the pressure inside a gas container was \(4.2\) atm (one atm is the standard pressure air at sea level). As the container was heated, the pressure increased by \(0.7\) atm per minute. The maximum pressure the container is rated to handle is \(21.7\) atm. Heating must be stopped once the pressure reaches \(21.7\) atm. Over what time interval was the container in a safe state (meaning the pressure was less than or equal to \(21.7\) atm)?
Explanation.
This is a situation where something had an initial value (the pressure starts at \(4.2\) atm) and then changed at a constant rate (it increased by \(0.7\) atm per minute). So we can use the rate model formula. Except we are not exactly interested in the pressure equaling the final value of \(21.7\) atm. Instead, we are asked about when the pressure was less than or equal to \(21.7\) atm. So we have the inequality:
\begin{align*}
0.7t+4.2\amp\leq21.7\\
0.7t+4.2\subtractright{4.2}\amp\leq21.7\subtractright{4.2}\\
0.7t\amp\leq17.5\\
\divideunder{0.7t}{0.7}\amp\leq\divideunder{17.5}{0.7}\\
t\amp\leq25
\end{align*}
In summary, the container was safe as long as \(t\leq25\text{.}\) Assuming that the time \(t\) also must be greater than or equal to zero, this means \(0\leq t\leq 25\text{.}\) We can write this as the time interval as \([0,25]\text{.}\) Thus the container was safe between \(0\) minutes and \(25\) minutes.
Example 2.2.7. Percent Problem.
The population of a certain country grew by \(7\%\) over the course of the past decade. One town in this country grew in population too, but not as fast as the country did overall. Its current population is \(22{,}341\text{.}\) What might its population have been ten years ago?
Explanation.
Let \(x\) be the townâs population from ten years ago. If the town had grown by \(7\%\text{,}\) then itâs population would be \(x + 0.07x\text{.}\) But since it actually grew less quickly than \(7\%\) per decade, \(x + 0.07x\) would work out to more than the townâs current population of \(22{,}341\text{.}\) So we have the inequality:
\begin{align*}
x + 0.07x\amp\gt22341\\
1.07x\amp\gt22341\\
\divideunder{1.07x}{1.07}\amp\gt\divideunder{22341}{1.07}\\
x\amp\gt20879.4\ldots
\end{align*}
So the townâs population from ten years ago was at least \(20880\text{.}\)
Reading Questions 2.2.3 Reading Questions
1.
When solving an inequality, what are the conditions when you have to reverse the direction of the inequality symbol?
2.
How is the solution set to a linear inequality different from the solution set to a linear equation?
3.
If you want to check your solution set to a linear inequality, what exactly are you going to do?
Exercises 2.2.4 Exercises
Review and Warmup
One-Step Inequalities.
Solve the inequality.
1.
\({D+16}\gt{13}\)
2.
\({5J}\lt{25}\)
3.
\({-\frac{P}{4}}\leq{-2}\)
4.
\({-2W}\leq{8}\)
Skills Practice
Solve the Inequality.
Solve the inequality. Graph the solution set, and write the solution set using both interval notation and set-builder notation.
5.
\({5b-2}\leq{23}\)
6.
\({2g+3}\lt{13}\)
7.
\({6+4m}\leq{-10}\)
8.
\({1+7s}\leq{29}\)
9.
\({-9y-4}\leq{5}\)
10.
\({-5D+9}\leq{34}\)
11.
\({4-2J}\leq{-2}\)
12.
\({-1-7P}\leq{6}\)
13.
\({-7+8V+2}\leq{-53}\)
14.
\({7+3b+1}\lt{17}\)
15.
\({5g+1+6g+3}\lt{-18}\)
16.
\({4m+1+m-7}\gt{-36}\)
17.
\({-9-\left(5s+1\right)}\leq{-20}\)
18.
\({5-\left(8x-1\right)}\lt{22}\)
19.
\({2D-1-5D+3}\leq{-16}\)
20.
\({-6I-1+5I-7}\leq{-10}\)
21.
\({-3P+3-2\mathopen{}\left(9P+5\right)}\gt{56}\)
22.
\({-3V-7-4\mathopen{}\left(4V+6\right)}\leq{-145}\)
23.
