Linear Inequalities and Absolute Value Alıştırma Soruları — GRE General Test

Soru 1Soru →

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.

Geçerli olan tümünü seçin

-2

0

3

6

9

Cevabı ve açıklamayı göster

Cevap: $-2$ ; $0$ ; $6$

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Soru 2Soru →

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

Cevabı ve açıklamayı göster

Cevap: $-21$

Cevap

-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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Linear Inequalities and Absolute Value Bounds

Tahmini Süre:1m 30s

Soru 3Soru →

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

Cevabı ve açıklamayı göster

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

Cevap

-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.

Adım Adım Çözüm

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.

Anahtar Kavram

Solving Linear Absolute Value Inequalities

Tahmini Süre:45s

Soru 4Soru →

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

Geçerli olan tümünü seçin

-3

-1

2

5

6

Cevabı ve açıklamayı göster

Cevap: $-1$ ; $2$ ; $5$

Cevap

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]

.

Adım Adım Çözüm

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.

Anahtar Kavram

Linear Inequalities and Absolute Value

Tahmini Süre:1m 0s

Soru 5Soru →

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

Geçerli olan tümünü seçin

-5

-3

0

4

8

Cevabı ve açıklamayı göster

Cevap: $-5$ ; $8$

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

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.

Tahmini Süre:1m 15s

Soru 6Soru →

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.

Geçerli olan tümünü seçin

x - y > -1

xy < 12

x + y > 0

|y| < 3

y - x > -3

Cevabı ve açıklamayı göster

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

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Linear Inequalities and Absolute Value Bounds

Soru 7Soru →

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

Cevabı ve açıklamayı göster

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

Cevap

-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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Linear Inequalities and Absolute Value Transformations

Tahmini Süre:1m 30s

Soru 8Soru →

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

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Tahmini Süre:2m 0s

Soru 9Soru →

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

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Linear Inequalities and Absolute Value

Soru 10Soru →

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 $M - m$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Linear Inequalities and Absolute Value

Tahmini Süre:2m 0s

Soru 11Soru →

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.

Geçerli olan tümünü seçin

-3

-1

2

4

7

Cevabı ve açıklamayı göster

Cevap: $-1$ ; $2$ ; $4$

Cevap

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]

.

Adım Adım Çözüm

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.

Anahtar Kavram

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.

Soru 12Soru →

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

Cevabı ve açıklamayı göster

Cevap: $-17$

Cevap

-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

.

Adım Adım Çözüm

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

).

Anahtar Kavram

Linear Inequalities and Absolute Value

Soru 13Soru →

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

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

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.

Adım Adım Çözüm

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

.

Anahtar Kavram

Linear Inequalities and Absolute Value Bounds

Tahmini Süre:1m 30s

Soru 14Soru →

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

10

12

15

17

18

Cevabı ve açıklamayı göster

Cevap: 15

Cevap

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.

Adım Adım Çözüm

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

.

Anahtar Kavram

Linear Inequalities and Absolute Value

Soru 15Soru →

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

Cevabı ve açıklamayı göster

Cevap: $-7$

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Linear Inequalities and Absolute Value

Tahmini Süre:2m 0s

Soru 16Soru →

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

Cevabı ve açıklamayı göster

Cevap: $-3 < x < 7$

Cevap

-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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Linear Inequalities and Absolute Value

Soru 17Soru →

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

Geçerli olan tümünü seçin

-4

-1

1

3

-5

Cevabı ve açıklamayı göster

Cevap: $-4$ ; $-1$ ; $1$

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Solving Absolute Value Inequalities with Variable Expressions

Soru 18Soru →

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.

Geçerli olan tümünü seçin

-5

-3

3

5

7

Cevabı ve açıklamayı göster

Cevap: $3$ ; $5$

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Soru 19Soru →

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}

Cevabı ve açıklamayı göster

Cevap: $x \ge 4$

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Solving systems of absolute value and linear inequalities with negative multipliers

Soru 20Soru →

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

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Linear Inequalities and Absolute Value

Tahmini Süre:2m 0s

Linear Inequalities and Absolute Value

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram