Linear Inequalities and Absolute Value Practice Questions — GRE General Test — Page 2

What is the least integer value of $x$ that satisfies the inequality $|4x - 5| \le 11$ ?

Show answer & explanation

Answer: -1

Answer

The least integer value of

x

that satisfies the inequality is

-1

.

Expanding the absolute value inequality

|4x - 5| \le 11

produces

-11 \le 4x - 5 \le 11

. Adding 5 across all sections yields

-6 \le 4x \le 16

, and dividing by 4 results in

-1.5 \le x \le 4

. The set of integers within this range is

\{-1, 0, 1, 2, 3, 4\}

, making

-1

the least integer value.

Step-by-Step Solution

1

Rewrite the absolute value inequality as a double inequality.

-11 \le 4x - 5 \le 11

An absolute value inequality of the form

|u| \le a

for

a \ge 0

is equivalent to

-a \le u \le a

.

2

Add 5 to all parts of the compound inequality.

-6 \le 4x \le 16

To isolate

x

, first add 5 to eliminate the constant term

-5

.

3

Divide all parts by 4.

-1.5 \le x \le 4

Dividing by a positive constant preserves the direction of the inequality signs.

4

Select the minimum integer contained within the solution interval

[-1.5, 4]

.

-1

The integer values satisfying

-1.5 \le x \le 4

are

-1, 0, 1, 2, 3, 4

. The smallest among these is

-1

.

Key Concept

Linear inequalities involving absolute value

Question 22Question →

If $x$ is a real number that satisfies both $|5 - 2x| \le 11$ and $-3x + 7 < 1$ , what is the least possible integer value of $x$ ?

-3

2

3

-8

8

Show answer & explanation

Answer: 3

Answer

The least possible integer value of

x

is 3.

Solving

|5 - 2x| \le 11

leads to

-11 \le 5 - 2x \le 11

. Subtracting 5 gives

-16 \le -2x \le 6

, and dividing by

-2

(flipping the inequalities) yields

-3 \le x \le 8

. Next, solving

-3x + 7 < 1

gives

-3x < -6

, which upon dividing by

-3

(and flipping the inequality sign) gives

x > 2

. Combining these two requirements yields

2 < x \le 8

. The integer values satisfying this inequality are 3, 4, 5, 6, 7, and 8. The least possible integer value among these is 3.

Step-by-Step Solution

1

Solve the absolute value inequality

|5 - 2x| \le 11

.

-11 \le 5 - 2x \le 11 \implies -16 \le -2x \le 6 \implies -3 \le x \le 8

.

An absolute value inequality

|u| \le k

expands to

-k \le u \le k

. Dividing by

-2

flips the inequality direction.

2

Solve the linear inequality

-3x + 7 < 1

.

-3x < -6 \implies x > 2

.

Subtract 7 from both sides, then divide by

-3

, remembering to reverse the inequality sign.

3

Determine the intersection of both solution sets.

2 < x \le 8

.

x

must be strictly greater than 2 and less than or equal to 8.

4

Identify the smallest integer within the range

2 < x \le 8

.

3

Since

x > 2

is strict, 2 is excluded, making 3 the smallest integer in the range.

Key Concept

Solving systems of linear inequalities involving absolute values and correctly applying sign-flipping rules when multiplying or dividing by negative quantities.

Alternative Method

Test integer candidates directly: for

x = 2

,

-3(2) + 7 = 1

, which is not strictly less than 1. For

x = 3

,

-3(3) + 7 = -2 < 1

(valid) and

|5 - 2(3)| = |-1| = 1 \le 11

(valid), confirming 3 is the smallest integer solution.

Estimated Time:1m 30s

Question 23Question →

If $-5x + 3 \le 18$ , which of the following inequalities represents all possible real values of $x$ ?

x \ge -3

x \le -3

x \ge 3

x \le 3

x \ge -\frac{21}{5}

Show answer & explanation

Answer: $x \ge -3$

Answer

x \ge -3

Subtracting 3 from both sides gives

-5x \le 15

. Dividing both sides by

-5

requires flipping the inequality sign from

\le

to

\ge

, resulting in

x \ge -3

.

Step-by-Step Solution

1

Subtract 3 from both sides of the inequality

-5x + 3 \le 18

.

-5x \le 15

Isolate the variable term on the left side.

2

Divide both sides by

-5

and reverse the inequality sign from

\le

to

\ge

.

x \ge -3

Dividing an inequality by a negative number reverses the direction of the inequality sign.

Key Concept

Linear Inequalities and Sign Reversal

Question 24Question →

If $x$ and $y$ are real numbers that satisfy the inequalities $|3 - 2x| \le 9$ and $|4y + 3| \le 13$ , what is the maximum possible value of the expression $|3x - 4y|$ ?

Show answer & explanation

Answer: 34

Answer

The maximum possible value of

|3x - 4y|

is

34

.

To find the maximum possible value of

|3x - 4y|

, we first solve for the ranges of

x

and

y

from their respective absolute value inequalities. From

|3 - 2x| \le 9

, we get

-9 \le 3 - 2x \le 9 \implies -3 \le x \le 6

. From

|4y + 3| \le 13

, we get

-13 \le 4y + 3 \le 13 \implies -4 \le y \le 2.5

. To maximize

3x - 4y

, we take the maximum value of

3x

(

3 \times 6 = 18

) and the maximum value of

-4y

(

-4 \times -4 = 16

), giving

18 + 16 = 34

. To minimize

3x - 4y

, we take the minimum value of

3x

(

3 \times -3 = -9

) and the minimum value of

-4y

(

-4 \times 2.5 = -10

), giving

-9 - 10 = -19

. The expression

3x - 4y

ranges from

-19

to

34

, so the maximum possible magnitude

|3x - 4y|

is

\max(|-19|, |34|) = 34

.

