Linear Inequalities and Absolute Value Practice Questions — GRE General Test

Question 1Question →

If $x$ is a real number that satisfies the compound absolute value inequality $||2x - 5| - 7| \le 4$ , which of the following values could be the value of $x$ ? Select all such values.

Select all that apply

-2

0

3

6

9

Show answer & explanation

Answer: $-2$ ; $0$ ; $6$

Answer

The real numbers

-2

,

0

, and

6

satisfy the inequality.

The solution set to the compound inequality

||2x - 5| - 7| \le 4

is the union of two intervals:

[-3, 1]

and

[4, 8]

. Among the options provided,

-2

and

0

fall into the interval

[-3, 1]

, and

6

falls into the interval

[4, 8]

. Therefore, these three values satisfy the original inequality.

Step-by-Step Solution

1

Unfold the outer absolute value inequality.

-4 \le |2x - 5| - 7 \le 4

For any expression

U

and constant

c \ge 0

,

|U| \le c

is equivalent to

-c \le U \le c

.

2

Isolate the inner absolute value term by adding

7

across all parts.

3 \le |2x - 5| \le 11

Adding a constant preserves inequality directions.

3

Break the compound inequality into two separate absolute value conditions.

Condition 1:

|2x - 5| \le 11

; Condition 2:

|2x - 5| \ge 3

Both conditions must hold simultaneously for

x

.

4

Solve Condition 1 (

|2x - 5| \le 11

).

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

Expanding the bounded absolute value inequality and solving for

x

.

5

Solve Condition 2 (

|2x - 5| \ge 3

).

2x - 5 \ge 3

or

2x - 5 \le -3 \implies 2x \ge 8

or

2x \le 2 \implies x \ge 4

or

x \le 1

For

|U| \ge c

,

U \ge c

or

U \le -c

.

6

Find the intersection of the two solution sets.

x \in [-3, 1] \cup [4, 8]

Intersecting

[-3, 8]

with

(-\infty, 1] \cup [4, \infty)

yields

[-3, 1] \cup [4, 8]

.

7

Test the provided choices against the solution set

[-3, 1] \cup [4, 8]

.

-2 \in [-3, 1]

(valid),

0 \in [-3, 1]

(valid),

3 \notin [-3, 1] \cup [4, 8]

(invalid),

6 \in [4, 8]

(valid),

9 \notin [-3, 1] \cup [4, 8]

(invalid).

Values lying within the solution intervals satisfy the original inequality.

Key Concept

Solving nested absolute value inequalities using multi-step compound interval intersections.

Question 2Question →

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

-21

-1

7

27

-27

Show answer & explanation

Answer: $-21$

Answer

-21

Solving

|5 - 2x| \le 7

yields

-7 \le 5 - 2x \le 7

. Subtracting

5

gives

-12 \le -2x \le 2

. Dividing by

-2

and flipping the inequality direction gives

-1 \le x \le 6

. To minimize

3 - 4x

, we choose the maximum value of

x

because the term

-4x

decreases as

x

increases. Substituting

x = 6

yields

3 - 4(6) = -21

.

Step-by-Step Solution

1

Unpack the absolute value inequality into a compound inequality

-7 \le 5 - 2x \le 7

For any real constant

k \ge 0

,

|u| \le k

is equivalent to

-k \le u \le k

.

2

Isolate the variable term by subtracting

5

from all parts of the inequality

-12 \le -2x \le 2

Subtracting a constant from all parts preserves inequality direction.

3

Divide all parts by

-2

and reverse the inequality signs

6 \ge x \ge -1

, which is equivalent to

-1 \le x \le 6

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

4

Find the minimum value of

3 - 4x

over the interval

[-1, 6]

The minimum value occurs at

x = 6

:

3 - 4(6) = 3 - 24 = -21

Because the linear expression

3 - 4x

has a negative coefficient for

x

, it is a decreasing function; its minimum occurs at the largest allowed value of

x

.

Key Concept

Linear Inequalities and Absolute Value Bounds

Estimated Time:1m 30s

Question 3Question →

If $|3x - 6| \le 12$ , which of the following inequality ranges represents all possible values of $x$ ?

-2 \le x \le 6

x \le 6

-6 \le x \le 2

-6 \le x \le 6

-2 \le x \le 2

Show answer & explanation

Answer: $-2 \le x \le 6$

Answer

-2 \le x \le 6

The expression

|3x - 6| \le 12

expands to the double inequality

-12 \le 3x - 6 \le 12

. Adding

6

to each part produces

-6 \le 3x \le 18

. Dividing all terms by

3

yields

-2 \le x \le 6

, which matches the correct solution range.

Step-by-Step Solution

1

Rewrite the absolute value inequality as a compound inequality.

-12 \le 3x - 6 \le 12

An absolute value inequality of the form

|u| \le k

(where

k \ge 0

) is equivalent to

-k \le u \le k

.

