**Inequalities**

**Inequalities** are similar to equations in that they compare two expressions, but an inequality only asserts that one side is greater than or equal to (or less than or equal to) the other side. Inequalities can be manipulated like equations, except that if you multiply or divide each side of an inequality by a negative number, then the direction of the inequality is reversed. Inequality problems, as you can see from these examples, often involve absolute value, because inequalities have to do with ranges of values as absolute value statements often do.

On this page we provide 3 examples of inequality problems.

**To practice more of these types of problems, click** here.

**Example 1)**

Let k be an integer. Then which of the following values for k makes this inequality

|k – 3/2| < 3

true?

I. -2

II. -1

III. 0

A. I only

B. II only

C. III only

D. II and III

E. I, II, and III

**Explanation:** It’s easy to substitute each of these 3 numbers into the expression, and see if it’s true. The only thing is, do we know how to interpret the expression? Let’s try each number, and see what happens. Remember, absolute value turns a negative number into a positive:

I

|k – 3/2| < 3

|-2 – 3/2| < 3

|-4/2 – 3/2| < 3

|-7/2| < 3

**3 1/2 < 3 **

II

|k – 3/2| < 3

|-1 – 3/2| < 3

|-2/2 – 3/2| < 3

|-5/2| < 3

**2 1/2 < 3**

III

|k – 3/2| < 3

|0 – 3/2| < 3

|-3/2| < 3

1 1/2 < 3

**1 1/2 < 3**

**Red indicates false**, and **blue true**. As you can see, only numbers II and III are true, so **the correct answer is D.**

**Example 2)**

If -3 ≤ x ≤ 7, then which of these is always true?

A. |x| ≤ 3

B. |x – 7| ≤ 0

C. |x – 7| ≤ -10

D. |x – 2| ≤ 5

E. |x + 3| ≤ 7

**Explanation:** Let’s look at each answer:

A. |x| ≤ 3 If x is 7, then its absolute value is > 3. **Answer is false.**

B. |x – 7| ≤ 0 If x is 5, then |5 – 7| = |-2| = 2 > 0. **Answer is false.**

C. |x – 7| ≤ -10 Absolute value is never negative. **Answer is false.**

D. |x – 2| ≤ 5 See below. **Answer is true.**

E. |x + 3| ≤ 7 If x is 7, then |x + 3| = 10.** Answer is false.**

Answer D explanation:

-3 ≤ x ≤ 7

-3 – 2 ≤ x – 2 ≤ 7 – 2

-5 ≤ x – 2 ≤ 5

So,

|x – 2| ≤ 5

**Example 3)**

100 members of a club each cast 2 ballots to elect 2 members from a slate of 10 candidates to its activities committee. Candidates B and C got by far the most votes, with B getting 50 votes and C getting 60. If at most 20 members voted for neither B nor C, the number of members who voted for both B and C is in what range?

A. From 20 to 30

B. From 10 to 40

C. From 20 to 40

D. From 12 to 28

E. From 10 to 30

**Explanation:** Considering just the two candidates B and C, define b as the number that voted for b only, c as the number that voted for c only, and x as the number that voted for both b and c. The problem asks us to determine the range for x. Notice that the problem says “at most 20 members voted for neither B nor C.” Thus, it’s possible that 0 voted for neither B nor C. We only know that there were not more than 20 who voted for neither B nor C. That being the case, we have these relatiionships:

100 ≥ b + c + x ≥ 80,

because there are 100 members in the club, and at least 80 voted for B or C; and

b + c + 2x = 110,

because B and C got a total of 110 (60 + 50) votes, and these came from the b people that voted for B only, the c people that voted for C only, and the x people who voted for both B and C, and who thus contributed 2 votes each to the total of 110 votes for B and C.

Using this last equation, we get

b + c = 110 – 2x

so substituting this into the inequality at the top, we get

100 ≥ 110 – 2x + x ≥ 80

100 ≥ 110 – x ≥ 80

These two inequalities can be handled separately as

100 ≥ 110 – x

-10 ≥ -x

10 ≤ x

and

110 – x ≥ 80

110 – 80 ≥ x

30 ≥ x

So, the range of possible values for x, the number of people who voted for both B and C, is

10 ≤ x ≤ 30

**Therefore, answer E is correct.**

**To practice more of these types of problems, click** here.