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## Section11.2Compound Inequalities

On the newest version of the SAT (an exam that often qualifies students for colleges) the minimum score that you can earn is $400$ and the maximum score that you can earn is $1600\text{.}$ This means that only numbers between $400$ and $1600\text{,}$ including these endpoints, are possible scores. To plot all of these values on a number line would look something like: Figure 11.2.1 Possible SAT Scores

Going back to the original statement, “the minimum score that you can earn is $400$ and the maximum score that you can earn is $1600\text{,}$” this really says two things. First, it says that $(\text{a SAT score})\geq400\text{,}$ and second, that $(\text{a SAT score})\le1600\text{.}$ When we combine two inequalities like this into a single problem, it becomes a compound inequality.

Our lives are often constrained by the compound inequalities of reality: you need to buy enough materials to complete your project, but you can only fit so much into your vehicle; you would like to finish your degree early, but only have so much money and time to put toward your courses; you would like a vegetable garden big enough to supply you with veggies all summer long, but your yard or balcony only gets so much sun. In the rest of the section we hope to illuminate how to think mathematically about problems like these.

Before continuing, a review on how notation for intervals works may be useful, and you may benefit from revisiting Section 1.6. Then a refresher on solving linear inequalities may also benefit you, which you can revisit in Section 3.2 and Section 3.3.

### Subsection11.2.1Unions of Intervals

###### Definition11.2.3

The union of two sets, $A$ and $B\text{,}$ is the set of all elements contained in either $A$ or $B$ (or both). We write $A\cup B$ to indicate the union of the two sets.

In other words, the union of two sets is what you get if you toss every number in both sets into a bigger set.

###### Example11.2.4

The union of sets $\{1,2,3,4\}$ and $\{3,4,5,6\}$ is the set of all elements from either set. So $\{\highlight{1,2,3,4}\}\cup\{3,4,\highlight{5,6}\}=\{\highlight{1,2,3,4,5,6}\}\text{.}$ Note that we don't write duplicates.

###### Example11.2.5

Let's visualize the union of the sets $(-\infty,4)$ and $[7,\infty)\text{.}$ First we make a number line with both intervals drawn to understand what both sets mean. Figure 11.2.6 A number line sketch of $(-\infty,4)$ as well as $[7,\infty)$

The two intervals should be viewed as a single object when stating the union, so here is the picture of the union. It looks the same, but now it is a graph of a single set. Figure 11.2.7 A number line sketch of $(-\infty,4)\cup[7,\infty)$

### Subsection11.2.2“Or” Compound Inequalities

###### Definition11.2.8

A compound inequality is a grouping of two or more inequalities into a larger inequality statement. These usually come in two flavors: “or” and “and” inequalities. For an example of an “or” compound inequality, you might get a discount at the movie theater if your age is less than $13$ or greater than $64\text{.}$ In general, compound inequalities of the “and” variety currently are beyond the scope of this book. However a special type of the “and” variety is covered later in Subsection 11.2.3.

In math, the technical term or means “either or both.” So, mathematically, if we asked if you would like “chocolate cake or apple pie” for dessert, your choices are either “chocolate cake,” “apple pie,” or “both chocolate cake and apple pie.” This is slightly different than the English “or” which usually means “one or the other but not both.”

“Or” shows up in math between equations (as in when solving a quadratic equation, you might end up with “$x=2$ or $x=-3$”) or between inequalities (which is what we're about to discuss).

###### Remark11.2.9

The definition of “or” is very close to the definition of a union where you combine elements from either or both sets together. In fact, when you have an “or” between inequalities in a compound inequality, to find the solution set of the compound inequality, you find the union of the the solutions sets of each of the pieces.

###### Example11.2.10

Solve the compound inequality.

\begin{equation*} x\le 1\text{ or } x\gt 4 \end{equation*}
Explanation

Writing the solution set to this compound inequality doesn't require any algebra beforehand because each of the inequalities is already solved for $x\text{.}$ The first thing we should do is understand what each inequality is saying using a graph. An “or” statement becomes a union of solution sets, so the solution set to the compound inequality must be:

\begin{equation*} (-\infty,1]\cup(4,\infty)\text{.} \end{equation*}
###### Example11.2.12

Solve the compound inequality.

\begin{equation*} 3-5x\gt-7\text{ or } 2-x\le-3 \end{equation*}
Explanation

First we need to do some algebra to isolate $x$ in each piece. Note that we are going to do algebra on both pieces simultaneously. Also note that the mathematical symbol “or” should be written on each line.

\begin{align*} 3-5x\amp\gt-7\amp\text{or}\amp\amp 2-x\amp\le-3\\ 3-5x\subtractright{3}\amp\gt-7\subtractright{3}\amp\text{or}\amp\amp 2-x\subtractright{2}\amp\le-3\subtractright{2}\\ -5x\amp\gt-10\amp\text{or}\amp\amp -x\amp\le-5\\ \divideunder{-5x}{-5}\amp\mathbin{\highlight{\lt}}\divideunder{-10}{-5}\amp\text{or}\amp\amp \divideunder{-x}{-1}\amp\mathbin{\highlight{\geq}}\divideunder{-5}{-1}\\ x\amp\lt2\amp\text{or}\amp\amp x\amp\geq5 \end{align*}

The solution set for the compound inequality $x\lt2$ is $(-\infty,2)$ and the solution set to $x\ge5$ is $[5,\infty)\text{.}$ To do the “or” portion of the problem, we need to take the union of these two sets. Let's first make a graph of the solution sets to visualize the problem. Figure 11.2.13 A number line sketch of $(-\infty,2)$ as well as $[5,\infty)$

The union combines both solution sets into one, and so

\begin{equation*} (-\infty,2)\cup[5,\infty) \end{equation*}

We have finished the problem, but for the sake of completeness, let's try to verify that our answer is reasonable.

• First, let's choose a number that is not in our proposed solution set. We will arbitrarily choose $\highlight{3}\text{.}$

\begin{align*} 3-5x\amp\gt-7\amp\text{or}\amp\amp 2-x\amp\le-3\\ 3-5(\highlight{3})\amp\stackrel{?}{\gt}-7\amp\text{or}\amp\amp 2-(\highlight{3})\amp\stackrel{?}{\le}-3\\ -9\amp\stackrel{\text{no}}{\gt}-7\amp\text{or}\amp\amp -1\amp\stackrel{\text{no}}{\le}-3 \end{align*}

This value made both inequalities false which is why $3$ isn't in our solution set.

• Next, let's choose a number that is in our solution region. We will arbitrarily choose $\highlight{1}\text{.}$

\begin{align*} 3-5x\amp\gt-7\amp\text{or}\amp\amp 2-x\amp\le-3\\ 3-5(\highlight{1})\amp\stackrel{?}{\gt}-7\amp\text{or}\amp\amp 2-(\highlight{1})\amp\stackrel{?}{\le}-3\\ -12\amp\stackrel{\checkmark}{\lt}-7\amp\text{or}\amp\amp -1\amp\stackrel{\text{no}}{\leq}-3 \end{align*}

This value made one of the inequalities true. Since this is an “or” statement, only one or the other piece has to be true to make the compound inequality true.

• Last, what will happen if we choose a value that was in the other solution region in Figure 11.2.13, like the number $\highlight{6}\text{?}$

\begin{align*} 3-5x\amp\gt-7\amp\text{or}\amp\amp 2-x\amp\le-3\\ 3-5(\highlight{6})\amp\stackrel{?}{\gt}-7\amp\text{or}\amp\amp 2-(\highlight{6})\amp\stackrel{?}{\le}-3\\ -27\amp\stackrel{\text{no}}{\gt}-7\amp\text{or}\amp\amp -4\amp\stackrel{\checkmark}{\le}-3 \end{align*}

This solution made the other inequality piece true.

This completes the check. Numbers from within the solution region make the compound inequality true and numbers outside the solution region make the compound inequality false.

###### Example11.2.14

Solve the compound inequality.

\begin{equation*} \frac{3}{4}t+2\le \frac{5}{2}\text{ or } -\frac{1}{2}(t-3)\lt -2 \end{equation*}
Explanation

First we will solve each inequality for $t\text{.}$ Recall that we usually try to clear denominators by multiplying both sides by the least common denominator.

\begin{align*} \frac{3}{4}t+2\amp\le \frac{5}{2}\amp\text{or}\amp\amp -\frac{1}{2}(t-3)\amp\lt -2\\ \multiplyleft{4}\left(\frac{3}{4}t+2\right)\amp\le \multiplyleft{4}\frac{5}{2}\amp\text{or}\amp\amp \multiplyleft{2}\left(-\frac{1}{2}(t-3)\right)\amp\lt \multiplyleft{2}(-2)\\ 3t+8\amp\le 10\amp\text{or}\amp\amp -t+3\amp\lt -4\\ 3t+8\subtractright{8}\amp\le 10\subtractright{8}\amp\text{or}\amp\amp -t+3\subtractright{3}\amp\lt -4\subtractright{3}\\ 3t\amp\le 2\amp\text{or}\amp\amp -t\amp\lt -7\\ \divideunder{3t}{3}\amp\le \divideunder{2}{3}\amp\text{or}\amp\amp \divideunder{-t}{-1}\amp\mathbin{\highlight{\gt}} \divideunder{-7}{-1}\\ t\amp\le\frac{2}{3}\amp\text{or}\amp\amp t\amp\gt 7 \end{align*}

The solution set to $t\le \frac{2}{3}$ is $\left(-\infty,\frac{2}{3}\right]$ and the solution set to $t\gt 7$ is $(7,\infty)\text{.}$ Figure 11.2.15 shows these two sets. Figure 11.2.15 A number line sketch of $\left(-\infty,\frac{2}{3}\right]$ and also $(7,\infty)$

Note that the two sets do not overlap so there will be no way to simplify the union. Thus the solution set to the compound inequality is:

\begin{equation*} \left(-\infty,\frac{2}{3}\right]\cup(7,\infty) \end{equation*}
###### Example11.2.16

Solve the compound inequality.

\begin{equation*} 3y-15\gt 6 \text{ or } 7-4y\ge y-3 \end{equation*}
Explanation

First we solve each inequality for $y\text{.}$

\begin{align*} 3y-15\amp\gt 6\amp\text{or}\amp\amp 7-4y\amp\ge y-3\\ 3y-15\addright{15}\amp\gt 6\addright{15} \amp\text{or}\amp\amp 7-4y\subtractright{7}\subtractright{y}\amp\ge y-3\subtractright{7}\subtractright{y}\\ 3y\amp\gt 21\amp\text{or}\amp\amp -5y\amp\ge -10\\ \divideunder{3y}{3}\amp\gt \divideunder{21}{3}\amp\text{or}\amp\amp \divideunder{-5y}{-5}\amp\mathbin{\highlight{\le}} \divideunder{-10}{-5}\\ y\amp\gt 7\amp\text{or}\amp\amp y\amp\le 2 \end{align*}

The solution set to $y\gt 7$ is $(7,\infty)$ and the solution set to $y\le 2$ is $(-\infty,2]\text{.}$ Figure 11.2.17 shows these two sets. Figure 11.2.17 A number line sketch of $(7,\infty)$ as well as $(-\infty,2]$

So the solution set to the compound inequality is:

\begin{equation*} (-\infty,2]\cup(7,\infty) \end{equation*}

### Subsection11.2.3Three-Part Inequalities

The inequality $1\leq 2\lt 3$ says a lot more than you might think. It actually says four different single inequalities which are highlighted for you to see.

