ORCCA Solving Multistep Linear Inequalities

Section 2.2 Solving Multistep Linear Inequalities

Objectives: PCC Course Content and Outcome Guide

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.

Figure 2.2.1. Alternative Video Lessons

Subsection 2.2.1 Solving Multistep Inequalities

When solving a linear inequality, we almost follow the exact same steps as we do in [cross-reference to target(s) "algorithm-solve-linear-equation" missing or not unique]. 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 term’s 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 “boundary number”. 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 shouldn’t 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.

We’ll 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{\highlight{\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, it’s 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 $0\text{.}$

$a number line with a mark at -2; a thick line overlays the number line to the left of -2 with an arrow pointing left; there is a right bracket at -2; three arrows point to -10, -2, and 0 on the number line, suggesting that these values will be checked as possible solutions$

\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, it’s 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 $0\text{.}$

$a number line with a mark at -5; a thick line overlays the number line to the right of -5 with an arrow pointing right; there is a left parenthesis at -5; three arrows point to -10, -5, and 0 on the number line, suggesting that these values will be checked as possible solutions$

\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} \amp\amp -2-2(2x+1)\amp\gt4-(3-x)\amp\amp\\ -2-2(2(\substitute{-2})+1)\amp\wonder{\gt}4-(3-(\substitute{-2}))\amp -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(-4+1)\amp\wonder{\gt}4-5\amp -2-2(-2+1)\amp\wonder{\gt}4-4 \amp -2-2(0+1)\amp\wonder{\gt}4-3\\ -2-2(-3)\amp\wonder{\gt}-1\amp -2-2(-1)\amp\wonder{\gt}0 \amp -2-2(1)\amp\wonder{\gt}1\\ -2+6\amp\wonder{\gt}-1\amp -2+2\amp\wonder{\gt}0 \amp -2-2\amp\wonder{\gt}1\\ 4\amp\confirm{\gt}-1\amp 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 of the air surrounding us). 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 equalling 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 town’s population from ten years ago. If the town had grown by $7\%\text{,}$ then it’s 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 town’s 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 town’s 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 Inequualities.

Solve the inquality.

1.

${h+8}\lt{11}$

2.

${16m}\geq{-16}$

3.

${-\frac{t}{2}}\geq{-2}$

4.

${-13y}\gt{-39}$

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.

${2D+9}\geq{5}$

6.

${5J+4}\geq{-26}$

7.

${-1+8P}\geq{15}$

8.

${-7+4W}\geq{-15}$

9.

${-4b+7}\geq{-17}$

10.

${-9h+2}\geq{-16}$

11.

${-3-5m}\leq{12}$

12.

${-9-2s}\leq{-21}$

13.

${5+4y+4}\lt{13}$

14.

${-1+8D+5}\leq{-20}$

15.

${3J+4+2J-7}\lt{22}$

16.

${9P+4+6P+3}\lt{22}$

17.

${3-\left(2W+4\right)}\lt{7}$

18.

${-3-\left(5b+4\right)}\leq{-32}$

19.

${4g-7-8g-2}\lt{-29}$

20.

${7m+3-5m+4}\gt{-1}$

21.

${4s-7-5\mathopen{}\left(2s+2\right)}\lt{-41}$

22.

${4y+3-2\mathopen{}\left(-5y+2\right)}\gt{-15}$

23.

${9D+2}\lt{2D-33}$

24.

${9J+1}\leq{4J+16}$

25.

${-4P+1}\lt{-2P+3}$

26.

${-7V+1}\leq{6V+79}$

27.

${-6b+1}\lt{8b-41}$

28.

${3g+3}\lt{4g+5}$

29.

${-7-9\mathopen{}\left(-4m-5\right)}\geq{-194-\left(2m-4\right)}$

30.

${4-\left(-9s-4\right)}\lt{35-\left(2s+5\right)}$

31.

${7x-2+5x-8}\lt{-3x+2-8x-58}$

32.

${-5D+2+2D-2}\leq{-2D+6+5D-42}$