Real Numbers, Number Line, and Absolute Value Alıştırma Soruları — GRE General Test

Soru 1Soru →

If $x$ and $y$ are real numbers such that $|2x - 5| \le 9$ and $|y + 3| \le 4$ , what is the maximum possible value of $|x - 2y + 1|$ ?

22

20

14

13

6

Cevabı ve açıklamayı göster

Cevap: $22$

Cevap

The maximum possible value of

|x - 2y + 1|

is 22.

The correct answer is 22 because solving the absolute value inequalities gives

-2 \le x \le 7

and

-7 \le y \le 1

. To maximize

x - 2y + 1

, we take the largest possible value of

x

(

7

), the smallest possible value of

y

(

-7

, making

-2y = 14

), and add

1

, obtaining

7 + 14 + 1 = 22

. Since the lower bound of

x - 2y + 1

is

-3

, the maximum absolute value is

|22| = 22

.

Adım Adım Çözüm

1

Find the range of possible values for

x

from the inequality

|2x - 5| \le 9

.

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

.

An absolute value inequality

|A| \le B

unwraps to

-B \le A \le B

.

2

Find the range of possible values for

y

from the inequality

|y + 3| \le 4

.

-4 \le y + 3 \le 4 \implies -7 \le y \le 1

.

Unwrapping the absolute value inequality gives the upper and lower bounds for

y

on the real number line.

3

Determine the range of possible values for

-2y

.

Multiplying

-7 \le y \le 1

by

-2

and reversing the inequality signs yields

-2(1) \le -2y \le -2(-7) \implies -2 \le -2y \le 14

.

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

4

Combine the ranges to find the minimum and maximum bounds for

z = x - 2y + 1

.

Minimum

z = (-2) + (-2) + 1 = -3

; Maximum

z = 7 + 14 + 1 = 22

. Thus,

-3 \le x - 2y + 1 \le 22

.

Adding the individual minimums gives the absolute minimum, and adding the individual maximums gives the absolute maximum.

5

Calculate the maximum value of the absolute value

|x - 2y + 1|

.

\\max(|x - 2y + 1|) = \\max(|-3|, |22|) = 22

.

The absolute value of a quantity ranging from

-3

to

22

reaches its maximum distance from zero at

22

.

Anahtar Kavram

Properties of Real Numbers, Number Line Inequalities, and Absolute Value Operations

Tahmini Süre:2m 30s

Soru 2Soru →

On the real number line, point $P$ represents the real number $x$ , point $Q$ represents $7$ , and point $R$ represents $-5$ . If the distance between $P$ and $Q$ is equal to $3$ times the distance between $P$ and $R$ , what is the sum of all possible values of $x$ ?

-13

-11

-7

-2

13

Cevabı ve açıklamayı göster

Cevap: $-13$

Cevap

The sum of all possible values of

x

is

-13

.

The distance between

x

and

7

is

|x - 7|

and the distance between

x

and

-5

is

|x + 5|

. Equating

|x - 7| = 3|x + 5|

leads to two equations:

x - 7 = 3(x + 5)

giving

x = -11

, and

x - 7 = -3(x + 5)

giving

x = -2

. Adding both solutions yields

(-11) + (-2) = -13

.

Adım Adım Çözüm

1

Set up the distance equation using absolute value notation

The distance between

P(x)

and

Q(7)

is

|x - 7|

, and the distance between

P(x)

and

R(-5)

is

|x - (-5)| = |x + 5|

. The problem specifies that

|x - 7| = 3|x + 5|

.

Distance between two points

a

and

b

on a real number line is expressed as

|a - b|

.

2

Solve Case 1 where

x - 7

and

x + 5

have the same sign

x - 7 = 3(x + 5) \implies x - 7 = 3x + 15 \implies -22 = 2x \implies x = -11

.

When both absolute value expressions have identical signs,

|x - 7| = 3|x + 5|

simplifies directly to

x - 7 = 3(x + 5)

.

3

Solve Case 2 where

x - 7

and

x + 5

have opposite signs

x - 7 = -3(x + 5) \implies x - 7 = -3x - 15 \implies 4x = -8 \implies x = -2

.

When the absolute value expressions have opposite signs,

|x - 7| = 3|x + 5|

simplifies to

x - 7 = -3(x + 5)

.

4

Calculate the sum of all possible values of

x

(-11) + (-2) = -13

.

Summing the two solutions gives the final required value.

Anahtar Kavram

Distance on a number line and absolute value equations

Tahmini Süre:1m 30s

Soru 3Soru →

If $x$ and $y$ are real numbers such that $|x + 2| \le 3$ and $|y - 5| \le 2$ , which of the following could be the value of $|x - y|$ ? Select all such values.