2

Add

6

to all three parts of the inequality.

-6 \le 3x \le 18

Isolate the term containing

x

.

3

Divide all three parts of the inequality by

3

.

-2 \le x \le 6

Since

3 > 0

, dividing by

3

isolates

x

without flipping the inequality signs.

Key Concept

Solving Linear Absolute Value Inequalities

Estimated Time:45s

Question 4Question →

If $|2x - 4| \le 6$ , which of the following values could be a solution for $x$ ? Select all that apply.

Select all that apply

-3

-1

2

5

6

Show answer & explanation

Answer: $-1$ ; $2$ ; $5$

Answer

The values

-1

,

2

, and

5

are solutions to the inequality.

Solving

|2x - 4| \le 6

yields

-6 \le 2x - 4 \le 6

. Adding

4

gives

-2 \le 2x \le 10

, and dividing by

2

gives

-1 \le x \le 5

. The options stating

-1

,

2

, and

5

are all within this interval

[-1, 5]

.

Step-by-Step Solution

1

Rewrite the absolute value inequality as a compound inequality.

-6 \le 2x - 4 \le 6

An inequality of the form

|u| \le c

(where

c \ge 0

) is equivalent to

-c \le u \le c

.

2

Add

4

to all three parts of the compound inequality.

-2 \le 2x \le 10

Isolate the variable term

2x

in the middle.

3

Divide all three parts by

2

.

-1 \le x \le 5

Isolate

x

. Since

2

is positive, the inequality signs remain unchanged.

4

Test which of the given choices fall within the interval

[-1, 5]

.

-1

,

2

, and

5

lie within the range

[-1, 5]

, while

-3

and

6

lie outside.

Values equal to or between

-1

and

5

inclusive are valid solutions.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:1m 0s

Question 5Question →

If $x$ satisfies the inequality $|3 - 2x| > 9$ , which of the following could be the value of $x$ ? Select all that apply.

Select all that apply

-5

-3

0

4

8

Show answer & explanation

Answer: $-5$ ; $8$

Answer

The values

-5

and

8

satisfy the inequality.

The absolute value inequality

|3 - 2x| > 9

splits into

3 - 2x > 9

or

3 - 2x < -9

. Solving these yields

x < -3

or

x > 6

. The value

-5

is less than

-3

and the value

8

is greater than

6

, so both are valid solutions.

Step-by-Step Solution

1

Set up the two separate linear inequalities based on the definition of absolute value.

3 - 2x > 9

or

3 - 2x < -9

An absolute value expression

|u| > c

(where

c > 0

) breaks into two disjoint cases:

u > c

or

u < -c

.

2

Solve the first case:

3 - 2x > 9

.

-2x > 6 \implies x < -3

Subtract 3 from both sides to get

-2x > 6

, then divide by

-2

and reverse the inequality sign.

3

Solve the second case:

3 - 2x < -9

.

-2x < -12 \implies x > 6

Subtract 3 from both sides to get

-2x < -12

, then divide by

-2

and reverse the inequality sign.

4

Combine the solution sets and evaluate the given options.

The valid solution set is

x < -3

or

x > 6

. Among the choices,

-5

(since

-5 < -3

) and

8

(since

8 > 6

) fall into the solution set.

Values between

-3

and

6

, inclusive, do not satisfy the original inequality.

Key Concept

Solving absolute value inequalities of the form

|u| > c

by splitting into compound inequalities and reversing the direction of inequality signs when dividing by negative numbers.

Estimated Time:1m 15s

Question 6Question →

If $x$ and $y$ are real numbers satisfying the inequalities $|2x - 7| \le 5$ and $|3y + 2| < 8$ , which of the following inequalities must be true? Select all that apply.

Select all that apply

x - y > -1

xy < 12

x + y > 0

|y| < 3

y - x > -3

Show answer & explanation

Answer: $x - y > -1$ ; $xy < 12$

Answer

The inequalities that must be true are

x - y > -1

and

xy < 12

.

Solving

|2x - 7| \le 5

yields

1 \le x \le 6

, and solving

|3y + 2| < 8

yields

-\frac{10}{3} < y < 2

. Combining

x \ge 1

with

-y > -2

gives

x - y > -1

, which is always true. Furthermore, since

x

is positive and bounded above by

6

while

y

is bounded above by

2

, the product

xy

must be strictly less than

12

.

Step-by-Step Solution

1

Solve the absolute value inequality for

x

.

-5 \le 2x - 7 \le 5 \implies 2 \le 2x \le 12 \implies 1 \le x \le 6

.

Unfolding the absolute value

|2x - 7| \le 5

gives a compound linear inequality.

2

Solve the absolute value inequality for

y

.

-8 < 3y + 2 < 8 \implies -10 < 3y < 6 \implies -\frac{10}{3} < y < 2

.

Unfolding the absolute value

|3y + 2| < 8

gives a strict compound linear inequality.

3

Evaluate the inequality

x - y > -1

.

