Inequalities and Absolute Value Equations Alıştırma Soruları — GMAT

Determine the sum of all real solutions to the equation $|x^2 - 4x| = 3x - 6$ .

7

8

9

10

12

Cevabı ve açıklamayı göster

Cevap: $9$

Cevap

The sum of all valid real solutions is

9

.

The correct answer is

9

. Setting up the two cases

x^2 - 4x = 3x - 6

and

x^2 - 4x = -(3x - 6)

yields candidate roots

x = 1, 6, 3,

and

-2

. Because the absolute value expression

|x^2 - 4x|

cannot be negative,

3x - 6

must be non-negative, requiring

x \ge 2

. Evaluating each candidate shows that

x = 1

and

x = -2

produce negative right-hand sides and are extraneous. The only valid solutions are

x = 3

and

x = 6

, whose sum is

3 + 6 = 9

.

Adım Adım Çözüm

1

Establish the domain condition for the right-hand side of the absolute value equation.

Since absolute values are non-negative,

|x^2 - 4x| \geq 0

requires

3x - 6 \geq 0 \implies x \geq 2

.

An absolute value expression cannot equal a negative number.

2

Solve Case 1 where

x^2 - 4x = 3x - 6

.

x^2 - 7x + 6 = 0 \implies (x - 1)(x - 6) = 0 \implies x = 1

or

x = 6

.

This corresponds to the positive branch of the absolute value.

3

Solve Case 2 where

x^2 - 4x = -(3x - 6)

.

x^2 - 4x = -3x + 6 \implies x^2 - x - 6 = 0 \implies (x - 3)(x + 2) = 0 \implies x = 3

or

x = -2

.

This corresponds to the negative branch of the absolute value.

4

Test all candidate solutions (

x = -2, 1, 3, 6

) against the domain constraint

x \geq 2

.

x = -2

yields

3(-2)-6 = -12 < 0

(extraneous).

x = 1

yields

3(1)-6 = -3 < 0

(extraneous).

x = 3

yields

|9-12| = 3 = 3(3)-6

(valid).

x = 6

yields

|36-24| = 12 = 3(6)-6

(valid).

Extraneous roots introduced by unconstrained case splitting must be eliminated.

5

Sum the valid real solutions.

3 + 6 = 9

.

The question asks specifically for the sum of all valid real solutions.

Anahtar Kavram

Absolute Value Equations and Extraneous Solution Verification

Soru 2Soru →

How many integer values of $y$ satisfy the inequality $|2y - 7| \le |y + 2|$ ?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The total number of integer values of

y

satisfying the inequality is 8.

Squaring both sides of

|2y - 7| \le |y + 2|

gives

3y^2 - 32y + 45 \le 0

, which factors into

(3y - 5)(y - 9) \le 0

. The solution range for

y

is

\frac{5}{3} \le y \le 9

. Since

\frac{5}{3} \approx 1.67

, the integer values of

y

satisfying this range are

2, 3, 4, 5, 6, 7, 8, 9

, giving a total of 8 integers.

Adım Adım Çözüm

1

Square both sides of the inequality

|2y - 7| \le |y + 2|

(2y - 7)^2 \le (y + 2)^2

Since both sides of an absolute value expression are non-negative, squaring both sides maintains the inequality direction.

2

Expand terms and move all terms to the left side

3y^2 - 32y + 45 \le 0

Expanding gives

4y^2 - 28y + 49 \le y^2 + 4y + 4

. Subtracting

(y^2 + 4y + 4)

from both sides produces the standard quadratic inequality.

3

Factor the quadratic expression to find critical points

(3y - 5)(y - 9) \le 0

, yielding

\frac{5}{3} \le y \le 9

The roots are

y = \frac{5}{3}

and

y = 9

. A quadratic with a positive leading coefficient is non-positive between its roots.

4

Determine all integers within the range

\left[\frac{5}{3}, 9\right]

2, 3, 4, 5, 6, 7, 8, 9

(8 integers total)

Because

\frac{5}{3} \approx 1.67

, the smallest integer within the range is 2 and the largest is 9.

Anahtar Kavram

Solving absolute value inequalities of the form

|A| \le |B|

by squaring both sides and determining integer solutions.

Soru 3Soru →

How many integer values of $x$ satisfy the inequality $||2x - 5| - 7| \leq 4$ ?

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The total number of integer values of

x

satisfying the inequality is 10.

To solve

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

, rewrite the inequality without the outer absolute value as

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

. Adding 7 across all parts yields

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

. The inequality

|2x - 5| \leq 11

simplifies to

-3 \leq x \leq 8

. The inequality

|2x - 5| \geq 3

simplifies to

x \leq 1

or

x \geq 4

. Intersecting these two regions gives the set of real numbers

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

. The integer solutions within

[-3, 1]

are

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

(5 integers), and within

[4, 8]

are

4, 5, 6, 7, 8

(5 integers). The total number of valid integer solutions is

5 + 5 = 10

.

Adım Adım Çözüm

1

Unfold the outer absolute value expression.

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

An inequality of the form

|U| \leq C

with

C > 0

is equivalent to

-C \leq U \leq C

.

2

Isolate the inner absolute value expression by adding 7 throughout.

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

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

3

Solve the upper bound

|2x - 5| \leq 11

.

-3 \leq x \leq 8

-11 \leq 2x - 5 \leq 11

adds 5 to give

-6 \leq 2x \leq 16

, which divides by 2 to yield

-3 \leq x \leq 8

.

4

Solve the lower bound

|2x - 5| \geq 3

.

x \leq 1

or

x \geq 4

An inequality

|U| \geq C

splits into

U \geq C

(

2x - 5 \geq 3 \implies x \geq 4

) or

U \leq -C

(

2x - 5 \leq -3 \implies x \leq 1

).

5

Find the intersection of the upper and lower bound conditions and count the integers.

10 integer solutions:

\{-3, -2, -1, 0, 1, 4, 5, 6, 7, 8\}

.

Combining

-3 \leq x \leq 8

with (

x \leq 1

or

x \geq 4

) produces two disjoint intervals

[-3, 1]

and

[4, 8]

, containing 5 integers each.

Anahtar Kavram

Solving nested absolute value inequalities using double inequalities and boundary region intersections.

Soru 4Soru →

What is the sum of all integer values of $x$ that satisfy the inequality $|x^2 - 6x + 5| < 2x - 2$ ?

Cevabı ve açıklamayı göster

Cevap: 15

Cevap

The sum of all integer values of

x

satisfying the inequality is

15

.

Factoring both sides gives

|(x - 1)(x - 5)| < 2(x - 1)

. Since the absolute value is non-negative, the right-hand side requires

2x - 2 > 0 \implies x > 1

. Under

x > 1

, the term

x - 1

is strictly positive, allowing us to simplify to

|x - 5| < 2

, which yields

3 < x < 7

. The integer solutions are

4, 5,

and

6

, and their sum is

15

.

Adım Adım Çözüm

1

Determine the necessary condition for the right-hand side of the inequality.

Because the left-hand side

|x^2 - 6x + 5|

is non-negative for all real

x

, the right-hand side must be strictly positive. Thus,

2x - 2 > 0 \implies x > 1

.

An absolute value expression cannot be strictly less than a zero or negative quantity.

2

Factor the quadratic expression inside the absolute value and the linear expression on the right.

|(x - 1)(x - 5)| < 2(x - 1)

.

Factoring reveals a common linear factor

(x - 1)

on both sides.

3

Simplify the inequality using the condition

x > 1

.

Since

x > 1

, we know

x - 1 > 0

, so

|x - 1| = x - 1

. Splitting the product inside the absolute value gives

(x - 1)|x - 5| < 2(x - 1)

. Dividing both sides by

(x - 1)

yields

|x - 5| < 2

.