Geçerli olan tümünü seçin

1

4

8

12

15

Cevabı ve açıklamayı göster

Cevap: 4; 8; 12

Cevap

The possible values for

|x - y|

are 4, 8, and 12.

Solving the inequality

|x + 2| \le 3

gives

-5 \le x \le 1

, and solving

|y - 5| \le 2

gives

3 \le y \le 7

. The minimum possible value of

x - y

occurs at

-5 - 7 = -12

, and the maximum value occurs at

1 - 3 = -2

. Thus,

x - y

lies entirely in the interval

[-12, -2]

. Taking absolute values shows that

|x - y|

must lie in the interval

[2, 12]

. The numbers 4, 8, and 12 all fall within this interval and are valid solutions.

Adım Adım Çözüm

1

Solve the absolute value inequality for

x

.

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

Unpack

|x + 2| \le 3

into a compound inequality and isolate

x

.

2

Solve the absolute value inequality for

y

.

-2 \le y - 5 \le 2 \implies 3 \le y \le 7

Unpack

|y - 5| \le 2

into a compound inequality and isolate

y

.

3

Determine the minimum and maximum possible values of

x - y

.

Minimum

x - y = -5 - 7 = -12

; Maximum

x - y = 1 - 3 = -2

.

To minimize

x - y

, take the smallest

x

and largest

y

. To maximize

x - y

, take the largest

x

and smallest

y

.

4

Find the range of

|x - y|

.

2 \le |x - y| \le 12

Since

-12 \le x - y \le -2

, taking the absolute value yields values in the closed interval

[2, 12]

.

5

Evaluate the choices against the interval

[2, 12]

.

4, 8, and 12 lie inside

[2, 12]

, whereas 1 and 15 do not.

Any real number in

[2, 12]

is a achievable value for

|x - y|

.

Anahtar Kavram

Real Numbers, Number Line, and Absolute Value Inequalities

Soru 4Soru →

On the real number line, the set of all real numbers $x$ that satisfy the inequality $|3x - 7| \le 11$ forms a closed interval $[a, b]$ . What is the value of $|a + b|$ ?

\frac{14}{3}

\frac{22}{3}

6

\frac{10}{3}

\frac{26}{3}

Cevabı ve açıklamayı göster

Cevap: $\frac{14}{3}$

Cevap

The value of

|a + b|

is

\frac{14}{3}

.

Rewriting the inequality

|3x - 7| \le 11

as

-11 \le 3x - 7 \le 11

and solving yields

-\frac{4}{3} \le x \le 6

. Thus, the endpoints are

a = -\frac{4}{3}

and

b = 6

. Summing these values gives

a + b = \frac{14}{3}

, and taking the absolute value yields

\frac{14}{3}

.

Adım Adım Çözüm

1

Express the absolute value inequality as a compound inequality.

-11 \le 3x - 7 \le 11

An inequality of the form

|u| \le k

for

k \ge 0

is equivalent to

-k \le u \le k

.

2

Isolate

3x

by adding

7

to all parts of the inequality.

-11 + 7 \le 3x \le 11 + 7 \implies -4 \le 3x \le 18

Adding a constant to all parts preserves the direction of the inequality.

3

Divide all parts by

3

to solve for

x

.

-\frac{4}{3} \le x \le 6

Dividing by a positive number isolates

x

without flipping inequality signs.

4

Identify interval bounds

a

and

b

, then calculate

|a + b|

.

a = -\frac{4}{3}

,

b = 6 \implies a + b = -\frac{4}{3} + \frac{18}{3} = \frac{14}{3} \implies \left|\frac{14}{3}\right| = \frac{14}{3}

The question asks for the absolute value of the sum of the endpoints of interval

[a, b]

.

Anahtar Kavram

Absolute value inequalities on the number line and interval endpoints

Soru 5Soru →

If $x$ is a real number such that $|2x - 5| = |x + 4|$ , which of the following could be the value of $x^2$ ? Select all such values.

Geçerli olan tümünü seçin

\frac{1}{9}

\frac{1}{3}

1

9

81

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{9}$ ; $81$

Cevap

The possible values of

x^2

are

\frac{1}{9}

and

81

.

To solve

|2x - 5| = |x + 4|

, set

2x - 5

equal to both

x + 4

and

-(x + 4)

. Solving

2x - 5 = x + 4

gives

x = 9

, which leads to

x^2 = 81

. Solving

2x - 5 = -x - 4

gives

3x = 1

, so

x = \frac{1}{3}

, which leads to

x^2 = \frac{1}{9}

. Therefore, the options representing

\frac{1}{9}

and

81

are correct.

Adım Adım Çözüm

1

Set up equations to remove the absolute value bars

Two linear equations:

2x - 5 = x + 4

or

2x - 5 = -(x + 4)

.

For real numbers

A

and

B

,

|A| = |B|

implies

A = B

or

A = -B

.