Step-by-Step Solution

1

Solve the inequality

|3 - 2x| \le 9

for

x

.

-3 \le x \le 6

Unpacking the absolute value yields

-9 \le 3 - 2x \le 9

. Subtracting

3

gives

-12 \le -2x \le 6

. Dividing by

-2

and reversing the inequality signs produces

-3 \le x \le 6

.

2

Solve the inequality

|4y + 3| \le 13

for

y

.

-4 \le y \le 2.5

Unpacking the absolute value yields

-13 \le 4y + 3 \le 13

. Subtracting

3

gives

-16 \le 4y \le 10

. Dividing by

4

gives

-4 \le y \le 2.5

.

3

Find the range of possible values for

3x

and

-4y

.

-9 \le 3x \le 18

and

-10 \le -4y \le 16

Multiplying

-3 \le x \le 6

by

3

gives

-9 \le 3x \le 18

. Multiplying

-4 \le y \le 2.5

by

-4

and flipping the signs gives

-10 \le -4y \le 16

.

4

Combine the bounds for

3x

and

-4y

to find the range for

3x - 4y

.

-19 \le 3x - 4y \le 34

The minimum value of

3x - 4y

is

(-9) + (-10) = -19

. The maximum value of

3x - 4y

is

18 + 16 = 34

.

5

Determine the maximum absolute value

|3x - 4y|

over the interval

[-19, 34]

.

34

The absolute value of any number in the interval

[-19, 34]

ranges from

0

to

\max(|-19|, |34|) = 34

.

Key Concept

Absolute Value Inequalities and Expression Bounding

Question 25Question →

If $x$ is a real number that satisfies both $|4 - 3x| > 7$ and $\frac{1 - 2x}{3} \ge -3$ , which of the following represents the complete set of all possible values of $x$ ?

x < -1

or

\frac{11}{3} < x \le 5

-1 < x \le 5

x \ge 5

-1 < x < \frac{11}{3}

x < -1

or

x > \frac{11}{3}

Show answer & explanation

Answer: $x < -1$ or $\frac{11}{3} < x \le 5$

Answer

x < -1

or

\frac{11}{3} < x \le 5

The correct option correctly solves

|4 - 3x| > 7

to yield

x < -1

or

x > \frac{11}{3}

, solves

\frac{1 - 2x}{3} \ge -3

to yield

x \le 5

, and takes their intersection to produce

x < -1

or

\frac{11}{3} < x \le 5

.

Step-by-Step Solution

1

Solve the absolute value inequality

|4 - 3x| > 7

.

4 - 3x > 7

or

4 - 3x < -7

. Solving

4 - 3x > 7

gives

-3x > 3 \implies x < -1

. Solving

4 - 3x < -7

gives

-3x < -11 \implies x > \frac{11}{3}

. Thus,

x \in (-\infty, -1) \cup (\frac{11}{3}, \infty)

.

An absolute value inequality of the form

|u| > c

splits into

u > c

or

u < -c

. Dividing by a negative number reverses the inequality direction.

2

Solve the linear inequality

\frac{1 - 2x}{3} \ge -3

.

Multiply both sides by

3

:

1 - 2x \ge -9

. Subtract

1

:

-2x \ge -10

. Divide by

-2

and flip the inequality sign:

x \le 5

.

Isolating the variable

x

requires reversing the inequality sign when dividing by the negative constant

-2

.

3

Find the intersection of the solution sets from Step 1 and Step 2.

We require

x

to satisfy

(x < -1 \text{ or } x > \frac{11}{3})

AND

x \le 5

. Case 1:

x < -1

automatically satisfies

x \le 5

, giving

x < -1

. Case 2:

x > \frac{11}{3}

combined with

x \le 5

gives

\frac{11}{3} < x \le 5

. Combining both cases yields

x < -1 \text{ or } \frac{11}{3} < x \le 5

.

The word 'both' in the stem indicates a logical AND (intersection) between the two conditions.

Key Concept

Solving systems involving absolute value inequalities and linear inequalities requires handling disjunctions (OR) for absolute values greater than a positive constant, reversing inequality signs when multiplying or dividing by negative numbers, and taking the intersection (AND) of all valid regions.

Estimated Time:2m 15s

Question 26Question →

If $x$ is a real number that satisfies the inequality $-2|3x - 4| + 5 \ge -7$ , which of the following inequalities represents all possible values of $x$ ?

-\frac{2}{3} \le x \le \frac{10}{3}

x \le -\frac{2}{3}

or

x \ge \frac{10}{3}

-\frac{10}{3} \le x \le \frac{2}{3}

x \le \frac{10}{3}

-\frac{1}{3} \le x \le 3

Show answer & explanation

Answer: $-\frac{2}{3} \le x \le \frac{10}{3}$

Answer

-\frac{2}{3} \le x \le \frac{10}{3}

Isolating

|3x - 4|

requires dividing

-2|3x - 4| \ge -12

by

-2

, which reverses the inequality to

|3x - 4| \le 6

. Expanding this into the compound inequality

-6 \le 3x - 4 \le 6

and solving for

x

yields the interval

-\frac{2}{3} \le x \le \frac{10}{3}

.

