Linear Inequalities and Absolute Value Practice Questions — GRE General Test — Page 3

Question 41Question →

If $x$ is an integer that satisfies both $|2x - 3| \le 9$ and $\frac{5 - 3x}{-2} > 4$ , how many possible values of $x$ exist?

2

3

5

8

10

Show answer & explanation

Answer: 2

Answer

There are 2 possible integer values of

x

(specifically

x = 5

and

x = 6

).

Solving

|2x - 3| \le 9

yields

-3 \le x \le 6

. Solving

\frac{5 - 3x}{-2} > 4

requires reversing the inequality sign twice (first when multiplying by

-2

, giving

5 - 3x < -8

, and second when dividing by

-3

, giving

x > \frac{13}{3} \approx 4.33

). The integers satisfying both

-3 \le x \le 6

and

x > 4.33

are

5

and

6

, which gives exactly 2 possible values.

Step-by-Step Solution

1

Solve the absolute value inequality

|2x - 3| \le 9

.

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

.

An absolute value inequality of the form

|u| \le k

unfolds into the compound inequality

-k \le u \le k

.

2

Solve the linear inequality

\frac{5 - 3x}{-2} > 4

.

5 - 3x < -8 \implies -3x < -13 \implies x > \frac{13}{3} \approx 4.33

.

Multiplying or dividing an inequality by a negative number requires reversing the direction of the inequality sign at each step.

3

Find the intersection of the two solution sets for integer values of

x

.

The range of valid values is

4.33 < x \le 6

. The integers in this interval are

x = 5

and

x = 6

.

Combining

-3 \le x \le 6

and

x > 4.33

restricts

x

to integer values strictly greater than

4.33

and less than or equal to

6

.

4

Count the valid integer values.

There are 2 integers (

5

and

6

).

Direct counting of the valid integer set

\{5, 6\}

.

Key Concept

Solving absolute value inequalities and compound linear inequalities with negative multipliers

Question 42Question →

If $x$ is a real number that satisfies both $|4 - x| < 3$ and $\frac{2x - 15}{-3} \le 1$ , which of the following inequalities expresses all possible values of $x$ ?

6 \le x < 7

1 < x \le 6

1 < x < 7

x \ge 6

6 < x < 7

Show answer & explanation

Answer: $6 \le x < 7$

Answer

The inequality expressing all possible values of

x

is

6 \le x < 7

.

The correct inequality is determined by finding the intersection of both given inequalities. The absolute value inequality

|4 - x| < 3

simplifies to

1 < x < 7

. The linear inequality

\frac{2x - 15}{-3} \le 1

simplifies to

x \ge 6

after reversing the inequality sign when multiplying by

-3

. Combining

1 < x < 7

and

x \ge 6

yields the compound inequality stating

x

is greater than or equal to

6

and strictly less than

7

.

Step-by-Step Solution

1

Solve the absolute value inequality

|4 - x| < 3

.

-3 < 4 - x < 3 \implies -7 < -x < -1 \implies 1 < x < 7

.

An absolute value inequality

|u| < k

translates to

-k < u < k

. Multiplying or dividing by

-1

flips the inequality signs.

2

Solve the linear inequality

\frac{2x - 15}{-3} \le 1

.

2x - 15 \ge -3 \implies 2x \ge 12 \implies x \ge 6

.

Multiplying 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.

1 < x < 7

and

x \ge 6

combine to give

6 \le x < 7

.

Values of

x

must satisfy both inequalities simultaneously.

Key Concept

Solving systems of absolute value inequalities and linear inequalities, including sign reversal rules for negative multipliers.

Question 43Question →

If $x$ is an integer that satisfies both $|3x - 12| \le 15$ and $|2x + 4| \ge 6$ , what is the sum of the maximum possible value of $x$ and the minimum possible value of $x$ ?

Show answer & explanation

Answer: 10

Answer

The correct answer is 10.

Solving the first inequality

|3x - 12| \le 15

yields

-1 \le x \le 9

. Solving the second inequality

|2x + 4| \ge 6

yields

x \ge 1

or

x \le -5

. The values of

x

that satisfy both constraints are those in the interval

[1, 9]

. Within this interval, the minimum integer value is

1

and the maximum integer value is

9

. Adding these values together gives

1 + 9 = 10

.

Step-by-Step Solution

1

Solve the bounded absolute value inequality

|3x - 12| \le 15

.

-15 \le 3x - 12 \le 15 \implies -3 \le 3x \le 27 \implies -1 \le x \le 9.