Dividing an inequality by a strictly positive number preserves the direction of the inequality sign.

4

Solve the simplified absolute value inequality and sum the integer solutions.

|x - 5| < 2 \implies -2 < x - 5 < 2 \implies 3 < x < 7

. The integer solutions strictly within this range are

x = 4, 5,

and

6

. Their sum is

4 + 5 + 6 = 15

.

The integers strictly between

3

and

7

are

4

,

5

, and

6

.

Anahtar Kavram

Solving Quadratic Absolute Value Inequalities via Domain Constraints and Factoring

Tahmini Süre:2m 0s

Soru 5Soru →

If $|2x - 5| = 9$ , what is the sum of all possible values of $x$ ?

5

7

-2

9

14

Cevabı ve açıklamayı göster

Cevap: $5$

Cevap

The sum of all possible values of

x

is

5

.

To solve

|2x - 5| = 9

, split the equation into two linear cases:

2x - 5 = 9

and

2x - 5 = -9

. Solving

2x - 5 = 9

gives

2x = 14

, so

x = 7

. Solving

2x - 5 = -9

gives

2x = -4

, so

x = -2

. Summing these two solutions gives

7 + (-2) = 5

.

Adım Adım Çözüm

1

Set up the two equations based on the definition of absolute value.

2x - 5 = 9

and

2x - 5 = -9

For any expression

A

and real number

B \ge 0

,

|A| = B

implies

A = B

or

A = -B

.

2

Solve the first equation for

x

.

2x = 14 \implies x = 7

Add

5

to both sides and divide by

2

.

3

Solve the second equation for

x

.

2x = -4 \implies x = -2

Add

5

to both sides and divide by

2

.

4

Calculate the sum of all valid solutions.

7 + (-2) = 5

The question asks for the sum of all possible solutions for

x

.

Anahtar Kavram

Solving Absolute Value Equations

Soru 6Soru →

How many integer values of $n$ satisfy the inequality $|n^2 - 9| \leq 2n + 6$ ?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

There are 6 integer values of

n

that satisfy the given inequality.

Because

|n^2 - 9| \geq 0

for all real numbers,

2n + 6

must be non-negative, requiring

n \geq -3

. At

n = -3

, the inequality reduces to

0 \leq 0

, which is valid. For

n > -3

, the term

n + 3

is positive, so factoring

|n^2 - 9|

into

|n - 3|(n + 3)

and dividing by

n + 3

simplifies the inequality to

|n - 3| \leq 2

. This yields the integer range

1 \leq n \leq 5

, containing 5 integers. Including

n = -3

gives a total of 6 integer solutions.

Adım Adım Çözüm

1

Determine the domain constraint from the non-negativity of the absolute value expression.

2n + 6 \geq 0 \implies n \geq -3

The absolute value expression

|n^2 - 9|

is non-negative for all real

n

, so the right-hand side must also be non-negative.

2

Evaluate the boundary case

n = -3

.

|(-3)^2 - 9| = 0

and

2(-3) + 6 = 0 \implies 0 \leq 0

(True)

When

n = -3

, both sides equal 0, making

n = -3

a valid integer solution.

3

Factor the quadratic inside the absolute value for

n > -3

and simplify.

|n - 3||n + 3| \leq 2(n + 3) \implies |n - 3| \leq 2

Since

n > -3

, the term

n + 3

is strictly positive, allowing division of both sides by

n + 3

without reversing the inequality.

4

Solve the linear absolute value inequality

|n - 3| \leq 2

.

-2 \leq n - 3 \leq 2 \implies 1 \leq n \leq 5

Removing the absolute value creates a compound inequality bounded between

-2

and

2

.

5

List all valid integer solutions and count them.

n \in \{-3, 1, 2, 3, 4, 5\} \implies 6

integer solutions

Combining the boundary root

n = -3

with the five consecutive integers from

1

to

5

gives 6 integer solutions.