2

Solve the first equation

2x - 5 = x + 4

x = 9

.

Subtract

x

and add

5

to both sides.

3

Solve the second equation

2x - 5 = -x - 4

3x = 1 \implies x = \frac{1}{3}

.

Add

x

and add

5

to both sides.

4

Compute

x^2

for each possible value of

x

If

x = 9

, then

x^2 = 81

. If

x = \frac{1}{3}

, then

x^2 = \frac{1}{9}

.

The question asks for the values of

x^2

, not

x

.

Anahtar Kavram

Solving equations with absolute values on both sides requires considering positive and negative case equivalences.

Soru 6Soru →

If $x$ and $y$ are real numbers such that $|x - 4| \le 3$ and $|y + 2| \le 5$ , what is the maximum possible value of $|x - y|$ ?

Cevabı ve açıklamayı göster

Cevap: 14

Cevap

The maximum possible value of

|x - y|

is 14.

Solving

|x - 4| \le 3

gives the closed interval

[1, 7]

for

x

. Solving

|y + 2| \le 5

gives the closed interval

[-7, 3]

for

y

. The maximum possible value of

|x - y|

is the maximum distance between a point in

[1, 7]

and a point in

[-7, 3]

, which is

7 - (-7) = 14

.

Adım Adım Çözüm

1

Determine the range of possible values for

x

.

1 \le x \le 7

The inequality

|x - 4| \le 3

represents all numbers within distance 3 of 4 on the number line.

2

Determine the range of possible values for

y

.

-7 \le y \le 3

The inequality

|y + 2| \le 5

represents all numbers within distance 5 of -2 on the number line.

3

Find the maximum distance between any point

x

in

[1, 7]

and any point

y

in

[-7, 3]

.

14

The maximum absolute difference

|x - y|

occurs between the upper endpoint of the

x

-interval (

7

) and the lower endpoint of the

y

-interval (

-7

).

Anahtar Kavram

Absolute value inequalities as distance intervals on the real number line and maximizing differences between bounded variables

Soru 7Soru →

On the real number line, $x$ is a negative real number such that $|5 - 2x| = 11$ , and $y$ is a positive real number such that $|3y + 1| = 13$ . What is the distance on the real number line between $x$ and $y$ ?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The distance between

x

and

y

on the real number line is 7.

Solving

|5 - 2x| = 11

with

x < 0

yields

x = -3

. Solving

|3y + 1| = 13

with

y > 0

yields

y = 4

. The distance between

-3

and

4

on the number line is

|4 - (-3)| = 7

.

Adım Adım Çözüm

1

Solve for the negative real number

x

using the equation

|5 - 2x| = 11

.

x = -3

The equation splits into

5 - 2x = 11

(yielding

x = -3

) and

5 - 2x = -11

(yielding

x = 8

). Since

x

must be negative,

x = -3

is selected.

2

Solve for the positive real number

y

using the equation

|3y + 1| = 13

.

y = 4

The equation splits into

3y + 1 = 13

(yielding

y = 4

) and

3y + 1 = -13

(yielding

y = -\frac{14}{3}

). Since

y

must be positive,

y = 4

is selected.

3

Compute the distance between

x

and

y

on the number line.

7

The distance between two points on the number line is given by

|y - x| = |4 - (-3)| = 7

.

Anahtar Kavram

Distance on the real number line between two points

a

and

b

is given by

|a - b|

, solved by evaluating absolute value equations under given sign constraints.

Alternatif Yöntem

Plot the candidate solutions for

x

(

x = -3

and

x = 8

) and

y

(

y = 4

and

y = -14/3

) on a number line, then directly count units between the valid points

x = -3

and

y = 4

.

Tahmini Süre:1m 15s

Soru 8Soru →

Let $x$ and $y$ be real numbers such that $|x^2 - 4| = 2y$ and $|y - 3| \le 1$ . Which of the following could be the value of $x$ ? Select all such values.

Geçerli olan tümünü seçin

-3

0

2

3

4

Cevabı ve açıklamayı göster

Cevap: $-3$ ; $0$ ; $3$

Cevap

The values that

x

could take are

-3

,

0

, and

3

.

The inequality

|y - 3| \le 1

restricts

y

to the closed interval

[2, 4]

. Consequently,

2y

lies in

[4, 8]

, which means

4 \le |x^2 - 4| \le 8

. Breaking this compound inequality into cases yields

x = 0

(from

x^2 = 0

) or

x \in [-\sqrt{12}, -\sqrt{8}] \cup [\sqrt{8}, \sqrt{12}]

. Since

\sqrt{8} \approx 2.83

and

\sqrt{12} \approx 3.46

, the integer values

-3

and

3

fall in these intervals, alongside

0

. Thus,

-3

,

0

, and

3

are all valid choices.