Since

x \ge 1

and

y < 2 \implies -y > -2

, adding the inequalities gives

x + (-y) > 1 + (-2) = -1

.

This establishes that

x - y > -1

is always true.

4

Evaluate the inequality

xy < 12

.

Since

1 \le x \le 6

(all positive) and

y < 2

, if

y > 0

,

xy < 6 \cdot 2 = 12

. If

y \le 0

,

xy \le 0 < 12

.

In all cases within the domain,

xy < 12

holds strictly.

5

Test counterexamples for the remaining statements.

For

x + y > 0

,

x=1, y=-3 \implies x+y=-2 \ngtr 0

. For

|y| < 3

,

y=-3.2 \implies |-3.2|=3.2 \nless 3

. For

y - x > -3

,

x=6, y=0 \implies y-x=-6 \ngtr -3

.

Counterexamples disprove that these remaining statements must be true.

Key Concept

Linear Inequalities and Absolute Value Bounds

Question 7Question →

If $x$ is a real number that satisfies the inequality $|2x - 7| < 5$ , which of the following represents all possible values of the expression $1 - 3x$ ?

-17 < 1 - 3x < -2

-2 < 1 - 3x < 17

-17 < 1 - 3x < 1

-17 < 1 - 3x < 17

-5 < 1 - 3x < 5

Show answer & explanation

Answer: $-17 < 1 - 3x < -2$

Answer

-17 < 1 - 3x < -2

Solving

|2x - 7| < 5

gives

-5 < 2x - 7 < 5

. Adding

7

to all three parts yields

2 < 2x < 12

, which simplifies to

1 < x < 6

. Multiplying this inequality by

-3

reverses the direction of the inequalities, resulting in

-18 < -3x < -3

. Adding

1

to each part gives

-17 < 1 - 3x < -2

. Therefore, the range of possible values for the expression is strictly between

-17

and

-2

.

Step-by-Step Solution

1

Express the absolute value inequality as a compound inequality.

-5 < 2x - 7 < 5

An inequality of the form

|u| < k

for

k > 0

is equivalent to

-k < u < k

.

2

Isolate

x

in the compound inequality.

1 < x < 6

Add

7

to all parts to get

2 < 2x < 12

, then divide all parts by

2

to obtain

1 < x < 6

.

3

Multiply the compound inequality by

-3

.

-18 < -3x < -3

Multiplying an inequality by a negative number reverses the direction of the inequality signs:

-3(6) < -3(x) < -3(1)

.

4

Add

1

to all parts of the compound inequality.

-17 < 1 - 3x < -2

Adding a constant to an inequality preserves the inequality direction.

Key Concept

Linear Inequalities and Absolute Value Transformations

Estimated Time:1m 30s

Question 8Question →

If $x$ is a real number that satisfies the inequality $||2x - 5| - 7| \le 4$ , and $y = |3 - x|$ , what is the difference between the maximum possible value and the minimum possible value of $y$ ?

3

4

5

6

7

Show answer & explanation

Answer: 5

Answer

The difference between the maximum possible value and the minimum possible value of

y

is 5.

Solving

||2x - 5| - 7| \le 4

gives

3 \le |2x - 5| \le 11

, which restricts

x

to the disconnected domain

[-3, 1] \cup [4, 8]

. Evaluating

y = |3 - x|

across these intervals gives a maximum value of 6 (at

x = -3

) and a minimum value of 1 (at

x = 4

). The difference between the maximum and minimum values is

6 - 1 = 5

.

Step-by-Step Solution

1

Unpack the outer absolute value inequality

||2x - 5| - 7| \le 4

.

-4 \le |2x - 5| - 7 \le 4 \implies 3 \le |2x - 5| \le 11

.

An absolute value inequality

|u| \le a

(for

a \ge 0

) is equivalent to

-a \le u \le a

.

2

Solve the double inequality

3 \le |2x - 5| \le 11

by splitting it into two conditions.

Condition 1:

|2x - 5| \le 11 \implies -11 \le 2x - 5 \le 11 \implies -3 \le x \le 8

.
Condition 2:

|2x - 5| \ge 3 \implies 2x - 5 \ge 3

or

2x - 5 \le -3 \implies x \ge 4

or

x \le 1

.

The quantity

|2x - 5|

must simultaneously satisfy upper and lower absolute value bounds.

3

Intersect Condition 1 and Condition 2 to determine the complete domain of

x

.

x \in [-3, 1] \cup [4, 8]

.

Values in the open interval

(1, 4)

make

|2x - 5| < 3

and must be excluded from the domain.

4

Evaluate the range of

y = |3 - x|

over the valid domain of

x

.

On

[-3, 1]

,

y = 3 - x

decreases from

3 - (-3) = 6

to

3 - 1 = 2

, giving

y \in [2, 6]

.
On

[4, 8]

,

y = x - 3

increases from

4 - 3 = 1

to

8 - 3 = 5

, giving

y \in [1, 5]

.
The complete range of

y

is

[1, 6]

.