Anahtar Kavram

Absolute Value Inequalities with Variable RHS and Boundary Factors

Soru 7Soru →

How many integer values of $x$ satisfy the inequality $\frac{|x - 2| - 3}{|x + 1| - 4} \leq 0$ ?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The correct answer is 6.

Analyzing the signs of the numerator

|x - 2| - 3

and denominator

|x + 1| - 4

shows that the quotient is non-positive when the numerator and denominator have opposite signs or when the numerator is zero. This yields two intervals:

(-5, -1]

and

(3, 5]

. The integers contained in these intervals are

-4, -3, -2, -1, 4, 5

, which totals 6 integer values.

Adım Adım Çözüm

1

Analyze the sign of the numerator

|x - 2| - 3

The numerator is zero at

x = -1

and

x = 5

, negative for

-1 < x < 5

, and positive for

x < -1

or

x > 5

.

Solving

|x - 2| = 3

yields

x - 2 = 3 \implies x = 5

and

x - 2 = -3 \implies x = -1

.

2

Analyze the sign of the denominator

|x + 1| - 4

The denominator is zero at

x = -5

and

x = 3

, negative for

-5 < x < 3

, and positive for

x < -5

or

x > 3

. Exclude

x = -5

and

x = 3

.

Solving

|x + 1| = 4

yields

x + 1 = 4 \implies x = 3

and

x + 1 = -4 \implies x = -5

. Denominators cannot be zero.

3

Find intervals where numerator and denominator have opposite signs or numerator is zero

The solution set is the union of

(-5, -1]

and

(3, 5]

.

A fraction

\frac{N}{D} \leq 0

requires

N \ge 0, D < 0

or

N \le 0, D > 0

.

4

Count the total number of integer solutions

The valid integers are

-4, -3, -2, -1, 4, 5

, giving a total count of 6.

Listing integers in

(-5, -1]

yields

-4, -3, -2, -1

, and in

(3, 5]

yields

4, 5

.

Anahtar Kavram

Solving Rational Inequalities with Absolute Values

Soru 8Soru →

If $x$ is a real number satisfying the equation $|2x - 3| = 5x - 12$ , what is the sum of all valid real solutions for $x$ ?

3

\frac{15}{7}

\frac{36}{7}

-3

\frac{45}{7}

Cevabı ve açıklamayı göster

Cevap: $3$

Cevap

The sum of all valid real solutions is

3

.

The option specifying

3

correctly identifies

x = 3

as the only valid solution after discarding the extraneous root

x = \frac{15}{7}

, which produces a negative right-hand side in the original equation.

Adım Adım Çözüm

1

Set up the two algebraic cases for the absolute value equation

|2x - 3| = 5x - 12

.

Case 1:

2x - 3 = 5x - 12

Case 2:

2x - 3 = -(5x - 12) = -5x + 12

By definition,

|a| = b

implies

a = b

or

a = -b

, provided

b \ge 0

.

2

Solve Case 1 for

x

.

2x - 3 = 5x - 12 \implies 3x = 9 \implies x = 3

Isolate the variable

x

algebraically.

3

Solve Case 2 for

x

.

2x - 3 = -5x + 12 \implies 7x = 15 \implies x = \frac{15}{7}

Isolate the variable

x

algebraically.

4

Check candidate solutions for extraneous roots in the original equation.

For

x = 3

:

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

and

5(3) - 12 = 3

. Valid.
For

x = \frac{15}{7}

:

|2(\frac{15}{7}) - 3| = |\frac{9}{7}| = \frac{9}{7}

, but

5(\frac{15}{7}) - 12 = -\frac{9}{7}

. Since

\frac{9}{7} \neq -\frac{9}{7}

,

x = \frac{15}{7}

is extraneous.

The output of an absolute value expression cannot be negative, so any candidate root making the right side negative must be discarded.