Step-by-Step Solution

1

Isolate the absolute value expression by subtracting 5 from both sides of the inequality.

-2|3x - 4| \ge -12

Before removing absolute value bars, the term

|3x - 4|

must be isolated on one side.

2

Divide both sides by

-2

and reverse the inequality sign.

|3x - 4| \le 6

Dividing an inequality by a negative number reverses the direction of the inequality sign.

3

Express the absolute value inequality

|u| \le k

as a double inequality

-k \le u \le k

.

-6 \le 3x - 4 \le 6

The distance of

3x - 4

from 0 on the number line must be at most 6 units.

4

Add 4 to all three parts of the inequality and divide by 3.

-2 \le 3x \le 10 \implies -\frac{2}{3} \le x \le \frac{10}{3}

Solving for

x

isolates the variable in the center of the double inequality.

Key Concept

Solving linear absolute value inequalities requires reversing the inequality sign when multiplying or dividing by negative numbers, and expressing

|A| \le B

as

-B \le A \le B

.

Question 27Question →

If $|x - 3| < 5$ , which of the following values could be the value of $x$ ? Select all such values.

Select all that apply

-3

-1

0

4

8

Show answer & explanation

Answer: $-1$ ; $0$ ; $4$

Answer

The possible values of

x

are

-1

,

0

, and

4

.

The absolute value inequality

|x - 3| < 5

represents all real numbers

x

whose distance from

3

on the number line is strictly less than

5

. Expressed as a compound inequality, this means

-5 < x - 3 < 5

. Adding

3

across the entire inequality yields

-2 < x < 8

. The values

-1

,

0

, and

4

are the only options that fall strictly within the interval

(-2, 8)

.

Step-by-Step Solution

1

Rewrite the absolute value inequality as a compound inequality.

-5 < x - 3 < 5

An absolute value inequality of the form

|u| < k

(where

k > 0

) is equivalent to

-k < u < k

.

2

Solve for

x

by adding

3

to all parts of the inequality.

-5 + 3 < x < 5 + 3 \implies -2 < x < 8

Adding a positive constant to all parts of an inequality isolates

x

while preserving inequality directions.

3

Test each given option against the range

-2 < x < 8

.

The values

-1

,

0

, and

4

lie within

(-2, 8)

, whereas

-3

and

8

lie outside.

-3

is less than or equal to

-2

, and

8

is not strictly less than

8

.

Key Concept

Solving linear absolute value inequalities

Estimated Time:45s

Question 28Question →

What is the minimum integer value of $x$ that satisfies the inequality $|2x - 7| \le 5$ ?

Show answer & explanation

Answer: 1

Answer

The minimum integer value of

x

that satisfies the inequality is 1.

To solve

|2x - 7| \le 5

, write it as the compound inequality

-5 \le 2x - 7 \le 5

. Adding 7 across all parts gives

2 \le 2x \le 12

. Dividing by 2 yields

1 \le x \le 6

. The integer solutions are 1, 2, 3, 4, 5, and 6. The minimum integer among these is 1.

Step-by-Step Solution

1

Convert the absolute value inequality into a compound inequality.

-5 \le 2x - 7 \le 5

An inequality of the form

|u| \le a

(where

a \ge 0

) is equivalent to

-a \le u \le a

.

2

Add 7 to all three parts of the inequality.

2 \le 2x \le 12

Adding a constant to an inequality preserves the direction of the inequality signs.

3

Divide all three parts by 2.

1 \le x \le 6

Dividing by a positive constant isolates

x

without reversing the inequality signs.

4

Determine the minimum integer within the solution set

[1, 6]

.

1

The solution set contains integers

\{1, 2, 3, 4, 5, 6\}

, making 1 the smallest integer value.

Key Concept

Solving absolute value inequalities using equivalent compound linear inequalities

Question 29Question →

Let $S$ be the set of all real numbers $x$ that satisfy the nested absolute value inequality $|3 - |2x - 1|| \le 4$ . Which of the following inequalities MUST be satisfied by every value of $x$ in $S$ ? Select all such inequalities.

Select all that apply

|x| \le 4

(x + 3)(x - 4) \le 0

|2x - 1| \le 4

-1 \le x \le 7

x \ge -2

Show answer & explanation

Answer: $|x| \le 4$ ; $(x + 3)(x - 4) \le 0$

Answer

The inequalities that must be satisfied by every value of

x

in

S

are

|x| \le 4

and

(x + 3)(x - 4) \le 0

.

The solution set to the nested inequality is

S = [-3, 4]

. For any value of

x

in

[-3, 4]

, the absolute value

|x|

ranges from

0

to

4

, so the inequality stating that the magnitude of

x

is at most 4 is satisfied. Additionally, the quadratic expression factored as

(x + 3)(x - 4)

has roots at

x = -3

and

x = 4

and opens upward, meaning it takes on values less than or equal to zero for all

x

between

-3

and

4

.

Step-by-Step Solution

1

Unfold the outer absolute value inequality

-4 \le 3 - |2x - 1| \le 4

By definition,

|u| \le k

(with

k \ge 0

) is equivalent to

-k \le u \le k

.