Combining the output ranges of both disjoint intervals yields all possible values for

y

.

5

Calculate the difference between the maximum and minimum values of

y

.

Maximum

y = 6

, Minimum

y = 1

, Difference =

6 - 1 = 5

.

Subtracting the minimum value 1 from the maximum value 6 gives the required difference.

Key Concept

Solving compound nested absolute value inequalities and finding the extreme values of a transformed function over disconnected solution intervals.

Estimated Time:2m 0s

Question 9Question →

If $x$ is an integer that satisfies both $|5 - 2x| \le 9$ and $|x + 1| > 3$ , what is the least possible value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The least possible value of

x

is 3.

Solving

|5 - 2x| \le 9

gives

-2 \le x \le 7

. Solving

|x + 1| > 3

gives

x > 2

or

x < -4

. Taking the intersection of both regions yields

2 < x \le 7

. The integer values satisfying this combined inequality are

3, 4, 5, 6,

and

7

. The smallest among these integer values is

3

.

Step-by-Step Solution

1

Solve the first absolute value inequality

|5 - 2x| \le 9

-2 \le x \le 7

Remove the absolute value bars to set up the compound inequality

-9 \le 5 - 2x \le 9

. Subtracting

5

from all parts gives

-14 \le -2x \le 4

. Dividing all parts by

-2

and reversing the inequality signs yields

7 \ge x \ge -2

, or

-2 \le x \le 7

.

2

Solve the second absolute value inequality

|x + 1| > 3

x > 2

or

x < -4

Remove the absolute value bars to create two separate cases:

x + 1 > 3 \implies x > 2

, or

x + 1 < -3 \implies x < -4

.

3

Determine the set of values that satisfy both inequalities simultaneously

2 < x \le 7

The intersection of the interval

[-2, 7]

and

(-\infty, -4) \cup (2, \infty)

is

(2, 7]

, because

x < -4

does not overlap with

[-2, 7]

.

4

Find the smallest integer within the interval

(2, 7]

3

The integers included in the interval

(2, 7]

are

3, 4, 5, 6,

and

7

. Note that

2

is excluded due to the strict inequality

x > 2

. Therefore, the least possible integer value is

3

.

Key Concept

Linear Inequalities and Absolute Value

Question 10Question →

If $x$ is a real number that satisfies both $|3x - 12| \le 9$ and $|2 - x| \ge 4$ , the maximum possible value of the expression $5 - 2x$ is $M$ and the minimum possible value is $m$ . What is the value of $M - m$ ?

Show answer & explanation

Answer: 2

Answer

The value of

M - m

is 2.

Solving

|3x - 12| \le 9

yields

1 \le x \le 7

. Solving

|2 - x| \ge 4

yields

x \le -2

or

x \ge 6

. The set of

x

-values satisfying both conditions is the intersection

[6, 7]

. Since

5 - 2x

is a linear expression with a negative coefficient, its maximum value

M

occurs at the smallest value of

x

(

x = 6

), giving

M = 5 - 2(6) = -7

. Its minimum value

m

occurs at the largest value of

x

(

x = 7

), giving

m = 5 - 2(7) = -9

. Thus,

M - m = -7 - (-9) = 2

.

Step-by-Step Solution

1

Solve the inequality

|3x - 12| \le 9

1 \le x \le 7

Expanding absolute value yields

-9 \le 3x - 12 \le 9

. Adding 12 gives

3 \le 3x \le 21

, then dividing by 3 yields

1 \le x \le 7

.

2

Solve the inequality

|2 - x| \ge 4

x \le -2

or

x \ge 6

Absolute value inequality

|u| \ge a

splits into

u \ge a

or

u \le -a

. Here

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

, and

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

.

3

Determine the overlapping domain for

x

6 \le x \le 7

Combining

1 \le x \le 7

with

x \le -2

or

x \ge 6

leaves only the interval

6 \le x \le 7

.

4

Evaluate maximum

M

and minimum

m

of

5 - 2x

on

6 \le x \le 7

M = -7

and

m = -9

Since

-2x

decreases as

x

increases, the maximum occurs at

x = 6

(

M = 5 - 12 = -7

) and the minimum occurs at

x = 7

(

m = 5 - 14 = -9

).

5

Calculate the difference

M - m

2

Subtracting

m

from

M

gives

-7 - (-9) = 2

.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:2m 0s

Question 11Question →

If $x$ is an integer that satisfies the inequality $|6 - 3x| - 4 \le 5$ , which of the following could be the value of $x$ ? Select all such values.

Select all that apply

-3

-1

2

4

7

Show answer & explanation

Answer: $-1$ ; $2$ ; $4$

Answer

The correct values of

x

are

-1

,

2

, and

4

.