\begin{equation*} \highlight{1}\mathbin{\highlight{\leq}}\highlight{2}\mathbin{\lowlight{\lt}}\lowlight{3}\quad \highlight{1}\mathbin{\highlight{\leq}}\lowlight{2}\mathbin{\lowlight{\lt}}\highlight{3}\quad \highlight{1}\mathbin{\lowlight{\leq}}\lowlight{2}\mathbin{\highlight{\lt}}\highlight{3}\quad \lowlight{1}\mathbin{\lowlight{\leq}}\highlight{2}\mathbin{\highlight{\lt}}\highlight{3} \end{equation*}

This might seem trivial at first, but if you are presented with an inequality like $-1\lt3\geq2\text{,}$ at first it might look sensible; however, in reality, you need to check that all four linear inequalities make sense. Those are highlighted here.

\begin{equation*} \highlight{-1}\mathbin{\highlight{\lt}}\highlight{3}\mathbin{\lowlight{\geq}}\lowlight{2}\quad \highlight{-1}\mathbin{\highlight{\lt}}\lowlight{3}\mathbin{\lowlight{\geq}}\highlight{2}\quad \highlight{-1}\mathbin{\lowlight{\lt}}\lowlight{3}\mathbin{\highlight{\geq}}\highlight{2}\quad \lowlight{-1}\mathbin{\lowlight{\lt}}\highlight{3}\mathbin{\highlight{\geq}}\highlight{2} \end{equation*}

One of these inequalities is false: $-1\ngeq2\text{.}$ This implies that the entire original inequality, $-1\lt3\geq2\text{,}$ is nonsense.

###### Example11.2.18

Decide whether or not the following inequalities are true or false.

1. True or False: $-5\lt7\le12\text{?}$

2. True or False: $-7\le-10\lt4\text{?}$

3. True or False: $-2\le0\geq1\text{?}$

4. True or False: $5\gt-3\geq-9\text{?}$

5. True or False: $3\lt3\le5\text{?}$

6. True or False: $9\gt1\lt5\text{?}$

7. True or False: $3\lt8\le-2\text{?}$

8. True or False: $-9\lt-4\le-2\text{?}$

Explanation

We need to go through all four single inequalities for each. If the inequality is false, for simplicity's sake, we will only highlight the one single inequality that makes the inequality false.

1. True: $-5\lt7\le12\text{.}$

2. False: $\highlight{-7\stackrel{\text{no}}{\leq}-10}\mathbin{\lowlight{\lt}}\lowlight{4}\text{.}$

3. False: $\highlight{-2}\mathbin{\lowlight{\le}}\lowlight{0}\mathbin{\highlight{\stackrel{\text{no}}{\geq}}}\highlight{1}\text{.}$

4. True: $5\gt-3\geq-9\text{.}$

5. False: $\highlight{3\stackrel{\text{no}}{\lt}3}\mathbin{\lowlight{\le}}\lowlight{5}\text{.}$

6. False: $\highlight{9}\mathbin{\lowlight{\gt}}\lowlight{1}\mathbin{\highlight{\stackrel{\text{no}}{\lt}}}\highlight{5}\text{.}$

7. False: $\lowlight{3}\mathbin{\lowlight{\lt}}\highlight{8\stackrel{\text{no}}{\leq}-2}\text{.}$

8. True: $-9\lt-4\le-2\text{.}$

As a general hint, no (nontrivial) three-part inequality can ever be true if the inequality signs are not pointing in the same direction. So no matter what numbers $a\text{,}$ $b\text{,}$ and $c$ are, both $a\lt b\geq c$ and $a\geq b \lt c$ cannot be true! Soon you will be writing inequalities like $2\lt x \le 4$ and you need to be sure to check that your answer is feasible. You will know that if you get $2\gt x \le 4$ or $2\lt x \geq 4$ that something went wrong in the solving process. The only exception is that something like $1\le1\geq1$ is true because $1=1=1\text{,}$ although this shouldn't come up very often!