An inequality of the form

|A| \le B

(where

B \ge 0

) unwraps to the compound inequality

-B \le A \le B

.

2

Solve the boundary absolute value inequality

|2x + 4| \ge 6

.

2x + 4 \ge 6 \implies 2x \ge 2 \implies x \ge 1

, or

2x + 4 \le -6 \implies 2x \le -10 \implies x \le -5$.

An inequality of the form

|A| \ge B

(where

B > 0

) splits into two separate cases:

A \ge B

or

A \le -B

.

3

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

The intersection of

[-1, 9]

and

(-\infty, -5] \cup [1, \infty)

is

[1, 9]

.

To satisfy both inequalities simultaneously,

x

must belong to both set

[-1, 9]

and set

(-\infty, -5] \cup [1, \infty)

.

4

Identify the extreme integer values of

x

and compute their sum.

\text{Minimum integer } x = 1, \quad \text{Maximum integer } x = 9; \quad 1 + 9 = 10.

The range of valid integer values for

x

is all integers from 1 to 9 inclusive.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:2m 0s

Question 44Question →

If $x$ is a real number that satisfies both $|3 - x| \le 5$ and $\frac{2x - 8}{-3} \le -2$ , which of the following could be the value of $x$ ? Indicate all such values.

Select all that apply

-1

4

7

7.5

8

Show answer & explanation

Answer: $7$ ; $7.5$ ; $8$

Answer

The values

7

,

7.5

, and

8

satisfy both inequalities.

Solving

|3 - x| \le 5

gives

-2 \le x \le 8

. Solving

\frac{2x - 8}{-3} \le -2

by multiplying by

-3

and reversing the inequality sign gives

2x - 8 \ge 6

, or

x \ge 7

. Taking the intersection of both conditions yields

7 \le x \le 8

. Among the choices, the values

7

,

7.5

, and

8

fall within this range.

Step-by-Step Solution

1

Solve the absolute value inequality

|3 - x| \le 5

.

-2 \le x \le 8

Rewrite as a compound inequality

-5 \le 3 - x \le 5

. Subtracting

3

gives

-8 \le -x \le 2

. Multiplying by

-1

and reversing inequality signs yields

-2 \le x \le 8

.

2

Solve the linear inequality

\frac{2x - 8}{-3} \le -2

.

x \ge 7

Multiply both sides by

-3

, making sure to flip the inequality sign:

2x - 8 \ge 6

. Adding

8

yields

2x \ge 14

, so

x \ge 7

.

3

Find the intersection of the two solution sets.

7 \le x \le 8

Combining

-2 \le x \le 8

and

x \ge 7

gives the range

7 \le x \le 8

.

4

Test the given options against the combined range

7 \le x \le 8

.

The values

7

,

7.5

, and

8

fall within

[7, 8]

, while

-1

and

4

do not.

Only numbers greater than or equal to

7

and less than or equal to

8

satisfy both conditions.

Key Concept

Solving compound linear and absolute value inequalities, ensuring inequality signs are reversed when multiplying or dividing by negative numbers.

Question 45Question →

If $k$ is a real constant such that the inequality $|2x - 3| + |x + 4| \le k$ has no real solutions for $x$ , which of the following inequality statements expresses all possible values of $k$ ?

k < \frac{11}{2}

k \le \frac{11}{2}

k < 7

k < 11

k > \frac{11}{2}

Show answer & explanation

Answer: $k < \frac{11}{2}$

Answer

The statement expressing all possible values of

k

is

k < \frac{11}{2}

.

The function

f(x) = |2x - 3| + |x + 4|

represents a continuous piecewise linear curve. Evaluating

f(x)

at its critical points

x = -4

and

x = \frac{3}{2}

yields

f(-4) = 11

and

f\left(\frac{3}{2}\right) = 5.5 = \frac{11}{2}

. Since the slope is

-3

for

x < -4

,

-1

for

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

, and

+3

for

x > \frac{3}{2}

, the global minimum value of

f(x)

across all real numbers is

\frac{11}{2}

. Consequently, the inequality

f(x) \le k

has no real solutions if and only if

k

is strictly less than this minimum value, leading to

k < \frac{11}{2}

.

Step-by-Step Solution

1

Identify the critical points of the absolute value terms.

The terms

|2x - 3|

and

|x + 4|

change behavior at

x = \frac{3}{2}

and

x = -4

, respectively.

Absolute value functions

f(x) = |ax + b|

reach zero and change slope at their roots.