Isolating the absolute value yields

|6 - 3x| \le 9

, which expands to

-9 \le 6 - 3x \le 9

. Subtracting

6

gives

-15 \le -3x \le 3

. Dividing by

-3

and reversing the inequality signs results in

-1 \le x \le 5

. Among the given choices,

-1

,

2

, and

4

lie within this solution interval

[-1, 5]

.

Step-by-Step Solution

1

Isolate the absolute value expression

|6 - 3x| \le 9

Add

4

to both sides of the inequality to isolate the absolute value term.

2

Rewrite as a compound inequality

-9 \le 6 - 3x \le 9

The property

|u| \le a

(where

a \ge 0

) expands to

-a \le u \le a

.

3

Subtract 6 from all parts

-15 \le -3x \le 3

Isolate the variable term

-3x

by subtracting

6

across the compound inequality.

4

Divide by -3 and reverse inequality signs

5 \ge x \ge -1

, which is equivalent to

-1 \le x \le 5

Dividing an inequality by a negative quantity requires flipping the inequality direction.

Key Concept

Solving absolute value inequalities of the form

|ax + b| \le c

by expanding into a compound inequality

-c \le ax + b \le c

and correctly reversing inequality signs when multiplying or dividing by negative numbers.

Question 12Question →

If $x$ is a real number that satisfies the inequality $\left|\frac{5 - 2x}{3}\right| \le 3$ , what is the minimum possible value of $4 - 3x$ ?

-23

-17

-2

10

25

Show answer & explanation

Answer: $-17$

Answer

-17

To find the minimum possible value of

4 - 3x

, we first solve the inequality

\left|\frac{5 - 2x}{3}\right| \le 3

. Multiplying by

3

gives

|5 - 2x| \le 9

. This unfolds into the compound inequality

-9 \le 5 - 2x \le 9

. Subtracting

5

gives

-14 \le -2x \le 4

. Dividing all parts by

-2

requires flipping the inequality signs, resulting in

7 \ge x \ge -2

, or

-2 \le x \le 7

. Because

4 - 3x

decreases as

x

increases, the expression reaches its minimum when

x

is at its maximum value of

7

. Substituting

x = 7

yields

4 - 3(7) = -17

.

Step-by-Step Solution

1

Clear the denominator from the absolute value inequality

|5 - 2x| \le 9

Multiplying both sides of the inequality by the positive number 3 preserves the inequality direction.

2

Express the absolute value inequality as a compound inequality

-9 \le 5 - 2x \le 9

For any non-negative constant

c

,

|u| \le c

is equivalent to

-c \le u \le c

.

3

Isolate the variable term by subtracting 5 from all parts

-14 \le -2x \le 4

Subtracting a constant from all parts of a compound inequality maintains the inequality relationships.

4

Divide by -2 and reverse the inequality signs

7 \ge x \ge -2 \implies -2 \le x \le 7

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

5

Determine which bound of

x

minimizes

4 - 3x

and evaluate

Minimum value

= 4 - 3(7) = -17

The expression

4 - 3x

has a negative coefficient for

x

, making it a decreasing function. Therefore, the minimum value of

4 - 3x

occurs when

x

takes its maximum possible value (

x = 7

).

Key Concept

Linear Inequalities and Absolute Value

Question 13Question →

If $x$ is an integer that satisfies both $|2x - 5| \le 9$ and $3 - x < 5$ , how many possible values of $x$ exist?

4

8

9

10

0

Show answer & explanation

Answer: 9

Answer

There are 9 possible integer values for x.

Solving

|2x - 5| \le 9

gives

-2 \le x \le 7

. Solving

3 - x < 5

gives

x > -2

. Taking the intersection yields

-2 < x \le 7

. The integers in this interval are

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

, amounting to 9 values in total.

Step-by-Step Solution

1

Solve the absolute value inequality

|2x - 5| \le 9

-2 \le x \le 7

Unfold

|2x - 5| \le 9

as

-9 \le 2x - 5 \le 9

, add 5 to obtain

-4 \le 2x \le 14

, and divide by 2.

2

Solve the linear inequality

3 - x < 5

x > -2

Subtract 3 to get

-x < 2

, then divide by

-1

and reverse the inequality symbol.

3

Determine the intersection of both inequalities

-2 < x \le 7

Combine

-2 \le x \le 7

and

x > -2

on the real number line.

4

Count the integer values within the intersection

-2 < x \le 7

9 integer values

The valid integers are

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

, giving a total count of

7 - (-1) + 1 = 9

.

Key Concept

Linear Inequalities and Absolute Value Bounds

Estimated Time:1m 30s

Question 14Question →

How many integer values of $x$ satisfy the compound absolute value inequality $1 \le ||x - 4| - 3| \le 5$ ?

10

12

15

17

18

Show answer & explanation

Answer: 15

Answer

There are 15 integer values of

x

that satisfy the given compound inequality.