Adım Adım Çözüm

1

Determine the acceptable range for

y

from the given absolute value inequality.

Solving

|y - 3| \le 1

gives

-1 \le y - 3 \le 1

, which simplifies to

2 \le y \le 4

.

The inequality

|y - c| \le r

represents all values of

y

within distance

r

from

c

.

2

Relate the range of

y

to the absolute value expression

|x^2 - 4|

.

Since

2 \le y \le 4

, multiplying by

2

yields

4 \le 2y \le 8

. Therefore,

4 \le |x^2 - 4| \le 8

.

Substitute

2y = |x^2 - 4|

into the inequality derived for

y

.

3

Analyze the two cases for the absolute value equation

|x^2 - 4|

.

Case 1:

4 \le x^2 - 4 \le 8 \implies 8 \le x^2 \le 12

, which gives

x \in [-\sqrt{12}, -\sqrt{8}] \cup [\sqrt{8}, \sqrt{12}]

. Case 2:

4 \le -(x^2 - 4) \le 8 \implies 4 \le 4 - x^2 \le 8 \implies -4 \le -x^2 \le 4 \implies -4 \le x^2 \le 0

. Since

x^2 \ge 0

for all real numbers,

x^2 = 0 \implies x = 0

.

Absolute value

|A|

splits into

A

when

A \ge 0

and

-A

when

A < 0

.

4

Evaluate the test values against the valid domains for

x

.

For

x = -3

,

x^2 = 9

, which satisfies

8 \le 9 \le 12

. For

x = 0

,

x^2 = 0

, which satisfies

x^2 = 0

. For

x = 3

,

x^2 = 9

, which satisfies

8 \le 9 \le 12

. The values

x = 2

and

x = 4

give

x^2 = 4

and

x^2 = 16

, neither of which fall into the valid intervals.

Checking test options confirms which values fall within the solution intervals.

Anahtar Kavram

Absolute Value Equations and System Constraints on Real Numbers

Soru 9Soru →

On the real number line, the distance between two real numbers $x$ and $y$ is $d$ . If the midpoint of $x$ and $y$ is $7$ and $|x - 3| = 2|y - 3|$ , what is the maximum possible value of $d$ ?

12

16

24

28

32

Cevabı ve açıklamayı göster

Cevap: 24

Cevap

The maximum possible value of

d

is 24.

The midpoint condition dictates that

x + y = 14

, so

y = 14 - x

and the distance between them is

d = |x - y| = 2|x - 7|

. Substituting

y = 14 - x

into

|x - 3| = 2|y - 3|

gives

|x - 3| = 2|11 - x|

. Solving the positive case

x - 3 = 2(x - 11)

gives

x = 19

and

y = -5

, resulting in distance

d = |19 - (-5)| = 24

. Solving the negative case

x - 3 = -2(x - 11)

gives

x = 25/3

and

y = 17/3

, resulting in distance

d = 8/3

. Thus, 24 is the maximum possible value of

d

.

Adım Adım Çözüm

1

Express

y

in terms of

x

using the midpoint formula

Since the midpoint of

x

and

y

is

7

,

\frac{x + y}{2} = 7

, which gives

y = 14 - x

. The distance

d = |x - y| = |x - (14 - x)| = |2x - 14| = 2|x - 7|

.

Relating

y

to

x

reduces the problem to a single variable.

2

Substitute

y = 14 - x

into the absolute value equation

|x - 3| = 2|(14 - x) - 3| \implies |x - 3| = 2|11 - x| = 2|x - 11|

.

Setting up the single-variable absolute value equation allows finding all possible values for

x

.

3

Solve the absolute value equation for all possible cases

Case 1:

x - 3 = 2(x - 11) \implies x - 3 = 2x - 22 \implies x = 19

. Then

y = 14 - 19 = -5

.
Case 2:

x - 3 = -2(x - 11) \implies x - 3 = -2x + 22 \implies 3x = 25 \implies x = \frac{25}{3}

. Then

y = 14 - \frac{25}{3} = \frac{17}{3}

.

Absolute value equations

|A| = B

split into

A = B

and

A = -B

.

4

Calculate the distance

d

for each case and select the maximum

For Case 1 (

x = 19, y = -5

):

d = |19 - (-5)| = 24

.
For Case 2 (

x = 25/3, y = 17/3

):

d = |25/3 - 17/3| = 8/3

.
The maximum possible value of

d

is

24

.

Comparing the distance values determined in each case identifies the maximum distance.

Anahtar Kavram

Absolute Value as Distance and Multi-Case Equations on the Real Number Line

Tahmini Süre:2m 30s

Soru 10Soru →

If $x$ is a real number such that $|x^2 - 9| = 5|x - 3|$ , which of the following could be the value of $x$ ? Select all such values.