5

Calculate the sum of all valid real solutions.

Sum =

3

There is only one valid solution,

x = 3

.

Anahtar Kavram

Solving absolute value equations with a variable expression on the right-hand side requires checking candidate solutions to eliminate extraneous roots.

Soru 9Soru →

If $x$ is a real number satisfying the inequality $\frac{|2x - 5|}{x + 1} \le 1$ , which of the following represents the complete set of all possible values of $x$ ?

x < -1

or

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

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

-1 < x \le 6

x < -1

or

x \ge \frac{4}{3}

x \le -1

or

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

Cevabı ve açıklamayı göster

Cevap: $x < -1$ or $\frac{4}{3} \le x \le 6$

Cevap

x < -1

or

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

The correct answer accounts for both possible sign states of the denominator

x + 1

. When

x > -1

, multiplying gives

|2x - 5| \le x + 1

, yielding

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

. When

x < -1

, multiplying flips the inequality to

|2x - 5| \ge x + 1

, which is trivially satisfied by all

x < -1

because an absolute value is non-negative and

x + 1

is negative. Combining both cases yields

x < -1

or

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

.

Adım Adım Çözüm

1

Determine domain restrictions and break the inequality into cases based on the denominator's sign.

The expression is undefined at

x = -1

, so

x \neq -1

. We evaluate Case 1 (

x > -1

) and Case 2 (

x < -1

).

Multiplying an inequality by an algebraic expression requires knowing whether that expression is positive or negative to maintain or flip the inequality sign.

2

Solve Case 1 where

x + 1 > 0

(

x > -1

).

|2x - 5| \le x + 1 \implies -(x + 1) \le 2x - 5 \le x + 1

. Solving

-x - 1 \le 2x - 5

yields

x \ge \frac{4}{3}

. Solving

2x - 5 \le x + 1

yields

x \le 6

. Combining gives

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

.

For positive denominators, multiplying both sides by

x + 1

preserves the inequality direction.

3

Solve Case 2 where

x + 1 < 0

(

x < -1

).

|2x - 5| \ge x + 1

. Since

|2x - 5| \ge 0

for all real

x

and

x + 1 < 0

, the left-hand side is non-negative while the right-hand side is strictly negative, which is always true for all

x < -1

.

Multiplying by a negative expression flips the inequality sign, and any non-negative quantity is strictly greater than any negative quantity.

4

Combine solutions from both cases.

The final solution set is

x < -1

or

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

.

Taking the union of valid solution regions from both disjoint cases gives the complete solution.

Anahtar Kavram

Solving Rational Inequalities with Absolute Values by Case Analysis

Soru 10Soru →

If $n$ is an integer such that $|2n + 1| \le |n - 4|$ , how many distinct integer values of $n$ satisfy the inequality?

5

6

7

8

9

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

There are 7 distinct integer values of n that satisfy the inequality.

Squaring both non-negative sides of the absolute value inequality

|2n + 1| \le |n - 4|

eliminates the absolute value bars without requiring multiple case splits. Expanding

(2n + 1)^2 \le (n - 4)^2

yields

4n^2 + 4n + 1 \le n^2 - 8n + 16

, which simplifies to

3n^2 + 12n - 15 \le 0

. Dividing the inequality by 3 gives

n^2 + 4n - 5 \le 0

, which factors as

(n + 5)(n - 1) \le 0

. The inequality is satisfied for all values in the closed interval

[-5, 1]

. Listing the integers in this range yields

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

, which comprises exactly 7 integers.

Adım Adım Çözüm

1

Square both sides of the inequality since absolute values are non-negative.

(2n + 1)^2 \le (n - 4)^2

Since

|A| \le |B|

is equivalent to

A^2 \le B^2

for all real numbers.

2

Expand both algebraic expressions and move all terms to one side.

4n^2 + 4n + 1 \le n^2 - 8n + 16 \implies 3n^2 + 12n - 15 \le 0

Standard quadratic inequality form requires comparing the quadratic polynomial to zero.