Solving

1 \le ||x - 4| - 3| \le 5

requires breaking the nested absolute value into its boundary constraints. The upper bound

\|x - 4| - 3| \le 5

restricts

x

to

[-4, 12]

. The lower bound

||x - 4| - 3| \ge 1

requires either

|x - 4| \ge 4

(giving

x \le 0

or

x \ge 8

) or

|x - 4| \le 2

(giving

2 \le x \le 6

). Taking the intersection produces three distinct inclusive integer intervals:

[-4, 0]

,

[2, 6]

, and

[8, 12]

. Each interval contains 5 integers, yielding a total of 15 integer solutions.

Step-by-Step Solution

1

Decompose the double inequality into two separate absolute value inequalities:

||x - 4| - 3| \le 5

and

||x - 4| - 3| \ge 1

.

Two simultaneous inequalities to solve for

x

.

A double inequality

a \le |u| \le b

requires satisfying both

|u| \le b

and

|u| \ge a

.

2

Solve the upper bound inequality

||x - 4| - 3| \le 5

.

-5 \le |x - 4| - 3 \le 5 \implies -2 \le |x - 4| \le 8

. Since

|x - 4| \ge 0 \ge -2

is always true, this simplifies to

|x - 4| \le 8 \implies -4 \le x \le 12

.

Absolute value is non-negative, so the lower bound of

-2

imposes no extra constraint.

3

Solve the lower bound inequality

||x - 4| - 3| \ge 1

.

This splits into two cases:

|x - 4| - 3 \ge 1

OR

|x - 4| - 3 \le -1

.
Case A:

|x - 4| \ge 4 \implies x - 4 \ge 4

or

x - 4 \le -4 \implies x \ge 8

or

x \le 0

.
Case B:

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

.

The absolute value inequality

|u| \ge 1

holds when

u \ge 1

or

u \le -1

.

4

Intersect the solution set from the upper bound

[-4, 12]

with the solution set from the lower bound

(-\infty, 0] \cup [2, 6] \cup [8, \infty)

.

The valid solution set is

x \in [-4, 0] \cup [2, 6] \cup [8, 12]

.

Both conditions must hold simultaneously.

5

Count the integer values in each of the three valid intervals.

Interval

[-4, 0]

has 5 integers:

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

.
Interval

[2, 6]

has 5 integers:

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

.
Interval

[8, 12]

has 5 integers:

\{8, 9, 10, 11, 12\}

.
Total integer solutions =

5 + 5 + 5 = 15

.

The number of integers in an inclusive integer range

[a, b]

is

b - a + 1

.

Key Concept

Linear Inequalities and Absolute Value

Question 15Question →

If $x$ is a real number such that $|4 - 3x| \leq 13$ , and $y$ is an integer such that $-5 < \frac{1 - 2y}{3} \leq 3$ , what is the least possible integer value of $x^2 - y$ ?

-8

-7

-4

2

9

Show answer & explanation

Answer: $-7$

Answer

The least possible integer value of

x^2 - y

is

-7

.

To find the minimum value of

x^2 - y

, we must minimize

x^2

and maximize

y

. The absolute value inequality

|4 - 3x| \leq 13

simplifies to

-3 \leq x \leq \frac{17}{3}

, which contains

0

, so the minimum of

x^2

is

0

. The inequality

-5 < \frac{1 - 2y}{3} \leq 3

simplifies to

-4 \leq y < 8

. Since

y

is an integer, its maximum value is

7

. Therefore, the minimum possible value of

x^2 - y

is

0 - 7 = -7

.

Step-by-Step Solution

1

Solve the absolute value inequality

|4 - 3x| \leq 13

for

x

.

-13 \leq 4 - 3x \leq 13 \implies -17 \leq -3x \leq 9 \implies \frac{17}{3} \geq x \geq -3 \implies -3 \leq x \leq \frac{17}{3}

.

An absolute value inequality

|u| \leq k

expands to

-k \leq u \leq k

. Dividing by a negative number flips the inequality signs.

2

Determine the minimum possible value of

x^2

.

Minimum

x^2 = 0

.

Since

x

can take any real value in the interval

[-3, \frac{17}{3}]

, which contains

0

, the minimum square of any real number in this interval is

0

at

x = 0

.

3

Solve the double inequality

-5 < \frac{1 - 2y}{3} \leq 3

for

y

.

-15 < 1 - 2y \leq 9 \implies -16 < -2y \leq 8 \implies 8 > y \geq -4 \implies -4 \leq y < 8

.

Multiplying by

3

preserves inequalities, subtracting

1

preserves inequalities, and dividing by

-2

reverses all inequality signs.

4

Find the maximum integer value of

y

.

Maximum integer

y = 7

.

The solution set for

y

is the half-open interval

[-4, 8)

. Since

y

is constrained to be an integer, the largest integer strictly less than

8

is

7

.

5

Minimize the expression

x^2 - y

.

\text{Minimum } (x^2 - y) = \text{Minimum } (x^2) - \text{Maximum } (y) = 0 - 7 = -7

.