2

Evaluate

f(x) = |2x - 3| + |x + 4|

at the critical points and analyze its piecewise behavior.

At

x = -4

,

f(-4) = |-11| + |0| = 11

. At

x = \frac{3}{2}

,

f\left(\frac{3}{2}\right) = |0| + |\frac{11}{2}| = \frac{11}{2}

. For

x < -4

,

f(x) = (3 - 2x) - (x + 4) = -3x - 1

. For

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

,

f(x) = (3 - 2x) + (x + 4) = -x + 7

. For

x > \frac{3}{2}

,

f(x) = (2x - 3) + (x + 4) = 3x + 1

.

Because

f(x)

is a convex piecewise linear function that grows to

\infty

as

x \to \pm\infty

, its global minimum must occur at one of its critical points.

3

Determine the global minimum value of

f(x)

.

Comparing values,

f\left(\frac{3}{2}\right) = \frac{11}{2}

is smaller than

f(-4) = 11

, so the minimum value of

|2x - 3| + |x + 4|

for all real

x

is

\frac{11}{2}

.

The function output is always greater than or equal to

\frac{11}{2}

for any real number

x

.

4

Apply the condition for no real solutions.

For

|2x - 3| + |x + 4| \le k

to have no solutions,

k

must be strictly less than the absolute minimum value of the expression, so

k < \frac{11}{2}

.

If

k \ge \frac{11}{2}

, there is at least one

x

value (such as

x = \frac{3}{2}

) satisfying the inequality.

Key Concept

Minimizing Sums of Absolute Values and Boundary Conditions of Inequalities

Question 46Question →

If $x$ is a real number that satisfies $|3 - 2x| \le 9$ , what is the maximum possible value of $|x - 5|$ ?

Show answer & explanation

Answer: 8

Answer

The maximum possible value of

|x - 5|

is 8.

Solving

|3 - 2x| \le 9

yields

-9 \le 3 - 2x \le 9

. Subtracting 3 gives

-12 \le -2x \le 6

. Dividing by

-2

and reversing inequality signs results in

-3 \le x \le 6

. To maximize

|x - 5|

, find the point in

[-3, 6]

that is farthest from 5. Evaluating at the endpoints gives

|-3 - 5| = 8

and

|6 - 5| = 1

. The maximum possible value is 8.

Step-by-Step Solution

1

Solve the absolute value inequality

|3 - 2x| \le 9

.

-9 \le 3 - 2x \le 9

By definition of absolute value,

|u| \le k

means

-k \le u \le k

for

k \ge 0

.

2

Isolate

-2x

by subtracting 3 from each part.

-12 \le -2x \le 6

Subtracting a constant from all parts preserves the inequality direction.

3

Isolate

x

by dividing each part by

-2

.

-3 \le x \le 6

Dividing an inequality by a negative number flips the inequality signs.

4

Determine the maximum value of

|x - 5|

over the interval

[-3, 6]

.

Maximum occurs at

x = -3

, where

|-3 - 5| = 8

.

The expression

|x - 5|

represents the distance from

x

to 5 on the number line. The endpoint

x = -3

is furthest from 5.

Key Concept

Linear inequalities and absolute value as distance on a number line

Question 47Question →

If $x$ is an integer that satisfies both $|2x + 1| \le 7$ and $\frac{5 - x}{-2} < -1$ , how many possible values of $x$ are there?

0

3

7

8

11

Show answer & explanation

Answer: 7

Answer

There are 7 possible integer values of

x

.

Solving

|2x + 1| \le 7

yields

-4 \le x \le 3

. Solving

\frac{5 - x}{-2} < -1

requires flipping the inequality symbol when multiplying by

-2

, resulting in

5 - x > 2

, which simplifies to

x < 3

. Taking the intersection of

-4 \le x \le 3

and

x < 3

gives the interval

-4 \le x < 3

. The integers in this range are

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

, totaling 7 integers.

Step-by-Step Solution

1

Solve the absolute value inequality

|2x + 1| \le 7

.

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

.

An absolute value inequality of the form

|u| \le k

(for

k \ge 0

) expands to

-k \le u \le k

.

2

Solve the rational inequality

\frac{5 - x}{-2} < -1

.

5 - x > (-1)(-2) \implies 5 - x > 2 \implies -x > -3 \implies x < 3

.

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

3

Find the intersection of the two solution sets and count integer solutions.

-4 \le x < 3

. The integer values satisfying this inequality are

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

, which gives a total of 7 integer values.

The solution must satisfy both conditions simultaneously.