2

Isolate the inner absolute value expression by subtracting 3 and multiplying by -1

-1 \le |2x - 1| \le 7

Subtracting 3 yields

-7 \le -|2x - 1| \le 1

. Multiplying by

-1

flips the inequality signs, giving

-1 \le |2x - 1| \le 7

.

3

Simplify the compound absolute value bound and solve for

x

x \in [-3, 4]

Since an absolute value is non-negative,

|2x - 1| \ge -1

is satisfied for all real

x

. Thus, we only need

|2x - 1| \le 7

, which gives

-7 \le 2x - 1 \le 7 \implies -6 \le 2x \le 8 \implies -3 \le x \le 4

.

4

Test the solution set

S = [-3, 4]

against each given statement

Statements

|x| \le 4

and

(x + 3)(x - 4) \le 0

hold for all

x \in [-3, 4]

.

For

x \in [-3, 4]

, the extreme values of

x

yield

|x| \le 4

. Furthermore, a quadratic with roots at

-3

and

4

is non-positive on

[-3, 4]

.

Key Concept

Solving nested absolute value inequalities by systematic expansion and isolating valid intervals.

Question 30Question →

For all real numbers $x$ that satisfy the absolute value inequality $|3 - 2x| - 4 \le x$ , the rational expression $y = \frac{|x - 6|}{2 - x}$ is defined. Which of the following intervals represents the complete set of all possible real values of $y$ ?

(-\infty, 0] \cup [\frac{19}{7}, \infty)

(-\infty, -\frac{1}{5}] \cup [\frac{19}{7}, \infty)

(-\infty, 0) \cup (1, \infty)

[-\frac{1}{5}, \frac{19}{7}]

(-\infty, 1] \cup [\frac{12}{7}, \infty)

Show answer & explanation

Answer: $(-\infty, 0] \cup [\frac{19}{7}, \infty)$

Answer

The complete set of all possible real values of

y

is

(-\infty, 0] \cup [\frac{19}{7}, \infty)

.

The correct answer is

(-\infty, 0] \cup [\frac{19}{7}, \infty)

. Solving

|3 - 2x| - 4 \le x

gives

-\frac{1}{3} \le x \le 7

. Excluding

x = 2

where the denominator is zero, evaluating

y = \frac{|x - 6|}{2 - x}

on

[-\frac{1}{3}, 2)

yields

y \ge \frac{19}{7}

, on

(2, 6)

yields

y < 0

, and on

[6, 7]

yields

-\frac{1}{5} \le y \le 0

. Taking the union of these intervals gives

(-\infty, 0] \cup [\frac{19}{7}, \infty)

.

Step-by-Step Solution

1

Isolate the absolute value expression in the inequality.

|3 - 2x| \le x + 4

Adding

4

to both sides prepares the inequality for standard double-inequality solving.

2

Set up the compound inequality and solve for

x

.

-(x + 4) \le 3 - 2x \le x + 4

, yielding

x \le 7

from

-x - 4 \le 3 - 2x

, and

x \ge -\frac{1}{3}

from

3 - 2x \le x + 4

(remembering to flip the inequality sign when dividing by

-3

). Thus,

x \in [-\frac{1}{3}, 7]

.

An absolute value inequality

|A| \le B

(with

B \ge 0

) is equivalent to

-B \le A \le B

.

3

Identify domain restrictions for

y = \frac{|x - 6|}{2 - x}

.

The expression is undefined at

x = 2

. Therefore, the domain of

x

is divided into three sub-intervals:

[-\frac{1}{3}, 2)

,

(2, 6)

, and

[6, 7]

.

The denominator cannot be zero, and the absolute value

|x - 6|

changes definition at

x = 6

.

4

Analyze

y

on the first interval

[-\frac{1}{3}, 2)

.

For

x < 6

,

|x - 6| = 6 - x

. So

y = \frac{6 - x}{2 - x} = 1 + \frac{4}{2 - x}

. As

x

increases from

-\frac{1}{3}

towards

2

,

2 - x

decreases from

\frac{7}{3}

to

0^+

, so

y

increases from

1 + \frac{4}{7/3} = \frac{19}{7}

to

+\infty

. Hence

y \in [\frac{19}{7}, \infty)

.

As the positive denominator approaches zero from above, the positive fraction grows without bound towards

+\infty

.

5

Analyze

y

on the second interval

(2, 6)

.

Here

x < 6

, so

|x - 6| = 6 - x

and

y = 1 + \frac{4}{2 - x}

. As

x

increases from

2^+

to

6

,

2 - x

increases from

0^-

to

-4

. Thus

y

increases from

-\infty

up to

1 + \frac{4}{-4} = 0

. Hence

y \in (-\infty, 0)

.

As the negative denominator moves away from zero towards

-4

, the expression increases from

-\infty

to

0

.

6

Analyze

y

on the third interval

[6, 7]

.

For

x \ge 6

,

|x - 6| = x - 6

. So

y = \frac{x - 6}{2 - x} = -1 - \frac{4}{2 - x}

. At

x = 6

,

y = 0

. At

x = 7

,

y = \frac{1}{-5} = -\frac{1}{5}

. As

x

increases from

6

to

7

,

y

decreases continuously from

0

down to

-\frac{1}{5}

. Hence

y \in [-\frac{1}{5}, 0]

.

Combining

(-\infty, 0)

from the second interval and

[-\frac{1}{5}, 0]

from the third interval gives

(-\infty, 0]

.

7

Combine the ranges from all intervals.

y \in (-\infty, 0] \cup [\frac{19}{7}, \infty)

.

Taking the union of all output values across the valid domain yields the total range.

Key Concept

Solving absolute value inequalities and finding the range of rational expressions with absolute values over restricted domains.

Estimated Time:3m 0s

Question 31Question →

Which of the following inequalities represents all real values of $x$ that satisfy $-3x + 14 \le 2$ ?

x \ge 4

x \le 4

x \ge -4

x \le -4

x \ge \frac{16}{3}

Show answer & explanation

Answer: $x \ge 4$

Answer

x \ge 4

Subtracting 14 from both sides of

-3x + 14 \le 2

yields

-3x \le -12

. Dividing both sides by

-3

requires reversing the inequality sign from

\le

to

\ge

, giving

x \ge 4

. Thus, the inequality

x \ge 4

correctly represents all solution values.

Step-by-Step Solution

1

Isolate the variable term on the left side of the inequality.

Subtract 14 from both sides:

-3x \le 2 - 14

, which simplifies to

-3x \le -12

.

To solve for

x

, constant terms must first be removed from the variable side using inverse operations.

2

Divide both sides by the coefficient of

x

and apply the inequality rule for negative multipliers.

Divide by

-3

and reverse the inequality sign:

x \ge \frac{-12}{-3}

, which simplifies to

x \ge 4

.

Dividing or multiplying an inequality by a negative quantity changes the direction of the inequality sign.

Key Concept

Solving linear inequalities with negative coefficients

Estimated Time:45s

Question 32Question →

If $x$ is a real number such that $|2x - 7| \le 5$ , what is the minimum possible value of $|x - 8|$ ?

Show answer & explanation

Answer: 2

Answer

2

Solving the given inequality

|2x - 7| \le 5

yields the compound inequality

-5 \le 2x - 7 \le 5

. Adding 7 across the inequality gives

2 \le 2x \le 12

, which simplifies to

1 \le x \le 6

. Geometrically,

|x - 8|

represents the distance between

x

and 8 on the real number line. To minimize this distance for any

x

in the closed interval

[1, 6]

, we select the point in

[1, 6]

closest to 8, which is

x = 6

. Evaluating at

x = 6

produces

|6 - 8| = 2

.

Step-by-Step Solution

1

Unpack the absolute value inequality

1 \le x \le 6

The inequality

|2x - 7| \le 5

is equivalent to

-5 \le 2x - 7 \le 5

. Adding 7 gives

2 \le 2x \le 12

, and dividing by 2 yields

1 \le x \le 6

.

2

Determine the value in the domain

[1, 6]

that minimizes

|x - 8|

x = 6

The expression

|x - 8|

measures the distance from

x

to 8 on the number line. The value within

[1, 6]

nearest to 8 is

x = 6

.

3

Evaluate the expression at

x = 6

2

Substituting

x = 6

into

|x - 8|

gives

|6 - 8| = |-2| = 2

.

Key Concept

Properties of Linear Inequalities and Absolute Value as Distance

Estimated Time:1m 30s

Question 33Question →

For all real numbers $x$ satisfying the absolute value inequality $|4x - 12| \le 20$ , the maximum possible value of the expression $|2 - 3x|$ is $M$ . What is the value of $M$ ?

Show answer & explanation

Answer: 22

Answer

The maximum possible value

M

of the expression

|2 - 3x|

on the domain

-2 \le x \le 8

is 22.

Solving

|4x - 12| \le 20

yields

-20 \le 4x - 12 \le 20

, which simplifies to

-2 \le x \le 8

. Evaluating

|2 - 3x|

over this interval gives a minimum of

0

(at

x = 2/3

) and endpoint values of

|2 - 3(-2)| = 8

and

|2 - 3(8)| = |-22| = 22

. Thus, the maximum value

M

is 22.

Step-by-Step Solution

1

Unfold the given absolute value inequality into a compound linear inequality.

-20 \le 4x - 12 \le 20

The inequality

|u| \le k

for

k \ge 0

is equivalent to

-k \le u \le k

.

2

Isolate the variable

x

by adding 12 and dividing by 4.

-2 \le x \le 8

Adding 12 gives

-8 \le 4x \le 32

. Dividing by positive 4 preserves inequality signs, yielding

-2 \le x \le 8

.

3

Evaluate the target expression

|2 - 3x|

at the boundary points of the interval

[-2, 8]

.

For

x = -2

:

|2 - 3(-2)| = |8| = 8

. For

x = 8

:

|2 - 3(8)| = |-22| = 22

.

The expression

f(x) = |2 - 3x|

is convex and non-negative, reaching its local minimum of 0 at

x = \frac{2}{3}

. Its maximum over a closed interval must occur at one of the endpoints.

4

Compare the evaluated values to find the maximum

M

.

M = \max(8, 22) = 22

Comparing 8 and 22 shows that 22 is the absolute maximum value achievable within the domain.

Key Concept

Solving linear absolute value inequalities to determine variable bounds and evaluating extreme values of absolute value expressions.

Estimated Time:2m 0s

Question 34Question →

How many integer values of $x$ satisfy the inequality $|3 - 2x| + |x + 4| \le 9$ ?

3

4

5

6

7

Show answer & explanation

Answer: 5

Answer

5 integer values (specifically -2, -1, 0, 1, and 2)

Solving the absolute value inequality using piecewise intervals yields the continuous solution set

[-2, \frac{8}{3}]

. The integers contained in this range are

-2, -1, 0, 1,

and

2

, giving a total of 5 integer values.

Step-by-Step Solution

1

Identify the critical points of the absolute value expressions.

The critical points are

x = \frac{3}{2}

and

x = -4

. These split the real number line into three intervals:

x < -4

,

-4 \le x \le \frac{3}{2}

, and

x > \frac{3}{2}

.

Absolute value expressions change definition at their zeroes.

2

Analyze Case 1:

x < -4

.

Here

|3 - 2x| = 3 - 2x

and

|x + 4| = -(x + 4)

. The inequality becomes

(3 - 2x) - (x + 4) \le 9 \implies -3x - 1 \le 9 \implies -3x \le 10 \implies x \ge -\frac{10}{3} \approx -3.33

. Since there is no overlap between

x < -4

and

x \ge -3.33

, no solutions exist in this interval.

Evaluating expressions according to the sign of terms when

x < -4

.

3

Analyze Case 2:

-4 \le x \le \frac{3}{2}

.

Here

|3 - 2x| = 3 - 2x

and

|x + 4| = x + 4

. The inequality becomes

(3 - 2x) + (x + 4) \le 9 \implies 7 - x \le 9 \implies -x \le 2 \implies x \ge -2

. Combining with the case interval gives

[-2, \frac{3}{2}]

.

Determining valid values of

x

within the middle interval.

4

Analyze Case 3:

x > \frac{3}{2}

.

Here

|3 - 2x| = 2x - 3

and

|x + 4| = x + 4

. The inequality becomes

(2x - 3) + (x + 4) \le 9 \implies 3x + 1 \le 9 \implies 3x \le 8 \implies x \le \frac{8}{3} \approx 2.67

. Combining with the case interval gives

(\frac{3}{2}, \frac{8}{3}]

.

Determining valid values of

x

within the upper interval.

5

Combine solution intervals and count integer solutions.

The total solution set is

[-2, \frac{8}{3}]

. The integer values within this interval are

-2, -1, 0, 1,

and

2

. Total count = 5.

Identifying all integer values within the bounded set

[-2, 2.67]

.

Key Concept

Solving absolute value inequalities with multiple absolute value terms using critical points and case analysis.

Estimated Time:2m 0s

Question 35Question →

What is the sum of all integer values of $x$ that satisfy both $|2x - 5| \le 9$ and $|x + 2| \ge 4$ ?

Show answer & explanation

Answer: 27

Answer

The sum of all integer values of

x

that satisfy both inequalities is 27.

First, solving

|2x - 5| \le 9

yields

-9 \le 2x - 5 \le 9 \implies -4 \le 2x \le 14 \implies -2 \le x \le 7

. Second, solving

|x + 2| \ge 4

yields

x + 2 \ge 4 \implies x \ge 2

or

x + 2 \le -4 \implies x \le -6

. Taking the intersection of

-2 \le x \le 7

and

(x \ge 2 \text{ or } x \le -6)

gives the solution set

2 \le x \le 7

. The integer values satisfying this condition are 2, 3, 4, 5, 6, and 7. Summing these integers gives

2 + 3 + 4 + 5 + 6 + 7 = 27

.

Step-by-Step Solution

1

Solve the inequality

|2x - 5| \le 9

-2 \le x \le 7

An inequality of the form

|u| \le a

(where

a \ge 0

) expands to

-a \le u \le a

. Adding 5 gives

-4 \le 2x \le 14

, and dividing by 2 yields

-2 \le x \le 7

.

2

Solve the inequality

|x + 2| \ge 4

x \ge 2 \text{ or } x \le -6

An inequality of the form

|u| \ge a

(where

a > 0

) expands to

u \ge a

or

u \le -a

. Subtracting 2 from both inequalities yields

x \ge 2

or

x \le -6

.

3

Determine the overlapping interval for both inequalities

2 \le x \le 7

Combining

-2 \le x \le 7

with

x \ge 2 \text{ or } x \le -6

eliminates

x \le -6

. The intersection of

[-2, 7]

and

[2, \infty)

is

[2, 7]

.

4

Identify the integer values in the solution interval and calculate their sum

27

The integers in the closed interval

[2, 7]

are 2, 3, 4, 5, 6, and 7. Adding them together gives

2 + 3 + 4 + 5 + 6 + 7 = 27

.

Key Concept

System of Linear Absolute Value Inequalities

Question 36Question →

For how many integer values of $k$ does the inequality $|2x - k| + |x + 3| \le 7$ have at least one real solution $x$ such that $x \ge 1$ ?

8

9

10

11

13

Show answer & explanation

Answer: 10

Answer

10 integer values of

k

satisfy the given condition.

The correct answer is 10 because analyzing the condition

x \ge 1

simplifies the inequality to

|2x - k| \le 4 - x

, requiring

x \in [1, 4]

. The double inequality

k - 4 \le x \le \frac{k + 4}{3}

yields solutions overlapping with

[1, 4]

if and only if

-1 \le k \le 8

, which contains exactly 10 integers.

Step-by-Step Solution

1

Simplify the absolute value term

|x + 3|

using the given condition

x \ge 1

.

Since

x \ge 1

,

x + 3 > 0

, so

|x + 3| = x + 3

. The inequality becomes

|2x - k| + x + 3 \le 7

, which simplifies to

|2x - k| \le 4 - x

.

Establishing the sign of

x + 3

allows eliminating one set of absolute value bars.

2

Determine the valid range for

x

.

Since

|2x - k| \ge 0

, it must hold that

4 - x \ge 0

, which implies

x \le 4

. Combined with

x \ge 1

, any solution

x

must lie in the interval

[1, 4]

.

An absolute value quantity cannot be less than a negative number.

3

Unwrap the absolute value inequality

|2x - k| \le 4 - x

.

-(4 - x) \le 2x - k \le 4 - x

. Splitting into two linear inequalities:
1)

2x - k \le 4 - x \implies 3x \le k + 4 \implies x \le \frac{k + 4}{3}

.
2)

2x - k \ge x - 4 \implies x \ge k - 4

.
Thus,

k - 4 \le x \le \frac{k + 4}{3}

.

Rewriting absolute value inequalities as compound inequalities defines explicit bounds on

x

in terms of

k

.

4

Find the range of

k

for which

[k - 4, \frac{k + 4}{3}]

overlaps with

[1, 4]

.

For an overlapping solution to exist in

[1, 4]

:
1) The upper bound

\frac{k + 4}{3}

must be at least 1:

\frac{k + 4}{3} \ge 1 \implies k \ge -1

.
2) The lower bound

k - 4

must be at most 4:

k - 4 \le 4 \implies k \le 8

.
Combining these gives

-1 \le k \le 8

.

The solution interval for

x

must have a non-empty intersection with the allowed domain

[1, 4]

.

5

Count the total number of integer values of

k

in the interval

[-1, 8]

.

The integers are

-1, 0, 1, 2, 3, 4, 5, 6, 7, 8

, giving a total of

8 - (-1) + 1 = 10

integer values.

Counting inclusive integer endpoints gives the total count.

Key Concept

Solving absolute value inequalities involving parameters and restricted variable domains.

Estimated Time:2m 0s

Question 37Question →

If $x$ is a real number that satisfies the inequality $|3 - 2x| \le 7$ , what is the maximum possible value of $5 - 3x$ ?

-10

-1

5

11

20

Show answer & explanation

Answer: 11

Answer

The maximum possible value of

5 - 3x

is

11

.

Solving

|3 - 2x| \le 7

gives

-7 \le 3 - 2x \le 7

. Subtracting

3

yields

-10 \le -2x \le 4

. Dividing by

-2

and flipping the inequality signs gives

-2 \le x \le 5

. Because

5 - 3x

has a negative coefficient for

x

, it decreases as

x

increases. Therefore, the maximum value of

5 - 3x

occurs at the smallest value in the domain,

x = -2

. Substituting

x = -2

into

5 - 3x

gives

5 - 3(-2) = 11

.

Step-by-Step Solution

1

Rewrite the absolute value inequality

|3 - 2x| \le 7

as a compound inequality.

-7 \le 3 - 2x \le 7

An absolute value inequality of the form

|u| \le k

(where

k \ge 0

) is equivalent to

-k \le u \le k

.

2

Isolate

x

by subtracting

3

from all parts of the inequality and then dividing by

-2

.

-10 \le -2x \le 4 \implies 5 \ge x \ge -2

, which is equivalent to

-2 \le x \le 5

Dividing an inequality by a negative number reverses the direction of the inequality signs.

3

Determine which value of

x

in the interval

-2 \le x \le 5

maximizes the linear expression

5 - 3x

.

The expression reaches its maximum at the minimum bound

x = -2

.

Since the coefficient of

x

is negative (

-3

), smaller values of

x

result in larger values for

5 - 3x

.

4

Substitute

x = -2

into

5 - 3x

.

5 - 3(-2) = 5 + 6 = 11

Evaluating the linear expression at the lower bound yields its maximum value.

Key Concept

Solving linear absolute value inequalities and optimizing linear expressions over a closed interval.

Estimated Time:1m 30s

Question 38Question →

If $x$ is an integer that satisfies both $|2x - 9| \le 15$ and $\frac{5 - 2x}{-3} \le 3$ , what is the product of the smallest and largest possible values of $x$ ?

-21

84

35

-36

21

Show answer & explanation

Answer: $-21$

Answer

The product of the smallest and largest possible integer values of

x

is

-21

.

Solving

|2x - 9| \le 15

gives

-3 \le x \le 12

. Solving

\frac{5 - 2x}{-3} \le 3

requires reversing the inequality sign twice (first when multiplying by

-3

, then when dividing by

-2

), which yields

x \le 7

. Taking the intersection of both intervals gives

-3 \le x \le 7

. The smallest integer in this range is

-3

and the largest is

7

, giving a product of

(-3) \times 7 = -21

.

Step-by-Step Solution

1

Solve the absolute value inequality

|2x - 9| \le 15

.

-15 \le 2x - 9 \le 15 \implies -6 \le 2x \le 24 \implies -3 \le x \le 12

.

An absolute value inequality

|u| \le a

(for

a \ge 0

) is equivalent to the compound inequality

-a \le u \le a

.

2

Solve the linear inequality

\frac{5 - 2x}{-3} \le 3

.

