Inequalities | Algebra | Achievable CLT

Introduction

Inequalities questions typically show up once or twice per CLT math section, while there can also be an inequalities-related aspect to another question or two in each section. As we’ll see in the Approach Question below, inequalities are often combined with absolute value on the CLT. The good news for inequalities? They can be solved just like equations (with one exception)! If you burnish your equation-solving skills, you’ll be helping yourself with inequalities as well.

Approach Question

Which of the following is the solution set for the inequality $∣2 x + 7∣ < 12$ ?

A. $-19/2 < x < 5/2$
B. $-5/2 < x < 19/2$
C. $x < - 19/2$ or $x < 5/2$
D. $x < - 5/2$ or $x > 19/2$

Explanation

Many CLT inequalities also include absolute values. Since absolute values are not common on the test when not paired with inequalities, we will cover absolute value equations and inequalities in this explanation before dealing with the inequality aspect. We’ll begin with the definition of absolute value, which is the distance of a value from zero on the real number line. Since distance can never be negative, absolute values are always nonnegative (they can be equal to 0).

Let’s pretend for a moment that the Approach Question involved not an inequality but an equation. In that case, we are looking to make the left side of the equation equal to 12. But because the absolute value symbols tell us that a negative value in between them must turn positive, we can make a successful equation if we manage to equate the left side of the equation to negative 12. For this reason, all absolute value equations in which the variable is inside the absolute value bars result in two equations. In this case, the equations would look like this:

$2 x + 7 = 12$

$2 x + 7 = - 12$

Now let’s return to the fact that our question is an inequality, not an equation. A key truth throughout this lesson: Inequalities can be solved just like equations, with one crucial exception. The exception is that when multiplying or dividing by a negative number, you must flip the direction of the inequality symbol.

These statements suggest that we can reproduce the above equations as inequalities as long as we pay attention to the exception. In this case, the exception means that we flip the inequality symbol when we turn the 12 into -12 in the second case. So our inequalities look like this:

$2 x + 7 < 12$

$2 x + 7 > - 12$

We can solve these inequalities by subtracting by $7$ an dividing by $2$ in each case. The result is $x < 5/2$ and $x > - 19/2$ . The answer is A.

One final note: consider the differences between answers $A$ and $B$ on the one hand and answers $C$ and $D$ on the other. Choices $C$ and $D$ include the word “or”; we can refer to choices $A$ and $B$ as “and” answers. If you note the direction of the inequality symbols for each answer, you may see why it has to be “and” in one case and “or” in the other. For choices $A$ and $B$ , we’re talking about a range in between two values. With choices $C$ and $D$ , it’s a range outside of two values. Choice $C$ and $D$ have to be “or” because it’s impossible for x to be simultaneously less than $- 5/2$ (for example) and greater than $19/2$ . It has to be one or the other.

Although the following rule is not necessary for solving CLT inequalities, we note it here to assist your understanding: when an absolute value inequality begins with a “less than” symbol, the result is an “and” arrangement; when it begins with a “greater than” symbol, the result is an “or” arrangement.

Definitions

Inequality: A relationship between two quantities in which one is consistently greater than (or perhaps equal to) the other. Depending on which side of the inequality symbol the variable falls, the variable solved for is described as having a “greater than (or equal to)” or “less than (or equal to)” relationship to the constant on the other side.

Topics for Cross-Reference

Variations

The chief variation for an inequality problem is the addition of the equality symbol; “less than” becomes “less than or equal to” and “greater than” becomes “greater than or equal to.” The symbols look like this: $\leq$ and $\geq$ . This difference doesn’t tend to be significant with CLT problems, but make sure everything is consistent; if the question includes the words “or equal to,” the correct answer must include the line under the inequality.

Strategy Insights

By far, the most important strategy to consider with inequalities is plugging in numbers. CLT inequalities will be able to be solved algebraically, but if the process appears complicated or you are uncertain about your algebraic work, plug in numbers to discover or confirm the right answer.
When doing so, consider that zero is a great number to plug in for inequalities. Since zero will immediately cancel whatever term(s) in which the variable(s) is/are found, you will be left with numeric quantities on each side of the inequality symbol. If plugging in zero creates a true inequality, eliminate all answers whose ranges do not equal zero. If substituting zero creates a false statement, cross out answers that do include zero.