To minimize a difference

A - B

, one must minimize the minuend

A

and maximize the subtrahend

B

.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:2m 0s

Question 16Question →

If $-3|x - 2| + 4 > -11$ , which of the following inequalities represents all possible real values of $x$ ?

-3 < x < 7

x < -3 \text{ or } x > 7

x < 7

-7 < x < 3

-5 < x < 5

Show answer & explanation

Answer: $-3 < x < 7$

Answer

-3 < x < 7

Subtracting

4

from both sides gives

-3|x - 2| > -15

. Dividing by

-3

and reversing the inequality sign results in

|x - 2| < 5

. Converting to the compound inequality

-5 < x - 2 < 5

and adding

2

to each part yields the correct interval

-3 < x < 7

.

Step-by-Step Solution

1

Subtract 4 from both sides of the inequality

-3|x - 2| > -15

Isolate the absolute value term on the left side.

2

Divide both sides by

-3

and reverse the inequality sign

|x - 2| < 5

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

3

Express the absolute value inequality as a compound inequality

-5 < x - 2 < 5

An inequality of the form

|u| < c

(where

c > 0

) is equivalent to

-c < u < c

.

4

Add 2 to all three parts of the compound inequality

-3 < x < 7

Isolate

x

to find the complete range of solution values.

Key Concept

Linear Inequalities and Absolute Value

Question 17Question →

Which of the following values of $x$ satisfy the inequality $|2x - 1| \le 5 - x$ ? Select all that apply.

Select all that apply

-4

-1

1

3

-5

Show answer & explanation

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

Answer

The values of

x

that satisfy the inequality are

-4

,

-1

, and

1

.

Solving the compound inequality

-(5 - x) \le 2x - 1 \le 5 - x

yields the solution range

-4 \le x \le 2

. The candidate values

-4

,

-1

, and

1

all lie within this range, making them valid solutions.

Step-by-Step Solution

1

Determine the non-negativity constraint for the right-hand side expression.

Since an absolute value

|2x - 1|

must be non-negative, we must have

5 - x \ge 0

, which simplifies to

x \le 5

.

An absolute value cannot be less than a negative number.

2

Rewrite the absolute value inequality

|2x - 1| \le 5 - x

as a compound linear inequality.

-(5 - x) \le 2x - 1 \le 5 - x

, which expands to

-5 + x \le 2x - 1 \le 5 - x

.

For any non-negative expression

B

,

|A| \le B

is logically equivalent to

-B \le A \le B

.

3

Solve the left-hand inequality

-5 + x \le 2x - 1

.

Subtracting

x

from both sides gives

-5 \le x - 1

. Adding

1

to both sides yields

x \ge -4

.

Isolating

x

establishes the lower bound of the solution set.

4

Solve the right-hand inequality

2x - 1 \le 5 - x

.

Adding

x

to both sides gives

3x - 1 \le 5

. Adding

1

yields

3x \le 6

, so

x \le 2

.

Isolating

x

establishes the upper bound of the solution set.

5

Intersect all constraints to find the valid domain for

x

.

Combining

x \ge -4

,

x \le 2

, and

x \le 5

gives the closed interval

[-4, 2]

.

A valid value of

x

must satisfy all component inequalities simultaneously.

6

Evaluate the candidate choices against the solution interval

[-4, 2]

.

The values

-4

,

-1

, and

1

fall within

[-4, 2]

, whereas

3

and

-5

fall outside this interval.

Only numbers inside

[-4, 2]

satisfy the original inequality.

Key Concept

Solving Absolute Value Inequalities with Variable Expressions

Question 18Question →

If $x$ is a real number that satisfies both of the inequalities $|3x + 4| \ge 11$ and $|x - 1| < 6$ , which of the following could be the value of $x$ ? Select all that apply.

Select all that apply

-5

-3

3

5

7

Show answer & explanation

Answer: $3$ ; $5$

Answer

The values 3 and 5 satisfy both inequalities.

Solving the first inequality

|3x + 4| \ge 11

yields

x \le -5

or

x \ge \frac{7}{3}

. Solving the second inequality

|x - 1| < 6

yields

-5 < x < 7

. Intersecting these two regions, the interval

x \le -5

does not overlap with

-5 < x < 7

because

-5

is excluded from the second inequality. The overlap occurs only for

\frac{7}{3} \le x < 7

. Among the choices, 3 and 5 fall within this valid interval.

Step-by-Step Solution

1

Solve the absolute value inequality

|3x + 4| \ge 11

.

x \le -5

or

x \ge \frac{7}{3}

.

An absolute value inequality of the form

|u| \ge k

(where

k > 0

) splits into two separate inequalities:

u \ge k

or

u \le -k

. Solving

3x + 4 \ge 11

gives

3x \ge 7 \implies x \ge \frac{7}{3}

. Solving

3x + 4 \le -11

gives

3x \le -15 \implies x \le -5

.