Geçerli olan tümünü seçin

-8

-2

2

3

8

Cevabı ve açıklamayı göster

Cevap: $-8$ ; $2$ ; $3$

Cevap

The values of

x

that satisfy the equation are

-8

,

2

, and

3

.

Factoring the left side of

|x^2 - 9| = 5|x - 3|

yields

|x - 3||x + 3| = 5|x - 3|

. Setting the common factor

|x - 3| = 0

gives

x = 3

. Dividing both sides by the non-zero quantity

|x - 3|

leaves

|x + 3| = 5

, which splits into

x + 3 = 5 \implies x = 2

and

x + 3 = -5 \implies x = -8

. Therefore,

-8

,

2

, and

3

are all valid solutions.

Adım Adım Çözüm

1

Apply the product rule for absolute values to factor the left-hand side.

|x^2 - 9| = |(x - 3)(x + 3)| = |x - 3| \cdot |x + 3|

, rewriting the equation as

|x - 3| \cdot |x + 3| = 5|x - 3|

.

The absolute value of a product is equal to the product of the individual absolute values.

2

Evaluate the case where the shared factor is zero:

|x - 3| = 0

.

x - 3 = 0 \implies x = 3

. Both sides equal

0

, making

x = 3

a valid solution.

Dividing by

|x - 3|

without checking if it can be zero would cause the loss of the root

x = 3

.

3

Evaluate the case where

|x - 3| \neq 0

by dividing both sides of the equation by

|x - 3|

.

|x + 3| = 5

.

Since

|x - 3| > 0

, we can safely divide both sides by this non-zero quantity.

4

Solve the remaining absolute value equation

|x + 3| = 5

.

x + 3 = 5 \implies x = 2

, and

x + 3 = -5 \implies x = -8

.

An expression inside an absolute value equal to a positive number

k

can equal either

k

or

-k

.

Anahtar Kavram

Factoring absolute value expressions using

|ab| = |a||b|

and systematically considering all cases to avoid dropping zero-roots or negative solutions.

Soru 11Soru →

On the real number line, the point $P$ corresponds to the real number $x$ . If $|x - 3| = 7$ , which of the following is a possible value of $x + 5$ ?

-9

-4

1

9

12

Cevabı ve açıklamayı göster

Cevap: $1$

Cevap

The correct answer is

1

.

Solving

|x - 3| = 7

produces two possible values for

x

:

x = 10

and

x = -4

. Substituting

x = -4

into

x + 5

yields

-4 + 5 = 1

, which is listed among the choices.

Adım Adım Çözüm

1

Set up the two linear equations defined by the absolute value equation

|x - 3| = 7

.

The two cases are

x - 3 = 7

and

x - 3 = -7

.

By definition of absolute value,

|k| = 7

means

k = 7

or

k = -7

.

2

Solve each case for

x

.

From

x - 3 = 7

,

x = 10

. From

x - 3 = -7

,

x = -4

.

Adding

3

to both sides of each equation isolates

x

.

3

Evaluate

x + 5

for both solutions of

x

.

If

x = 10

,

x + 5 = 15

. If

x = -4

,

x + 5 = 1

.

Substitute each possible value of

x

into the expression

x + 5

.

Anahtar Kavram

Absolute Value as Distance on the Number Line

Tahmini Süre:45s

Soru 12Soru →

On the real number line, $x$ is a real number such that $|x + 2| \le 5$ . Which of the following could be the value of $x$ ? Select all such values.

Geçerli olan tümünü seçin

-8

-7

0

3

4

Cevabı ve açıklamayı göster

Cevap: $-7$ ; $0$ ; $3$

Cevap

The values that satisfy the inequality are

-7

,

0

, and

3

.

The inequality

|x + 2| \le 5

represents all real numbers

x

whose distance from

-2

on the number line is at most

5

. This gives the closed interval

[-7, 3]

. The values

-7

,

0

, and

3

all fall within this range.

Adım Adım Çözüm

1

Set up the compound inequality for the absolute value expression.

-5 \le x + 2 \le 5

For any real number expression

A

and non-negative constant

k

,

|A| \le k

is equivalent to

-k \le A \le k

.

2

Isolate

x

by subtracting

2

from all parts of the inequality.

-5 - 2 \le x \le 5 - 2 \implies -7 \le x \le 3

Subtracting a constant from all parts preserves the inequality signs.

3

Evaluate which given choices lie within the interval

[-7, 3]

.

-7

,

0

, and

3

lie within

[-7, 3]

, while

-8

and

4

lie outside.

Only values inside or on the boundaries of

[-7, 3]

satisfy the inequality.

Anahtar Kavram

Absolute Value Inequalities and Real Number Line Distance

Soru 13Soru →

On a number line, point $P$ corresponds to $-3$ and point $Q$ corresponds to $5$ . If point $R$ is the midpoint of line segment $PQ$ , what is the value of $|R - P|$ ?