###### Example11.2.19

Write the solution set to the compound inequality.

\begin{equation*} -7\lt x\le 5 \end{equation*}
Explanation

The solutions to the three-part inequality $-7\lt x\le 5$ are those numbers that are trapped between $-7$ and $5\text{,}$ including $5$ but not $-7\text{.}$ Keep in mind that there are infinitely many decimal numbers and irrational numbers that satisfy this inequality like $-2.781828$ and $\pi\text{.}$ We will write these numbers in interval notation as $(-7,5]$ or in set builder notation as $\{x\mid -7\lt x\le 5\}\text{.}$

###### Example11.2.20

Solve the compound inequality.

\begin{equation*} 4\le 9x+13\lt 20 \end{equation*}
Explanation

This is a three-part inequality which we can treat just as a regular inequality with three “sides.” The goal is to isolate $x$ in the middle and whatever you do to one “side,” you have to do to the other two “sides.”

The solutions to the three-part inequality $-1\le x\lt \frac{7}{9}$ are those numbers that are trapped between $-1$ and $\frac{7}{9}\text{,}$ including $-1$ but not $\frac{7}{9}\text{.}$ The solution set in interval notation is $\left[-1,\frac{7}{9}\right)\text{.}$

\begin{align*} 4\amp\le 9x+13\lt 20\\ 4\subtractright{13}\amp\le 9x+13\subtractright{13}\lt 20\subtractright{13}\\ -9\amp\le 9x\lt 7\\ \divideunder{-9}{9}\amp\le \divideunder{9x}{9}\lt \divideunder{7}{9}\\ -1\amp\le x\lt \frac{7}{9} \end{align*}
###### Example11.2.21

Solve the compound inequality.

\begin{equation*} -13\lt 7-\frac{4}{3}x\le 15 \end{equation*}
Explanation

This is a three-part inequality which we can treat just as a regular inequality with three “sides.” The goal is to isolate $x$ in the middle and whatever you do to one “side,” you have to do to the other two “sides.” We will begin by canceling the fraction by multiplying each part by the least common denominator.

At the end we reverse the entire statement to go from smallest to largest. The solution set is $[-6,15)\text{.}$

\begin{align*} -13\amp\lt 7-\frac{4}{3}x\le15\\ -13\multiplyright{3}\amp\lt \left(7-\frac{4}{3}x\right)\multiplyright{3}\le15\multiplyright{3}\\ -39\amp\lt 21-4x\le45\\ -39\subtractright{21}\amp\lt 21-4x\subtractright{21}\le45\subtractright{21}\\ -60\amp\lt -4x\le24\\ \divideunder{-60}{-4}\amp\mathbin{\highlight{\gt}} \divideunder{-4x}{-4}\mathbin{\highlight{\ge}}\divideunder{24}{-4}\\ 15\amp\gt x\ge-6\\ -6\amp\le x\lt 15 \end{align*}

### Subsection11.2.4Solving Compound Inequalities Graphically

So far we have focused on solving inequalities algebraically. Next, we will describe how to solve compound inequalities graphically.

###### Example11.2.22

Figure 11.2.23 shows a graph of $y=f(x)\text{.}$ Use the graph to solve the inequality $2\le f(x) \lt 6\text{.}$ To solve the inequality $2\le f(x) \lt 6$ means to find the $x$-values that give function values between $2$ and $6\text{,}$ not including $6\text{.}$ We draw the horizontal lines $y=2$ and $y=6\text{.}$ Then we look for the points of intersection and find their $x$-values. We see that when $x$ is between $-4$ and $4\text{,}$ not including $-4\text{,}$ the inequality will be true.

We have drawn the interval $(-4,4]$ along the $x$-axis, which is the solution set. Figure 11.2.24 Graph of $y=f(x)$ and the solution set to $2\le f(x) \lt 6$
###### Example11.2.25

Figure 11.2.26 shows a graph of $y=g(x)\text{.}$ Use the graph to solve the inequality $-4\lt g(x) \le 3\text{.}$ Figure 11.2.26 Graph of $y=g(x)$
Explanation

To solve $-4\lt g(x) \le 3\text{,}$ we first draw the horizontal lines $y=-4$ and $y=3\text{.}$ To solve this inequality we notice that there are two pieces of the function $g$ that are trapped between the $y$-values $-4$ and $3\text{.}$

The solution set is the compound inequality $(-2.1,0.7)\cup(2.4,3.2]\text{.}$ Figure 11.2.27 Graph of $y=g(x)$ and solution set to $-4\lt g(x) \le 3$

### Subsection11.2.5Applications of Compound inequalities

###### Example11.2.28

Raphael's friend is getting married and he's decided to give them some dishes from their registry. Raphael doesn't want to seem cheap but isn't a wealthy man either, so he wants to buy “enough” but not “too many.” He's decided that he definitely wants to spend at least $\150$ on his friend, but less than $\250\text{.}$ Each dish is $\21.70$ and shipping on an order of any size is going to be $\19.99\text{.}$ Given his budget, set up and algebraically solve a compound inequality to find out what his different options are for the number of dishes that he can buy.

Explanation

First, we should define our variable. Let $x$ represent the number of dishes that Raphael can afford. Next we should write a compound inequality that describes this situation. In this case, Raphael wants to spend between $\150$ and $\250$ and, since he's buying $x$ dishes, the price that he will pay is $21.70x+19.99\text{.}$ All of this translates to a triple inequality

\begin{equation*} 150 \lt 21.70x+19.99 \lt 250 \end{equation*}

Now we have to solve this inequality in the usual way.

\begin{align*} 150 \amp\lt 21.70x+19.99 \lt 250\\ 150\subtractright{19.99} \amp\lt 21.70x+19.99\subtractright{19.99} \lt 250\subtractright{19.99}\\ 130.01 \amp\lt 21.70x \lt 230.01\\ \divideunder{130.01}{21.70} \amp\lt \divideunder{21.70x}{21.70} \lt \divideunder{230.01}{21.70}\\ 5.991 \amp\lt x \lt 10.6\amp\text{(note: these values are approximate)} \end{align*}

The interpretation of this inequality is a little tricky. Remember that $x$ represents the number of dishes Raphael can afford. Since you cannot buy $5.991$ dishes (manufacturers will typically only ship whole number amounts of tableware) his minimum purchase must be $6$ dishes. We have a similar problem with his maximum purchase: clearly he cannot buy $10.6$ dishes. So, should we round up or down? If we rounded up, that would be $11$ dishes and that would cost $\21.70\cdot\highlight{11}+\19.99=\258.69\text{,}$ which is outside his price range. Therefore, we should actually round down in this case.

In conclusion, Raphael should buy somewhere between 6 and 10 dishes for his friend to stay within his budget.

###### Example11.2.29

Oak Ridge National Laboratory, a renowned scientific research facility, compiled some data in table 4.28 1 cta.ornl.gov/data/chapter4.shtml on fuel efficiency of a mid-size hybrid car versus the speed that the car was driven. A model for the fuel efficiency $e(x)$ (in miles per gallon, mpg) at a speed $x$ (in miles per hour, mph) is $e(x)=88-0.7x\text{.}$

1. Evaluate and interpret $e(60)$ in the context of the problem.

2. Note that this model only applies between certain speeds. The maximum fuel efficiency for which this formula applies is 55 mpg and the minimum fuel efficiency for which it applies is 33 mpg. Set up and algebraically solve a compound inequality to find the range of speeds for which this model applies.

Explanation
1. Let's evaluate $e(60)$ first.

\begin{align*} e(x)\amp=88-0.7x\\ e(\highlight{60})\amp=88-0.7(\highlight{60})\\ \amp=46 \end{align*}

So, when the hybrid car travels at a speed of 60 mph, it has a fuel efficiency of 46 mpg.