3

Simplify by dividing by 3 and factor the quadratic expression.

n^2 + 4n - 5 \le 0 \implies (n + 5)(n - 1) \le 0

Dividing by a positive constant preserves the inequality sign and allows factoring into linear binomials.

4

Determine the solution set for the inequality and count the integer solutions.

The inequality holds for

-5 \le n \le 1

. The integer solutions are

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

, giving a total of

1 - (-5) + 1 = 7

integers.

The quadratic expression is non-positive between its two real roots, inclusive of the endpoints.

Anahtar Kavram

Solving Absolute Value Inequalities via Squaring and Quadratic Factoring

Tahmini Süre:1m 30s

Soru 11Soru →

How many integer values of $x$ satisfy the inequality $|x^2 - 4x| \leq 3x - 6$ ?

3

4

5

6

8

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

4 integer values satisfy the inequality.

The solution requires breaking the inequality into cases based on the sign of

x^2 - 4x

while enforcing that the right-hand side

3x - 6

must be non-negative (meaning

x \geq 2

). Combining the valid sub-intervals

3 \leq x < 4

and

4 \leq x \leq 6

yields

3 \leq x \leq 6

. The integers satisfying this inequality are 3, 4, 5, and 6, giving a total of 4 integers.

Adım Adım Çözüm

1

Establish the domain restriction based on non-negativity.

Since the left side

|x^2 - 4x| \geq 0

for all real

x

, the right side must also be non-negative:

3x - 6 \geq 0 \implies x \geq 2

.

An absolute value quantity can never be less than a negative number.

2

Analyze Case 1 where the expression inside the absolute value is non-negative (

x^2 - 4x \geq 0

).

For

x \geq 2

,

x(x - 4) \geq 0

implies

x \geq 4

. The inequality becomes

x^2 - 4x \leq 3x - 6 \implies x^2 - 7x + 6 \leq 0 \implies (x - 1)(x - 6) \leq 0

, so

1 \leq x \leq 6

. Intersecting with

x \geq 4

yields

4 \leq x \leq 6

.

When

x \geq 4

,

|x^2 - 4x| = x^2 - 4x

.

3

Analyze Case 2 where the expression inside the absolute value is negative (

x^2 - 4x < 0

).

For

x \geq 2

,

x(x - 4) < 0

implies

2 \leq x < 4

. The inequality becomes

-(x^2 - 4x) \leq 3x - 6 \implies -x^2 + 4x \leq 3x - 6 \implies x^2 - x - 6 \geq 0 \implies (x - 3)(x + 2) \geq 0

, so

x \leq -2

or

x \geq 3

. Intersecting with

2 \leq x < 4

yields

3 \leq x < 4

.

When

2 \leq x < 4

,

|x^2 - 4x| = -(x^2 - 4x)

.

4

Combine the valid intervals from both cases and count the integer solutions.

Combining

[3, 4)

and

[4, 6]

gives the interval

[3, 6]

. The integer values in this range are

x = 3, 4, 5, 6

. Total count = 4.

The union of the solution intervals yields all real numbers between 3 and 6 inclusive.

Anahtar Kavram

Solving Absolute Value Inequalities with Variable RHS

Tahmini Süre:2m 30s

Soru 12Soru →

How many integer values of $x$ satisfy the inequality $\frac{|2x - 7| - 3}{|x + 1| - 4} \leq 0$ ?

7

8

9

10

11

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

9 integer values satisfy the given inequality.

The quotient of two algebraic expressions is non-positive (

\leq 0

) when the numerator is zero (and denominator non-zero) or when the numerator and denominator have opposite signs. Testing these cases yields two disjoint intervals:

-5 < x \leq 2

and

3 < x \leq 5

. Within these intervals, the integers

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

and

5

satisfy the inequality, yielding exactly 9 integer solutions.

