Lane ORCCA (2020–2021): Open Resources for Community College Algebra

To find the union of the sets \([-3,5)\) and \([2,8)\text{,}\) first we draw a number line with both intervals drawn to visualize both sets.

Recall that the union of two sets is the set of the combination of numbers from both sets. Looking at the graph, we need to include all of the numbers from both colored lines. Imagine pressing these two colored lines together as if they were long strands of modeling clay. The new combined strand would go from \(-3\) all the way to \(8\text{.}\) Note that \(-3\) would be included on this new line because it was in \([-3,5)\text{,}\) but \(8\) wouldn't be on this new line because it wasn't on either of the lines.

In conclusion,

To find the intersection of the sets \([-3,5)\) and \([2,8)\text{,}\) first we draw a number line with both intervals drawn to visualize the overlap of the sets.

Recall that the intersection of two sets is the set of the numbers in common to both sets. The lines have numbers in common on a long stretch where they overlap. How would we describe all of the numbers where the two lines overlap? In English, we might say that the lines overlap at every number between \(2\) and \(5\text{.}\) Note that the number \(2\) is in both sets, but since \(5\) is not in the set \([-3,5)\text{,}\) we should exclude \(5\) from the intersection. The description we have come up with is the same as the interval \([2,5)\text{.}\)

In conclusion,

To find the intersection of the sets \((-\infty,-1)\) and \((-\infty,3)\text{,}\) first we draw a number line with both intervals drawn to visualize where the sets overlap.

Recall that the intersection of two sets is the set of the numbers in common to both sets. In English, we might say that the lines overlap at every number below \(-1\text{.}\) This description is the same as the interval \((-\infty,-1)\text{.}\)

In conclusion,

1. \(\{2,4,6,8\}\cup\{1,2,3,4\}=\{1,2,3,4,6,8\}\text{.}\)

2. \(\{2,4,6,8\}\cap\{1,2,3,4\}=\{2,4\}\text{.}\)

3. \(\{2,4,6,8\}\cup\{1,3,5,7\}=\{1,2,3,4,5,6,7,8\}\text{.}\)

4. \(\{2,4,6,8\}\cap\{1,3,5,7\}=\emptyset\text{.}\)

5. \((-\infty,9]\cup(-\infty,-3]=(-\infty,9]\text{.}\)

6. \((-\infty,9]\cap(-\infty,-3]=(-\infty,-3]\text{.}\)

7. \((-\infty,12)\cup[-3,\infty)=\mathbb{R}\text{.}\)

8. \((-\infty,12)\cap[-3,\infty)=[-3,12)\text{.}\)

9. \((-\infty,-2]\cup[4,\infty)=(-\infty,-2]\cup[4,\infty)\text{.}\) This union cannot be simplified because the two sets have nothing in common.

10. \((-\infty,-2]\cap[4,\infty)=\emptyset\text{.}\)

11. \([-7,8)\cup(0,12]=[-7,12]\text{.}\)

12. \([-7,8)\cap(0,12]=(0,8)\text{.}\)

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\\ -12\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\\ -2\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 12.2.25, 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 12.2.27 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 12.2.29 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 solution set to \(t\lt3\) is \((-\infty,3)\) and the solution set to \(t\geq-1\) is \([-1,\infty)\text{.}\) Shown is a graph of these solution sets.

Recall that an “and” problem finds the intersection of the solution sets. Intersection finds the \(t\)-values where the two lines overlap, so the solution to the compound inequality must be

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

• First, choose a number within our solution region and test that it makes both original inequalities true. We will arbitrarily choose \(\highlight{1}\text{.}\)

  \begin{align*} 4-2t\amp\gt -2\amp\text{and}\amp\amp 3t+1\amp\geq -2\\ 4-2(\highlight{1})\amp\stackrel{?}{\gt} -2\amp\text{and}\amp\amp 3(\highlight{1})+1\amp\stackrel{?}{\geq} -2\\ 2\amp\stackrel{\checkmark}{\gt} -2\amp\text{and}\amp\amp 4\amp\stackrel{\checkmark}{\geq} -2 \end{align*}

• Next, choose a value outside the solution set and test that it makes at least one of the inequalities false. We will arbitrarily choose \(\highlight{4}\text{.}\)

  \begin{align*} 4-2t\amp\gt -2\amp\text{and}\amp\amp 3t+1\amp\geq -2\\ 4-2(\highlight{4})\amp\stackrel{?}{\gt} -2\amp\text{and}\amp\amp 3(\highlight{4})+1\amp\stackrel{?}{\geq} -2\\ -4\amp\stackrel{\text{no}}{\gt} -2\amp\text{and}\amp\amp 13\amp\stackrel{\checkmark}{\geq} -2 \end{align*}

  Since one of the inequalities is false and this is an “and” statement, the compound inequality is false for this value, which is what is expected by picking a number outside the solution set.

• Last, we should choose a number that is not a solution that is on the “other side” of the solution set. We will arbitrarily choose \(\highlight{-2}\text{.}\)

  \begin{align*} 4-2t\amp\gt -2\amp\text{and}\amp\amp 3t+1\amp\geq -2\\ 4-2(\highlight{-2})\amp\stackrel{?}{\gt} -2\amp\text{and}\amp\amp 3(\highlight{-2})+1\amp\stackrel{?}{\geq} -2\\ 8\amp\stackrel{\checkmark}{\gt} -2\amp\text{and}\amp\amp -5\amp\stackrel{\text{no}}{\geq} -2 \end{align*}

  Again, since one of the inequalities is false and this is an “and” statement, the compound inequality is false for \(-2\text{.}\)

So, numbers outside the proposed solution region make the compound inequality false, and numbers inside the region make the compound inequality true. We have verified our solution set.

The solutions to the double 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 double (or three-part) inequality, which we can treat just as a regular inequality, but 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 double 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 double inequality, which we can treat just as a regular inequality, but 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.

Notice that we had to reverse the direction of the inequality signs since we divided by a negative number.

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

The solution set to \(x\geq -3\) is \([-3,\infty)\) and the solution set to \(x\lt-5\) is \((-\infty,-5)\text{.}\) Shown is a graph of these solution sets on a number line.

Recall that “and” statements of inequalities become intersections of the solution sets. Since intersections refer to where the sets overlap, and these sets never overlap, there must be no solution to the compound inequality. Mathematically, we would say

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 double 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.