2. In this case, the minimum efficiency is 33 mpg and the maximum efficiency is 55 mpg. We need to trap our formula between these two values to solve for the respective speeds.

\begin{align*} 33 \amp\lt 88-0.7x \lt 55\\ 33\subtractright{88} \amp\lt 88-0.7x\subtractright{88} \lt 55\subtractright{88}\\ -55 \amp\lt -0.7x \lt -33\\ \divideunder{-55}{-0.7} \amp\mathbin{\highlight{\gt}} \divideunder{-0.7x}{-0.7} \mathbin{\highlight{\gt}} \divideunder{-33}{-0.7}\\ 78.57 \amp\gt x \gt 47.14\amp\text{(note: these values are approximate)} \end{align*}

This inequality says that our model is applicable when the car's speed is between about 47 mph and about 79 mph.

### Subsection11.2.6Exercises

###### 1

Here is an interval: Write the interval using set-builder notation.

Write the interval using interval notation.

###### 2

Here is an interval: Write the interval using set-builder notation.

Write the interval using interval notation.

###### 3

Here is an interval: Write the interval using set-builder notation.

Write the interval using interval notation.

###### 4

Here is an interval: Write the interval using set-builder notation.

Write the interval using interval notation.

###### 5

Solve this inequality.

$\displaystyle{ {5} > {x+10} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

###### 6

Solve this inequality.

$\displaystyle{ {1} > {x+8} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

###### 7

Solve this inequality.

$\displaystyle{ {-2x} \geq {4} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

###### 8

Solve this inequality.

$\displaystyle{ {-2x} \geq {8} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

###### 9

Solve this inequality.

$\displaystyle{ {4} \geq {-5x+4} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

###### 10

Solve this inequality.

$\displaystyle{ {2} \geq {-6x+2} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

###### 11

Solve this inequality.

$\displaystyle{ {8t+9} \lt {3t+34} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

###### 12

Solve this inequality.

$\displaystyle{ {9t+6} \lt {5t+18} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

###### 13

Decide whether the given value for the variable is a solution.

1. $x \gt 8$ and $x \leq 2 \qquad x = 4$

The given value

• is

• is not

a solution.

2. $x \lt 8$ or $x \geq 5 \qquad x = 4$

The given value

• is

• is not

a solution.

3. $x \geq -2$ and $x \leq 6 \qquad x = 8$

The given value

• is

• is not

a solution.

4. $-2 \leq x \leq 3 \qquad x = 1$

The given value

• is

• is not

a solution.

###### 14

Decide whether the given value for the variable is a solution.

1. $x \gt 9$ and $x \leq 7 \qquad x = 8$

The given value

• is

• is not

a solution.

2. $x \lt 6$ or $x \geq 5 \qquad x = 9$

The given value

• is

• is not

a solution.

3. $x \geq -1$ and $x \leq 9 \qquad x = -5$

The given value

• is

• is not

a solution.

4. $-1 \leq x \leq 2 \qquad x = 1$

The given value

• is

• is not

a solution.

###### 15

Solve the compound inequality. Write the solution set in interval notation.

$\displaystyle{-10 \lt x \leq 5}$

$x$ is in

###### 16

Solve the compound inequality. Write the solution set in interval notation.

$\displaystyle{-9 \lt x \leq 1}$

$x$ is in

###### 17

Solve the compound inequality. Write the solution set in interval notation.

$\displaystyle{-8 > x\text{ or } x \geq 8}$

$x$ is in

###### 18

Solve the compound inequality. Write the solution set in interval notation.

$\displaystyle{-7 > x\text{ or } x \geq 4}$

$x$ is in

###### 19

Express the following inequality using interval notation.

$\displaystyle{x \lt -6 \text{ or } x \leq 1}$

$x$ is in

###### 20

Express the following inequality using interval notation.

$\displaystyle{x \lt -5 \text{ or } x \leq 7}$

$x$ is in

###### 21

Solve the compound inequality algebraically.

$\displaystyle{-14 \lt 7 - x \leq -9}$

$x$ is in

###### 22

Solve the compound inequality algebraically.