Key Concept

Solving combined linear and absolute value inequalities while correctly reversing inequality signs when multiplying or dividing by negative values.

Estimated Time:1m 30s

Question 48Question →

If $x$ is a real number that satisfies the inequality $|3x - 3| \le 9$ , which of the following statements must be true? Select all that apply.

Select all that apply

-4 \le 2x \le 8

|x - 1| \le 3

x^2 \le 16

x \ge 0

1 - x \le 2

Show answer & explanation

Answer: $-4 \le 2x \le 8$ ; $|x - 1| \le 3$ ; $x^2 \le 16$

Answer

The statements that must be true are

-4 \le 2x \le 8

,

|x - 1| \le 3

, and

x^2 \le 16

.

Solving

|3x - 3| \le 9

gives

-9 \le 3x - 3 \le 9

, which simplifies to

-2 \le x \le 4

. Multiplying this range by 2 yields

-4 \le 2x \le 8

. Subtracting 1 gives

-3 \le x - 1 \le 3

, which is

|x - 1| \le 3

. Squaring values in

[-2, 4]

yields non-negative numbers up to 16, so

x^2 \le 16

is also true.

Step-by-Step Solution

1

Unfold the absolute value inequality into a compound inequality.

-9 \le 3x - 3 \le 9

By definition,

|u| \le k

(where

k \ge 0

) is equivalent to

-k \le u \le k

.

2

Add 3 to all parts of the compound inequality.

-6 \le 3x \le 12

Isolating the variable term

3x

.

3

Divide all parts by 3 to solve for

x

.

-2 \le x \le 4

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

4

Test each proposed statement against the interval

[-2, 4]

.

-4 \le 2x \le 8

is true;

|x - 1| \le 3

is true;

x^2 \le 16

is true;

x \ge 0

fails for

x = -1

;

1 - x \le 2

fails for

x = -2

.

Determining which properties hold for every real number in the solution set.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:1m 30s

Question 49Question →

If $x$ is a real number that satisfies the inequality $|4 - 3x| \le 11$ , what is the maximum possible value of the expression $\frac{2 - x}{-3}$ ?

1

-\frac{13}{9}

-1

\frac{5}{3}

3

Show answer & explanation

Answer: $1$

Answer

1

Expanding the absolute value inequality

|4 - 3x| \le 11

gives

-11 \le 4 - 3x \le 11

. Subtracting

4

from all sides yields

-15 \le -3x \le 7

. Dividing by

-3

and reversing the inequality signs gives

-\frac{7}{3} \le x \le 5

. Simplifying the target expression gives

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

. Because

\frac{x - 2}{3}

increases with

x

, substituting the maximum allowed value

x = 5

gives

\frac{5 - 2}{3} = 1

.

Step-by-Step Solution

1

Unfold the absolute value inequality into a compound inequality

-11 \le 4 - 3x \le 11

For any real number

k \ge 0

,

|A| \le k

is equivalent to

-k \le A \le k

.

2

Isolate the variable term by subtracting 4 from all parts

-15 \le -3x \le 7

Subtracting a constant maintains the direction of the inequality.

3

Divide by -3 and flip the inequality signs

5 \ge x \ge -\frac{7}{3}

, or equivalently

-\frac{7}{3} \le x \le 5

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

4

Simplify the target expression and maximize it over the valid domain of x

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

. Maximum occurs at

x = 5

:

\frac{5 - 2}{3} = 1

Since

\frac{x - 2}{3}

increases as

x

increases, its maximum occurs at the largest possible value of

x

.

Key Concept

Linear Inequalities and Absolute Value Bounds

Question 50Question →

If $x$ is an integer that satisfies both $|7 - 2x| \le 9$ and $\frac{1 - x}{-2} > 2$ , what is the sum of all possible values of $x$ ?

Show answer & explanation

Answer: 21

Answer

The sum of all possible integer values of

x

is 21.

To solve the system of inequalities, analyze each part separately. For

|7 - 2x| \le 9

, express it as

-9 \le 7 - 2x \le 9

. Subtracting

7

yields

-16 \le -2x \le 2

. Dividing all parts by

-2

requires reversing the inequality symbols, giving

8 \ge x \ge -1

, or

-1 \le x \le 8

. For

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

, multiplying by

-2

reverses the inequality symbol to

1 - x < -4

. Subtracting

1

gives

-x < -5

, and dividing by

-1

reverses the symbol again to

x > 5

. The integer values satisfying both

-1 \le x \le 8

and

x > 5

are

6, 7,

and

8

. Summing these integer values yields

6 + 7 + 8 = 21

.