Adım Adım Çözüm

1

Determine where the numerator and denominator equal zero to set critical points and domain restrictions.

Numerator is zero at

|2x - 7| = 3 \implies x = 2

or

x = 5

. Denominator is zero at

|x + 1| = 4 \implies x = -5

or

x = 3

, which are excluded from the domain.

Division by zero is undefined, so

x \neq -5

and

x \neq 3

.

2

Analyze Case 1: Numerator

\geq 0

and Denominator

< 0

.

Numerator

\geq 0 \implies x \leq 2

or

x \geq 5

. Denominator

< 0 \implies -5 < x < 3

. Intersecting these gives

-5 < x \leq 2

.

A fraction is non-positive when the numerator and denominator have opposite signs.

3

Analyze Case 2: Numerator

\leq 0

and Denominator

> 0

.

Numerator

\leq 0 \implies 2 \leq x \leq 5

. Denominator

> 0 \implies x < -5

or

x > 3

. Intersecting these gives

3 < x \leq 5

.

A fraction is non-positive when the numerator and denominator have opposite signs.

4

Combine solution intervals and count integer solutions.

Combined solution set is

(-5, 2] \cup (3, 5]

. The integer values in this set are

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

, giving a total count of 9.

Integrate all valid cases while excluding values that make the denominator zero.

Anahtar Kavram

Solving Rational Inequalities with Absolute Values

Tahmini Süre:2m 30s

Soru 13Soru →

How many integer values of $x$ satisfy both $|2x - 3| < 9$ and $|x + 1| \ge 3$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

There are 4 integer values of

x

that satisfy both inequalities.

Solving

|2x - 3| < 9

gives

-3 < x < 6

, so the set of possible integer values is

\{-2, -1, 0, 1, 2, 3, 4, 5\}

. Solving

|x + 1| \ge 3

gives

x \ge 2

or

x \le -4

. Taking the intersection of these two conditions gives

x \in \{2, 3, 4, 5\}

, for a total of 4 integer values.

Adım Adım Çözüm

1

Unfold the first absolute value inequality

|2x - 3| < 9

-9 < 2x - 3 < 9, which simplifies to -3 < x < 6

An absolute value inequality of the form |A| < B is equivalent to -B < A < B.

2

Unfold the second absolute value inequality

|x + 1| \ge 3

x + 1 \ge 3 or x + 1 \le -3, which simplifies to x \ge 2 or x \le -4

An absolute value inequality of the form |A| >= B is equivalent to A >= B or A <= -B.

3

List integer candidates and find the intersection

Candidates from first condition: {-2, -1, 0, 1, 2, 3, 4, 5}. Applying second condition (x >= 2 or x <= -4) leaves {2, 3, 4, 5}

The solution must satisfy both conditions simultaneously.

Anahtar Kavram

Solving systems of absolute value inequalities for integer solutions

Tahmini Süre:1m 30s

Soru 14Soru →

If $k$ is the product of all real solutions to the equation $|x^2 - 7| = 3x - 3$ , what is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The product of all real solutions is 8.

The equation

|x^2 - 7| = 3x - 3

requires that

3x - 3 \ge 0

, which gives the restriction

x \ge 1

. Solving the two cases

x^2 - 7 = 3x - 3

and

x^2 - 7 = -(3x - 3)

produces the candidate roots

x = 4, -1, 2, -5

. Eliminating the negative candidate roots leaves

x = 2

and

x = 4

as the only valid real solutions. Their product is

2 \times 4 = 8

.

Adım Adım Çözüm

1

Determine the domain constraint for valid solutions.

Since the absolute value expression

|x^2 - 7|

cannot be negative,

3x - 3 \ge 0

, which requires

x \ge 1

.

An absolute value quantity

|A|

is always non-negative, so any equation of the form

|A| = B

requires

B \ge 0

for real solutions.