\({9b-2}\leq{-4b+11}\)
24.
\({-4g-2}\lt{-8g-14}\)
25.
\({6l-2}\leq{7l-7}\)
26.
\({-6s-2}\lt{3s-11}\)
27.
\({3x-3}\leq{-5x-35}\)
28.
\({-9D-3}\lt{-8D-8}\)
29.
\({-3+4\mathopen{}\left(-4I+3\right)}\lt{29-\left(3I+7\right)}\)
30.
\({-8+7\mathopen{}\left(-2P+2\right)}\lt{91-\left(-5P+9\right)}\)
31.
\({-5V+3-7V-4}\lt{-3V-3+4V-50}\)
32.
\({8a+6-5a+4}\lt{-8a-3+4a+34}\)
Applications
33.
You hail a taxi and can only pay with cash. This cab service charges a flat fee of \({\$15.00}\) and then charges \({\$2.00}\) per mile. No tip is expected. You are carrying a total of \({\$43.00}\) in cash with you. You want to know how many miles you can afford.
(a)
Write an inequality to represent this situation, using \(x\) to represent how many miles you can afford.
(b)
Solve this inequality. At most how many miles can you afford?
(c)
Use interval notation to express the number of miles you might ride that day.
34.
Dayanara is driving on the highway, and presently has 18 gal of gasoline in her tank. Her car, under ideal conditions, uses gas at a rate of 0.046 gal/mi. When the tank reaches only one gallon of gas, the low gas light will turn on and Dayanara will start looking for a gas station. How far will she drive before this happens?
(a)
Write an inequality to represent this situation, using \(x\) to represent how many miles Dayanara might drive before the low gas light turns on.
(b)
Solve this inequality. At most how far will Dayanara drive before the low gas light turns on?
(c)
Use interval notation to express the number of miles Dayanara might drive before the low gas light turns on.
35.
Water straight out of the tap is \(47\) degrees Fahrenheit, and you fill a tea kettle. You place the tea kettle on an old stove burner and turn on the burner, but the heating element is old and not working well. At best, it begins raising the temperature inside the kettle at a rate of \(24\) degrees Fahrenheit per minute. How long will it take to boil the water? (Water boils at \(212\) degrees Fahrenheit.)
(a)
Write an inequality to represent this situation, using \(x\) to represent how many minutes it will take for the water to boil.
(b)
Solve this inequality. At best, how long will it take for the water to boil?
(c)
Use interval notation to express the number of minutes it might take for the water to boil.
36.
Isiah is a runner, and set out this morning on a jog with a goal to reach a nearby summit and return home in under one hour. He jogged a steady uphill path to the top of the nearby summit, and it took 32 min to get there. Now jogging home, he will run a distance of 2.2 mi. What will his rate (in minutes per mile) need to be in order to make the goal of returning home in under 60 minutes?
(a)
Write an inequality to represent this situation, using \(x\) to represent Isiahâs jogging rate in minutes per mile.
(b)
Solve this inequality. Isiahâs jogging rate needs to beat what rate?
(c)
Use interval notation to express what Isiahâs jogging rate might be.
37.
The population of a certain country grew by \(8\%\) over the course of the past decade. One city in this country grew in population too, but at an even faster rate than the country grew overall. The cityâs current population is \(94485\text{.}\) What might its population have been ten years ago? (Note: the population ten years ago is known to have been at least \(10{,}000\text{.}\))
(a)
Write an inequality to represent this situation, using \(x\) to represent the cityâs population from ten years ago.
(b)
Solve this inequality. Whatâs the most the population could have been ten years ago?
(c)
Use interval notation to express what the cityâs population could have been ten years ago.
38.
The bill at a restaurant came and Marshall offered to pay for the whole table. We are not sure what the bill was, but it was at least \(\$50\text{.}\) Marshall added a tip, and the total came to \(\$105\text{.}\) We are sure that he added at least \(18\%\) tip. What might the original bill total have been before tip?
(a)
Write an inequality to represent this situation, using \(x\) to represent the bill total before tip was added.
(b)
Solve this inequality. Whatâs the most the bill could have been before tip?
(c)
Use interval notation to express what the bill could have been before tip was added.