Step-by-Step Solution

1

Solve the absolute value inequality

|7 - 2x| \le 9

-1 \le x \le 8

Unfolding the absolute value gives

-9 \le 7 - 2x \le 9

. Subtracting 7 yields

-16 \le -2x \le 2

, and dividing by

-2

reverses the inequalities to give

-1 \le x \le 8

.

2

Solve the linear inequality \frac{1 - x}{-2} > 2

x > 5

Multiplying by

-2

flips the inequality sign to yield

1 - x < -4

. Subtracting 1 gives

-x < -5

, and dividing by

-1

flips the inequality sign again to yield

x > 5

.

3

Find the overlapping integer values for x

x \in \{6, 7, 8\}

The integers that are greater than 5 and less than or equal to 8 are 6, 7, and 8.

4

Sum the integer solutions

21

Calculating 6 + 7 + 8 = 21.

Key Concept

Solving absolute value inequalities and compound linear inequalities with sign reversal when dividing by negative numbers.

Question 51Question →

Which of the following values of $x$ satisfy the inequality $|6 - 3x| > 9$ ? Select all that apply.

Select all that apply

-4

-2

1

3

6

Show answer & explanation

Answer: $-4$ ; $-2$ ; $6$

Answer

The values

-4

,

-2

, and

6

satisfy the inequality

|6 - 3x| > 9

.

The absolute value inequality

|6 - 3x| > 9

is equivalent to

6 - 3x > 9

or

6 - 3x < -9

. Solving these yields

x < -1

or

x > 5

. Among the given choices,

-4

,

-2

, and

6

fall within these solution ranges.

Step-by-Step Solution

1

Set up the compound linear inequalities from the absolute value inequality

|6 - 3x| > 9

.

6 - 3x > 9

or

6 - 3x < -9

An absolute value inequality of the form

|u| > c

(where

c > 0

) splits into

u > c

or

u < -c

.

2

Solve the first inequality

6 - 3x > 9

.

-3x > 3 \implies x < -1

Subtracting

6

gives

-3x > 3

. Dividing both sides by

-3

reverses the inequality sign to yield

x < -1

.

3

Solve the second inequality

6 - 3x < -9

.

-3x < -15 \implies x > 5

Subtracting

6

gives

-3x < -15

. Dividing both sides by

-3

reverses the inequality sign to yield

x > 5

.

4

Combine the solution sets and evaluate the given options.

The solution set consists of all real numbers where

x < -1

or

x > 5

. Therefore,

-4

,

-2

, and

6

are valid solutions.

Values

-4

and

-2

are strictly less than

-1

, while

6

is strictly greater than

5

.

Key Concept

Solving absolute value inequalities and reversing inequality signs when multiplying or dividing by negative numbers.

Question 52Question →

If $x$ is an integer that satisfies both $|2x - 1| \le 9$ and $\frac{x + 2}{-3} > -2$ , how many possible values of $x$ exist?

1

4

7

8

9

Show answer & explanation

Answer: 8

Answer

There are 8 possible integer values for x.

Solving

|2x - 1| \le 9

yields

-4 \le x \le 5

. Solving

\frac{x + 2}{-3} > -2

requires flipping the inequality sign when multiplying by

-3

, which gives

x < 4

. Combining both inequalities yields

-4 \le x < 4

. The integers satisfying this compound inequality are

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

, making a total of 8 possible integer values.

Step-by-Step Solution

1

Solve the absolute value inequality

|2x - 1| \le 9

-9 \le 2x - 1 \le 9 \implies -8 \le 2x \le 10 \implies -4 \le x \le 5

An absolute value inequality of the form

|u| \le k

expands to

-k \le u \le k

.

2

Solve the linear inequality

\frac{x + 2}{-3} > -2

x + 2 < 6 \implies x < 4

Multiplying both sides of an inequality by a negative quantity (

-3

) requires reversing the inequality sign from

>

to

<

.

3

Find the intersection of the two solution sets

-4 \le x < 4

The integer

x

must satisfy both

-4 \le x \le 5

and

x < 4

simultaneously.

4

Count the integer values satisfying

-4 \le x < 4

The integers are

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

, which totals 8 values.

Counting all integers from

-4

up to (but not including)

4

gives 8 valid integers.

Key Concept

Solving systems of linear inequalities involving absolute values and negative multipliers

Estimated Time:1m 30s

Linear Inequalities and Absolute Value

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