Multiply by

-3

and flip the inequality direction:

5 - 2x \ge -9

. Subtract

5

:

-2x \ge -14

. Divide by

-2

and flip the inequality direction again:

x \le 7

.

Multiplying or dividing an inequality by a negative number reverses the direction of the inequality sign.

3

Determine the intersection of the two solution sets.

Combining

-3 \le x \le 12

and

x \le 7

gives

-3 \le x \le 7

.

The value of

x

must satisfy both conditions simultaneously.

4

Identify the smallest and largest integer values of

x

and calculate their product.

Smallest integer

= -3

, largest integer

= 7

. Product

= (-3) \times 7 = -21

.

Both endpoints

-3

and

7

are included in the closed interval

[-3, 7]

.

Key Concept

Solving compound linear inequalities involving absolute values and applying the rule for reversing inequality signs when multiplying or dividing by negative numbers.

Estimated Time:2m 0s

Question 39Question →

If $x$ is a real number that satisfies the inequality $3|x - 4| + 2|x + 5| \le 26$ , what is the maximum possible value of $|x - 7|$ ?

Show answer & explanation

Answer: 11

Answer

11

Solving the piecewise linear inequality

3|x - 4| + 2|x + 5| \le 26

yields the interval

[-4, 5.6]

. The distance function

|x - 7|

reaches its maximum at the endpoint farthest from

7

, which is

x = -4

. Evaluating

|-4 - 7|

yields 11.

Step-by-Step Solution

1

Identify the critical points of the absolute value terms.

The critical points are

x = -5

(where

x + 5 = 0

) and

x = 4

(where

x - 4 = 0

).

Critical points mark where the linear expressions inside the absolute values change sign.

2

Analyze the inequality piecewise across the three regions defined by the critical points.

For

x < -5

:

3(4 - x) + 2(-5 - x) \le 26 \implies 2 - 5x \le 26 \implies x \ge -4.8

. This produces no solution since

x

cannot be simultaneously

< -5

and

\ge -4.8

.
For

-5 \le x < 4

:

3(4 - x) + 2(x + 5) \le 26 \implies 22 - x \le 26 \implies x \ge -4

, yielding

-4 \le x < 4

.
For

x \ge 4

:

3(x - 4) + 2(x + 5) \le 26 \implies 5x - 2 \le 26 \implies x \le 5.6

, yielding

4 \le x \le 5.6

.

Expanding absolute value terms according to their regional sign definitions removes the absolute values.

3

Combine the valid regional solutions to establish the complete solution interval for

x

.

The set of all satisfying real numbers is

x \in [-4, 5.6]

.

Taking the union of the non-empty piecewise solution intervals yields the total solution set.

4

Find the maximum value of

|x - 7|

over

x \in [-4, 5.6]

.

At

x = -4

,

|-4 - 7| = |-11| = 11

. At

x = 5.6

,

|5.6 - 7| = |-1.4| = 1.4

. The maximum possible value is 11.

The expression

|x - 7|

measures distance from

7

. The maximum distance on a closed interval occurs at the endpoint farthest from

7

, which is

x = -4

.

Key Concept

Piecewise analysis of linear absolute value inequalities and optimization of absolute value distance functions.

Estimated Time:2m 30s

Question 40Question →

If $x$ and $y$ are real numbers that satisfy the absolute value inequalities $|2x - 6| \le 4$ and $|y + 3| \le 5$ , which of the following values could be equal to the product $xy$ ? Indicate all such values.

Select all that apply

-45

-35

0

8

15

Show answer & explanation

Answer: $-35$ ; $0$ ; $8$

Answer

The possible values for the product

xy

are

-35

,

0

, and

8

.

Solving

|2x - 6| \le 4

yields

1 \le x \le 5

, and solving

|y + 3| \le 5

yields

-8 \le y \le 2

. The product

xy

attains its minimum at

5 \times (-8) = -40

and its maximum at

5 \times 2 = 10

. Since

xy

can take any value in the continuous interval

[-40, 10]

, the values

-35

,

0

, and

8

are all valid choices.

Step-by-Step Solution

1

Solve the inequality

|2x - 6| \le 4

for

x

.

-4 \le 2x - 6 \le 4 \implies 2 \le 2x \le 10 \implies 1 \le x \le 5

.

Unfold the absolute value into a compound inequality to determine the valid range for

x

.

2

Solve the inequality

|y + 3| \le 5

for

y

.

-5 \le y + 3 \le 5 \implies -8 \le y \le 2

.

Unfold the absolute value into a compound inequality to determine the valid range for

y

.

3

Determine the minimum and maximum possible values of the product

xy

.

Evaluating the extreme product combinations of endpoints:

1 \times (-8) = -8

,

1 \times 2 = 2

,

5 \times (-8) = -40

, and

5 \times 2 = 10

. Thus,

-40 \le xy \le 10

.

The continuous product of two real intervals

[a, b]

and

[c, d]

spans from the minimum endpoint product to the maximum endpoint product.

4

Select all options that fall within the interval

[-40, 10]

.

The values

-35

,

0

, and

8

lie within

[-40, 10]

, while

-45

and

15

fall outside.

Any real number within the closed interval

[-40, 10]

can be formed by valid choices of

x

and

y

.

Key Concept

Determining the range of a product from two independent absolute value inequalities.

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