2

4

8

1

15

Cevabı ve açıklamayı göster

Cevap: $4$

Cevap

The correct distance between point P and midpoint R is 4.

The total distance between

P (-3)

and

Q (5)

on the real number line is

|5 - (-3)| = 8

. Since

R

is the midpoint of segment

PQ

, it splits the total length into two equal halves of length

4

. Therefore, the distance from

P

to

R

, represented by

|R - P|

, is

4

.

Adım Adım Çözüm

1

Calculate the total distance between points P and Q on the number line.

Distance

PQ = |5 - (-3)| = |5 + 3| = 8

.

The distance between any two points

a

and

b

on a number line is given by

|b - a|

.

2

Find the coordinate of the midpoint R.

Coordinate of

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

.

The midpoint coordinate is the average of the coordinates of the two endpoints.

3

Calculate the absolute distance between R and P.

|R - P| = |1 - (-3)| = |1 + 3| = 4

.

The absolute value gives the non-negative distance between the midpoint and point P.

Anahtar Kavram

Distance on a Number Line and Midpoint Formula

Soru 14Soru →

If $|x - 4| = 9$ and $x < 0$ , what is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: -5

Cevap

The value of

x

is

-5

.

The absolute value equation

|x - 4| = 9

specifies that the distance between

x

and

4

on the real number line is equal to

9

. Moving

9

units to the left of

4

gives

4 - 9 = -5

, and moving

9

units to the right gives

4 + 9 = 13

. Because

x

is specified to be negative (

x < 0

), the correct value of

x

is

-5

.

Adım Adım Çözüm

1

Set up the two linear equations representing the absolute value equation

|x - 4| = 9

.

x - 4 = 9

or

x - 4 = -9

By definition,

|A| = B

(where

B \ge 0

) implies

A = B

or

A = -B

.

2

Solve for

x

in both equations.

x = 13

or

x = -5

Adding

4

to both sides of

x - 4 = 9

gives

x = 13

, and adding

4

to both sides of

x - 4 = -9

gives

x = -5

.

3

Select the value of

x

that satisfies the given condition

x < 0

.

x = -5

The value

13

is positive, while

-5

is negative and fulfills

x < 0

.

Anahtar Kavram

Absolute Value as Distance and Solving Absolute Value Equations

Tahmini Süre:45s

Soru 15Soru →

On the real number line, $p$ and $q$ are real numbers such that $|3 - 2p| \le 5$ and $|q + 4| = 2$ . What is the maximum possible value of $|p^2 - q|$ ?

22

18

14

10

7

Cevabı ve açıklamayı göster

Cevap: 22

Cevap

22

Solving

|3 - 2p| \le 5

gives

-5 \le 3 - 2p \le 5 \implies -1 \le p \le 4

, so

p^2

can range from

0

up to

16

. Solving

|q + 4| = 2

yields

q = -2

or

q = -6

. To maximize

|p^2 - q|

, we combine the maximum possible value of

p^2

(

16

) with

q = -6

, obtaining

|16 - (-6)| = 22

.

Adım Adım Çözüm

1

Solve the absolute value inequality

|3 - 2p| \le 5

for

p

.

-1 \le p \le 4

-5 \le 3 - 2p \le 5 \implies -8 \le -2p \le 2

. Dividing by

-2

and reversing the inequality direction gives

-1 \le p \le 4

.

2

Determine the range of possible values for

p^2

.

0 \le p^2 \le 16

Since

p

spans from

-1

to

4

(which includes

0

), the minimum square is

0^2 = 0

and the maximum square is

4^2 = 16

.

3

Solve the absolute value equation

|q + 4| = 2

for

q

.

q = -2

or

q = -6

q + 4 = 2 \implies q = -2

, and

q + 4 = -2 \implies q = -6

.

4

Find the combination of

p^2

and

q

that maximizes

|p^2 - q|

.

Maximum value is 22

Pairing

p^2 = 16

with

q = -6

yields

|16 - (-6)| = 22

, which is greater than

|16 - (-2)| = 18

.

Anahtar Kavram

Absolute value inequalities, real number bounds, and distance optimization.

Soru 16Soru →

On a real number line, point $A$ has coordinate $-5$ and point $B$ has coordinate $11$ . Point $C$ has coordinate $x$ such that the distance between $C$ and the midpoint of line segment $AB$ is equal to $\frac{1}{3}$ of the distance between $C$ and point $B$ . What is the maximum possible value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The maximum possible value of

x

is

5

.