Flashcard Fodder

There are few flashcard-worthy elements in this module, but it might be worth making a flashcard about the rule concerning flipping the direction of the inequality symbol. As a reminder, that happens when you multiply or divide by a negative number in the course of solving.

Sample Questions

Difficulty 1

Which of the following is equivalent to $- 6 z > 66$ ?

A. $z > - 11$
B. $z < - 11$
C. $z > 11$
D. $z < 11$

(spoiler)

The answer is B. This inequality might look relatively simple, especially when you know that inequalities can be solved like equations. We divide both sides by $- 6$ , so the right side is $- 11$ . But wait! This is a problem where the “inequality exception” applies; because we are dividing by a negative value, we must change the direction of the inequality symbol. So the answer is not $A$ (trap answer), but $B$ .

Since this problem is a bit tricky due to its trap answer, it’s worth taking a few more seconds to plug in something like $- 12$ . Seeing how $- 12$ makes the inequality true confirms our confidence in our answer.

Difficulty 2

A car rental company charges a fixed fee of $$40$ plus $$0.15$ per mile driven. If your budget for renting a car is $$115$ and you choose this company, what is the maximum number of miles you can drive the rental car without exceeding your budget?

A. 400
B. 500
C. 600
D. 700

(spoiler)

The answer is B. If the CLT presents you with an inequality-based word problem, it will often be a linear model with a variable aspect and a constant part. We need to identify which part is variable (which one will change) so we can attach our variable to the number involved. In this case, the changing part of the equation is the $$0.15$ per mile cost, so we should plan to have a term of $0.15 x$ . Meanwhile, the $$40$ only gets charged once, so $40$ can be added, making $0.15 x + 40$ . We need to stay at or under a budget of $$115$ ; this sounds like an inequality but with the equals sign included (since spending exactly $$115$ is acceptable in this case). The inequality comes together as follows:

$0.15 x + 40 \leq 115$

Once we subtract $40$ from both sides, we can see we’ll need to divide $75$ by $0.15$ –not an easy task without a calculator. You can use long division, but always seek ways to use $10$ ’s to help you, since our math system is based on tens. One aspect of “ten-ness” in math is the decimal; since the decimal has to do only with the place value, we can ignore it for the moment and think about how many times $15$ goes into $75$ . If you see that the answer to that is $5$ , you should see the right answer already. But in case you want to make sure, the two zeroes in the right answer correspond to the two places we have to move the decimal in $0.15$ to make $15$ . A combination of math reasoning and common sense can often save you time in calculating!

For an alternate strategy, consider how well this problem works with the “B, then C” strategy of plugging in the answers. If we start with one of the answers in the middle, that answer will either be 1) correct, or 2) too large, or 3) too small. If it’s too large or too small, you can quickly arrive at the answer. Since $500$ (choice $B$ ) is the answer here, let’s pretend we started with choice $C$ instead. If we suppose that we drove the car 600 miles, that would accrue $(600) ($0.15) = $90$ in variable charges. Add the $$40$ fixed charge and we have $$130$ , which is over our budget. You can probably tell already that an answer a little smaller, like $500$ , is correct at this stage, but you could plug in either choice $A$ or choice $B$ to check.

Difficulty 3

The sum of three consecutive integers is greater than $50$ . Which of the following inequalities represents this situation?

A. $x + 3 < 50$
B. $x + 3 > 50$
C. $2 x + 3 > 50$
D. $3 x + 3 > 50$

(spoiler)

The answer is D. To understand this question, especially given how the answers are organized, it’s most helpful to envision how three consecutive integers might be described algebraically. We can always label a quantity as $x$ , so we might as well use $x$ to represent the smallest number in the sequence. Once we do that, it becomes apparent that the next highest number is $x + 1$ and the one after that is $x + 2$ . The sum of the three, then, must be $x + x + 1 + x + 2 = 3 x + 3$ . There’s only one answer that fits that, and the “greater than” symbol makes sense in that context.