$\displaystyle{-16 \lt 20 - x \leq -11}$

$x$ is in

###### 23

Solve the compound inequality algebraically.

$\displaystyle{19 \leq x+13 \lt 24}$

$x$ is in

###### 24

Solve the compound inequality algebraically.

$\displaystyle{1 \leq x+7 \lt 6}$

$x$ is in

###### 25

Solve the compound inequality algebraically.

$\displaystyle{ 4 \le \frac{5}{9}(F-32) \le 50 }$

$F$ is in

###### 26

Solve the compound inequality algebraically.

$\displaystyle{ 8 \le \frac{5}{9}(F-32) \le 43 }$

$F$ is in

###### 27

Solve the compound inequality algebraically.

$\displaystyle{-10x-11\leq-20 \quad \text{and} \quad -2x-18\lt -10}$

###### 28

Solve the compound inequality algebraically.

$\displaystyle{17x+6\leq9 \quad \text{and} \quad 18x-14\leq7}$

###### 29

Solve the compound inequality algebraically.

$\displaystyle{-9x-8\leq-5 \quad \text{or} \quad -4x-15\geq-13}$

###### 30

Solve the compound inequality algebraically.

$\displaystyle{-12x-13\geq-11 \quad \text{or} \quad -7x-1\lt -17}$

###### 31

Solve the compound inequality algebraically.

$\displaystyle{13x+13\leq-13 \quad \text{or} \quad -5x+11\leq-4}$

###### 32

Solve the compound inequality algebraically.

$\displaystyle{-9x+13\lt 2 \quad \text{or} \quad 16x-2\lt -1}$

###### 33

Solve the compound inequality algebraically.

$\displaystyle{-13x+10\lt -7 \quad \text{and} \quad -3x+4\lt 8}$

###### 34

Solve the compound inequality algebraically.

$\displaystyle{13x-14\lt -18 \quad \text{and} \quad 18x+8\leq-16}$

###### 35

Solve the compound inequality algebraically.

$\displaystyle{ {6} \lt \frac{2}{5}x \lt {20} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

###### 36

Solve the compound inequality algebraically.

$\displaystyle{ {15} \lt \frac{3}{2}x \lt {54} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

###### 37

Solve the compound inequality algebraically.

$\displaystyle{ {5} \gt -1-\frac{3}{7}x \geq {-10} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

###### 38

Solve the compound inequality algebraically.

$\displaystyle{ {20} \gt 4-\frac{4}{5}x \geq {-8} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

###### 39

A graph of $f$ is given. Use the graph alone to solve the compound inequalities. 1. $f(x)\gt 2$

2. $-2 \lt f(x) \le 2$

###### 40

A graph of $f$ is given. Use the graph alone to solve the compound inequalities. 1. $f(x)\gt -1$

2. $-1 \lt f(x) \le 3$

###### 41

A graph of $f$ is given. Use the graph alone to solve the compound inequalities. 1. $f(x)\gt 0$

2. $f(x)\leq 0$

###### 42

A graph of $f$ is given. Use the graph alone to solve the compound inequalities. 1. $f(x)\gt 5$

2. $f(x)\leq 5$

###### 43

A graph of $f$ is given. Use the graph alone to solve the compound inequalities. 1. $f(x)\gt -2$

2. $f(x)\leq -2$

###### 44

A graph of $f$ is given. Use the graph alone to solve the compound inequalities. 1. $f(x)\gt 5$

2. $f(x)\leq 5$

###### 45

As dry air moves upward, it expands. In so doing, it cools at a rate of about $1^{\circ}\text{C}$ for every $100\,\text{m}$ rise, up to about $12\,\text{km}\text{.}$

1. If the ground temperature is $18^{\circ}\text{C}\text{,}$ write a formula for the temperature at height $x\,\text{km}\text{.}$ $T(x)=$

2. What range of temperature will a plane be exposed to if it takes off and reaches a maximum height of $5\,\text{km}\text{?}$ Write answer in interval notation.

The range is