The midpoint of

A(-5)

and

B(11)

is

M = 3

. Setting up the distance equation

|x - 3| = \frac{1}{3}|x - 11|

leads to

3|x - 3| = |x - 11|

. Evaluating the two cases

3(x - 3) = x - 11

and

3(x - 3) = -(x - 11)

yields solutions

x = -1

and

x = 5

. The maximum possible value among these is 5.

Adım Adım Çözüm

1

Calculate the coordinate of the midpoint of line segment

AB

.

Midpoint coordinate

M = 3

.

The midpoint of two coordinates

a

and

b

on a number line is given by

\frac{a + b}{2} = \frac{-5 + 11}{2} = 3

.

2

Formulate the distance equation using absolute values.

3|x - 3| = |x - 11|

.

The distance between

x

and

3

is

|x - 3|

and the distance between

x

and

11

is

|x - 11|

. The given condition is

|x - 3| = \frac{1}{3}|x - 11|

.

3

Solve the absolute value equation for all possible values of

x

.

x = -1

and

x = 5

.

Splitting

3(x - 3) = x - 11

yields

x = -1

, and splitting

3(x - 3) = -(x - 11)

yields

x = 5

.

4

Select the maximum value among all valid solutions.

5

Comparing

x = -1

and

x = 5

, the maximum value is

5

.

Anahtar Kavram

Distance on a number line using absolute value and midpoint formula

Tahmini Süre:2m 0s

Soru 17Soru →

For how many integer values of $x$ does the inequality $||2x - 7| - 5| \le 4$ hold true?

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

10

To solve

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

, break the outer absolute value into

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

. Adding 5 to all parts yields

1 \le |2x - 7| \le 9

. This produces two simultaneous conditions:

|2x - 7| \le 9

, which gives

-1 \le x \le 8

, and

|2x - 7| \ge 1

, which gives

x \le 3

or

x \ge 4

. Intersecting these solution sets yields the real intervals

[-1, 3]

and

[4, 8]

. The integers contained within these intervals are

-1, 0, 1, 2, 3

and

4, 5, 6, 7, 8

, giving a total of 10 valid integer values.

Adım Adım Çözüm

1

Unpack the outer absolute value inequality

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

An inequality of the form

|u| \le k

for

k \ge 0

is equivalent to

-k \le u \le k

.

2

Isolate the inner absolute value term

1 \le |2x - 7| \le 9

Adding 5 across the compound inequality isolates the term

|2x - 7|

.

3

Solve the upper bound inequality

|2x - 7| \le 9

-1 \le x \le 8

-9 \le 2x - 7 \le 9 \implies -2 \le 2x \le 16 \implies -1 \le x \le 8

.

4

Solve the lower bound inequality

|2x - 7| \ge 1

x \le 3

or

x \ge 4

The inequality

|u| \ge 1

splits into

u \ge 1

or

u \le -1

, yielding

2x - 7 \ge 1 \implies x \ge 4

or

2x - 7 \le -1 \implies x \le 3

.

5

Determine the overlapping interval and count integer solutions

10 integer solutions

The intersection of

[-1, 8]

with

((-\infty, 3] \cup [4, \infty))

is

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

. The integers in this domain are

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

, which equals 10 integer values.

Anahtar Kavram

Solving nested absolute value inequalities on the real number line

Soru 18Soru →

On the real number line, $x$ is a real number such that its distance from $1$ is at most $6$ , and its distance from $-2$ is at least $4$ . Which of the following inequalities represents all possible values of $x$ ?

2 \le x \le 7

-5 \le x \le 7

-6 \le x \le 7

-5 \le x \le 2

-5 \le x \le -2 \text{ or } 2 \le x \le 7

Cevabı ve açıklamayı göster

Cevap: $2 \le x \le 7$

Cevap

The inequality

2 \le x \le 7

represents all possible values of

x

.

The distance condition 'at most 6 units from 1' translates to

|x - 1| \le 6

, yielding the interval

[-5, 7]

. The condition 'at least 4 units from -2' translates to

|x + 2| \ge 4

, yielding

x \le -6

or

x \ge 2

. Finding the values that belong to both conditions requires taking the intersection of

[-5, 7]

and

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

. Since

[-5, 7]

does not overlap with

(-\infty, -6]

, the only overlapping region is

[2, 7]

, which corresponds to

2 \le x \le 7

.

Adım Adım Çözüm

1

Express the geometric distance conditions as absolute value inequalities

Condition 1:

|x - 1| \le 6

. Condition 2:

|x - (-2)| = |x + 2| \ge 4

.

The distance between two numbers

a

and

b

on the number line is given by

|a - b|

.

2

Solve the first inequality

|x - 1| \le 6

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

.

An absolute value inequality of the form

|u| \le k

(for

k \ge 0

) unwraps to

-k \le u \le k

.

3

Solve the second inequality

|x + 2| \ge 4

x + 2 \ge 4

or

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

or

x \le -6

.

