ORCCAOpen Resources for Community College Algebra

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:

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.

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.

The union combines both solution sets into one, and so

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.

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.

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.

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:

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

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.

So the solution set to the compound inequality is:

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: \(\highlight{-7}\mathbin{\highlight{\nleq-10}}\mathbin{\lowlight{\lt}}\lowlight{4}\text{.}\)

3. False: \(\highlight{-2}\mathbin{\lowlight{\le}}\lowlight{0}\mathbin{\highlight{\ngeq}}\highlight{1}\text{.}\)

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

5. False: \(\highlight{3}\mathbin{\highlight{\nless}}\highlight{3}\mathbin{\lowlight{\le}}\lowlight{5}\text{.}\)

6. False: \(\highlight{9}\mathbin{\lowlight{\gt}}\lowlight{1}\mathbin{\highlight{\nless}}\highlight{5}\text{.}\)

7. False: \(\lowlight{3}\mathbin{\lowlight{\lt}}\highlight{8}\mathbin{\highlight{\nleq}}\highlight{-2}\text{.}\)

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

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 \(\{x\mid -7\lt x\le 5\}\text{.}\)

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 \(\left[-1,\frac{7}{9}\right)\text{.}\)

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 \((-6,15]\text{.}\)

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 \((-2.1,0.7)\cup(2.4,3.2]\text{.}\)

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

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

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.

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.