2

Solve the absolute value inequality

|x - 1| < 6

.

-5 < x < 7

.

An absolute value inequality of the form

|u| < k

is equivalent to the compound inequality

-k < u < k

. Thus,

-6 < x - 1 < 6

. Adding 1 to all parts yields

-5 < x < 7

.

3

Determine the intersection of the two solution sets.

\frac{7}{3} \le x < 7

.

The portion

x \le -5

has no overlap with

-5 < x < 7

because

-5

is excluded by the strict inequality in the second condition. The portion

x \ge \frac{7}{3}

overlaps with

-5 < x < 7

to yield the interval

[\frac{7}{3}, 7)

.

4

Evaluate which of the given options fall inside

[\frac{7}{3}, 7)

.

The numbers 3 and 5 belong to the interval, while

-5

,

-3

, and

7

do not.

Since

\frac{7}{3} \approx 2.33

, the values 3 and 5 fall strictly between 2.33 and 7.

Key Concept

Solving systems of linear absolute value inequalities by finding the intersection of compound solution intervals

Question 19Question →

If $x$ is a real number that satisfies both $|3 - 2x| \ge 5$ and $\frac{7 - 3x}{-2} > -1$ , which of the following expresses all possible values of $x$ ?

x \ge 4

x \le -1

\frac{5}{3} < x \le 4

-1 \le x \le 4

x > \frac{5}{3}

Show answer & explanation

Answer: $x \ge 4$

Answer

The condition is satisfied by all values of

x

such that

x \ge 4

.

Solving

|3 - 2x| \ge 5

yields two separate intervals:

x \le -1

or

x \ge 4

. Solving

\frac{7 - 3x}{-2} > -1

involves multiplying by

-2

and dividing by

-3

, both of which flip the inequality sign, leading to

x > \frac{5}{3}

. The values of

x

that satisfy both constraints are those in the overlap of

x \in (-\infty, -1] \cup [4, \infty)

and

x > \frac{5}{3}

, which simplifies directly to

x \ge 4

.

Step-by-Step Solution

1

Solve the absolute value inequality

|3 - 2x| \ge 5

.

Splitting into two cases:

3 - 2x \ge 5 \implies -2x \ge 2 \implies x \le -1

, or

3 - 2x \le -5 \implies -2x \le -8 \implies x \ge 4

. So

x \in (-\infty, -1] \cup [4, \infty)

.

An absolute value inequality

|u| \ge k

(for

k > 0

) decouples into

u \ge k

or

u \le -k

.

2

Solve the linear inequality

\frac{7 - 3x}{-2} > -1

.

Multiply both sides by

-2

(reversing the inequality):

7 - 3x < 2

. Subtract 7:

-3x < -5

. Divide by

-3

(reversing the inequality again):

x > \frac{5}{3}

.

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

3

Find the intersection of the two solution sets.

We require

x \in ((-\infty, -1] \cup [4, \infty)) \cap (\frac{5}{3}, \infty)

. Since

(-\infty, -1]

has no overlap with

(\frac{5}{3}, \infty)

, the intersection is

[4, \infty)

, or

x \ge 4

.

A real number must satisfy both inequalities simultaneously.

Key Concept

Solving systems of absolute value and linear inequalities with negative multipliers

Question 20Question →

If $x$ is an integer that satisfies both $|4x - 5| \le 13$ and $\frac{2x - 1}{-3} \le -1$ , what is the sum of all possible values of $x$ ?

-7

0

7

9

14

Show answer & explanation

Answer: 9

Answer

9

Solving the absolute value inequality

|4x - 5| \le 13

yields

-13 \le 4x - 5 \le 13

, which simplifies to

-2 \le x \le 4.5

. Next, solving

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

requires multiplying both sides by

-3

and reversing the inequality sign, giving

2x - 1 \ge 3

, which simplifies to

x \ge 2

. Combining

-2 \le x \le 4.5

and

x \ge 2

for integer

x

gives the set

\{2, 3, 4\}

. The sum of these values is

2 + 3 + 4 = 9

.

Step-by-Step Solution

1

Solve the absolute value inequality

|4x - 5| \le 13

-13 \le 4x - 5 \le 13 \implies -8 \le 4x \le 18 \implies -2 \le x \le 4.5

An absolute value inequality

|u| \le k

unfolds into the compound inequality

-k \le u \le k

.

2

Solve the linear inequality

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

2x - 1 \ge 3 \implies 2x \ge 4 \implies x \ge 2

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

3

Find the intersection of the two solution sets for integer

x

x \in \{2, 3, 4\}

The integers satisfying both

-2 \le x \le 4.5

and

x \ge 2

are 2, 3, and 4.

4

Calculate the sum of the possible integer values

2 + 3 + 4 = 9

Summing the valid integer solutions yields the final requested answer.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:2m 0s

Linear Inequalities and Absolute Value

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