An absolute value inequality of the form

|u| \ge k

(for

k \ge 0

) unwraps to

u \ge k

or

u \le -k

.

4

Find the intersection of the two solution sets

Combine

[-5, 7]

with

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

. The intersection of

[-5, 7]

and

(-\infty, -6]

is empty since

-5 > -6

. The intersection of

[-5, 7]

and

[2, \infty)

is

[2, 7]

, which means

2 \le x \le 7

.

The variable

x

must satisfy both conditions simultaneously.

Anahtar Kavram

Distance on a number line expressed as absolute value inequalities and finding overlapping intervals

Tahmini Süre:2m 0s

Soru 19Soru →

On the real number line, points $P$ , $Q$ , and $R$ have coordinates $x$ , $y$ , and $z$ , respectively, such that $|x - 2| = 5$ , $|y + 4| = 3$ , and $z$ is the midpoint of segment $PQ$ . If $x < y$ , what is the value of $z$ ?

Cevabı ve açıklamayı göster

Cevap: -2

Cevap

The coordinate of point

R

(the value of

z

) is

-2

.

Solving the two absolute value equations yields

x \in \{-3, 7\}

and

y \in \{-7, -1\}

. Testing the given constraint

x < y

across all four possible ordered pairs reveals that only

x = -3

and

y = -1

satisfy the condition, since

-3 < -1

is true while all other pairs fail. The midpoint of points with coordinates

-3

and

-1

is

\frac{-3 + (-1)}{2} = -2

.

Adım Adım Çözüm

1

Solve the absolute value equation

|x - 2| = 5

.

x = 7

or

x = -3

The equation

|x - 2| = 5

splits into

x - 2 = 5 \implies x = 7

and

x - 2 = -5 \implies x = -3

.

2

Solve the absolute value equation

|y + 4| = 3

.

y = -1

or

y = -7

The equation

|y + 4| = 3

splits into

y + 4 = 3 \implies y = -1

and

y + 4 = -3 \implies y = -7

.

3

Evaluate all candidate pairs

(x, y)

under the inequality constraint

x < y

.

The only valid pair is

x = -3

and

y = -1

.

Comparing all four combinations:

7 < -1

(false),

7 < -7

(false),

-3 < -7

(false), and

-3 < -1

(true).

4

Calculate the midpoint

z

of the segment

PQ

.

z = -2

The midpoint of coordinates

-3

and

-1

on a number line is given by their average:

\frac{-3 + (-1)}{2} = -2

.

Anahtar Kavram

Absolute Value Equations and Midpoint on the Real Number Line

Soru 20Soru →

Let $a$ and $b$ be non-zero real numbers such that $|a + b| = 3|a - b|$ . What is the value of $\left|\frac{a^2 + b^2}{ab}\right|$ ?

\frac{3}{2}

2

\frac{5}{2}

3

\frac{10}{3}

Cevabı ve açıklamayı göster

Cevap: $\frac{5}{2}$

Cevap

The value of

\left|\frac{a^2 + b^2}{ab}\right|

is

\frac{5}{2}

.

Squaring both sides of the given equation

|a + b| = 3|a - b|

eliminates the absolute values to give

(a + b)^2 = 9(a - b)^2

. Expanding and simplifying yields

2a^2 - 5ab + 2b^2 = 0

, which factors as

(2a - b)(a - 2b) = 0

. Thus,

b = 2a

or

a = 2b

. In either case, substituting into

\left|\frac{a^2 + b^2}{ab}\right|

reduces the expression to

\frac{5}{2}

.

Adım Adım Çözüm

1

Square both sides of the absolute value equality

(a + b)^2 = 9(a - b)^2

Since both sides of

|a + b| = 3|a - b|

are non-negative real numbers, squaring preserves equality and eliminates the absolute value signs.

2

Expand both algebraic expressions and collect like terms

a^2 + 2ab + b^2 = 9a^2 - 18ab + 9b^2 \implies 8a^2 - 20ab + 8b^2 = 0 \implies 2a^2 - 5ab + 2b^2 = 0

Expanding the binomial squares allows combining like terms to solve for the relationship between

a

and

b

.

3

Factor the quadratic expression in terms of

a

and

b

(2a - b)(a - 2b) = 0 \implies b = 2a \text{ or } a = 2b

Factoring determines the exact proportional relationship between

a

and

b

.

4

Substitute the relation into the target expression

\left|\frac{a^2 + (2a)^2}{a(2a)}\right| = \left|\frac{5a^2}{2a^2}\right| = \frac{5}{2}

Substituting

b = 2a

allows

a^2

to cancel completely, leaving a constant numerical value.

Anahtar Kavram

Properties of Real Numbers and Absolute Value Equations

Tahmini Süre:2m 0s

Real Numbers, Number Line, and Absolute Value

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