Plugging in could help in this case if you’re uncertain about mapping the algebra. Choose three numbers whose sum is greater than $50$ , like $20 - 21 - 22$ . Plugging in one of these numbers results in a false statement for choices $A$ , $B$ , and $C$ . This process can give you choice $D$ as an educated guess and may well deepen your understanding of the problem so that you become more certain of the answer.

Difficulty 4

Which of the following is a solution to the system of inequalities?

$6 - 9 y < - 30$

$14 + y^{3} < - 13$

A. $y < - 3$ or $y > 4$
B. $- 3 < y < 4$
C. $y < - 4$ or $y > 3$
D. $- 4 < y < 3$

(spoiler)

The answer is A. Our best bet seems to be solving both inequalities, but where to begin? The linear equality is more approachable than the cubic inequality (the one with $y^{3}$ ), so let’s begin there. Subtracting $6$ from both sides is the first step, but be careful: the negative remains in front of the 9y, so our result is this:

$- 9 y < - 36$

Dividing both sides by $- 9$ reverses the sign so we arrive at:

$y > 4$ .

Have you done the L step of the UnCLES method, looking carefully at the answers? If so, you might notice that we don’t need to deal with the other inequality at all!. There is only one answer in which $y > 4$ . However, here is a solution to the second inequality if you’re interested:

$14 + y^{3} < - 13$

$y^{3} < - 27$

$y < - 3$

(Note: we are allowed to take the cube root of both sides because a cube root of a negative number is legal; it’s only even powers, like the square root, where we run into problems with negative numbers.)

Difficulty 5

Which of the following is always true for the numbers $b$ and $c$ , if $b < c$ and $b < 1$ ?

I. $b^{2} < c$
II. $b^{3} < c^{2}$
III. $b^{4} < c^{3}$

A. II only
B. III only
C. I and II only
D. II and III only

(spoiler)

The answer is A. This is as much a number properties problem as it is an inequalities problem, so feel free to click on the cross-reference link to the Number Properties lesson. We have two inequalities; the more concrete of the two tells us that $b$ is a negative number unless it is either $0$ or a fraction less than $1$ . We should therefore consider all three of those possible number categories for $b$ (negative, zero, or small positive fraction) as we relate $b$ to $c$ . We should be careful not to assume too much about $c$ : just because it is greater than $b$ does not mean it is positive; $b$ could be negative while $c$ is just a bit less negative.

With this in mind, we can address the statements given with the Roman numerals. Statement I can be disproved by considering that a number squared is positive or zero; if we choose a “negative enough” number, we can make $b^{2}$ larger than $c$ . For example, if $b = - 10$ and $c = - 5$ , then $b < c$ but $b^{2} > c$ . Since Statement I is not always true, answer choice $C$ is eliminated.

Statement II changes the relationship because now the quantity that can never be negative ( $c^{2}$ ) is on the “greater than” side of the inequality. This tells us that as long as $b$ is negative, the inequality will be true because $b^{3}$ will also be negative. But what if $b$ is $0$ or $0 < b < 1$ ? If $b = 0$ , then $b^{3} = 0$ and in that instance $c$ , and therefore $c^{2}$ , must be positive. So that works. If $b$ is something like $\frac{1}{2}$ , $b^{3}$ will actually be smaller than that. If $c > b$ , then $c^{2}$ won’t decrease as much as $b^{3}$ does and therefore can never be smaller. Put in numerical terms, imagine that $b = \frac{1}{3}$ and $c = \frac{1}{2}$ . In that case, $b^{3} = \frac{1}{27}$ while $c^{2}$ is much larger: $\frac{1}{4}$ . Statement II always works.

Does Statement III also work? We can actually deal with this statement more quickly by applying the same logic as we did with Statement I. Since $b^{4}$ must always be at least $0$ and could in fact very positive, we cannot guarantee it will stay less than $c^{3}$ . Statement III is not always true, therefore, and the answer is $A$ .

For Reflection

How will you approach inequality questions on test day? Write down at least three takeaways from this module.
Rate the difficulty of these questions for you from 1 (no problem) to 5 (problem!). This will help you decide when to answer them and when to skip them on test day.
A small mistake can doom your work on CLT math questions. Are you certain in your knowledge of which symbol means “less than” and which means “greater than”? If not, make up a mnemonic device or put it on a flashcard.