2

Solve the positive case

x^2 - 7 = 3x - 3

and test for extraneous solutions.

x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0

, giving

x = 4

(valid, as

4 \ge 1

) and

x = -1

(extraneous, as

-1 < 1

).

Solutions must satisfy the non-negativity constraint of the right-hand side.

3

Solve the negative case

x^2 - 7 = -(3x - 3)

and test for extraneous solutions.

x^2 + 3x - 10 = 0 \implies (x + 5)(x - 2) = 0

, giving

x = 2

(valid, as

2 \ge 1

) and

x = -5

(extraneous, as

-5 < 1

).

Solutions must satisfy the non-negativity constraint of the right-hand side.

4

Compute the product of the valid real solutions.

k = 2 \times 4 = 8

.

The question asks for the product of all valid real solutions.

Anahtar Kavram

Absolute Value Equations with Variable Expressions and Extraneous Solution Elimination

Soru 15Soru →

How many integer values of $x$ satisfy the inequality $|x - 3| \leq 4$ ?

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

There are 9 integer values of

x

that satisfy the inequality.

Unfolding

|x - 3| \leq 4

yields

-4 \leq x - 3 \leq 4

. Adding 3 across all terms gives

-1 \leq x \leq 7

. The integer values satisfying this range are

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

, which totals 9 values.

Adım Adım Çözüm

1

Convert the absolute value inequality into a compound linear inequality.

-4 \leq x - 3 \leq 4

The absolute value inequality

|k| \leq c

(for

c \geq 0

) is equivalent to

-c \leq k \leq c

.

2

Isolate

x

by adding 3 across all parts of the inequality.

-1 \leq x \leq 7

Adding the same constant to all parts of an inequality maintains the inequality relationships.

3

Count the total number of integer values within the inclusive range

[-1, 7]

.

9

The number of integers in an inclusive range

[a, b]

is calculated as

b - a + 1

, which gives

7 - (-1) + 1 = 9

.

Anahtar Kavram

Solving absolute value inequalities and counting integer solutions within a bounded interval

Soru 16Soru →

If $|3 - 2x| = 7$ , what is the product of all possible real values of $x$ ?

-10

-3

3

5

10

Cevabı ve açıklamayı göster

Cevap: $-10$

Cevap

The product of all possible real values of

x

is

-10

.

Solving

|3 - 2x| = 7

yields two equations:

3 - 2x = 7

, which gives

x = -2

, and

3 - 2x = -7

, which gives

x = 5

. Multiplying these two values together gives

(-2) \times 5 = -10

.

Adım Adım Çözüm

1

Set up two linear equations corresponding to the positive and negative cases of the absolute value expression.

3 - 2x = 7

or

3 - 2x = -7

By definition,

|a| = b

(where

b \ge 0

) implies

a = b

or

a = -b

.

2

Solve the first linear equation for

x

.

-2x = 4 \implies x = -2

Subtract 3 from both sides and divide by

-2

.

3

Solve the second linear equation for

x

.

-2x = -10 \implies x = 5

Subtract 3 from both sides and divide by

-2

.

4

Calculate the product of the two solutions.

(-2) \times 5 = -10

The question asks for the product of all possible real values of

x

.

Anahtar Kavram

Absolute Value Equations

Soru 17Soru →

Let $S$ be the set of all real numbers $x$ that satisfy the inequality $x^2 - 6x \leq 0$ . How many integer values of $k$ are there such that the equation $||x - 4| - |x + 2|| = k$ has at least one solution $x \in S$ ?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The correct answer is 7.

Solving the quadratic inequality

x^2 - 6x \leq 0

gives the domain

S = [0, 6]

. Over this closed interval, the function

g(x) = ||x - 4| - |x + 2||

is continuous and attains its minimum value of 0 at

x = 1

and its maximum value of 6 at

x = 4

(and throughout

[4, 6]

). By the Intermediate Value Theorem,

g(x)

takes on all real values in the interval

[0, 6]

. The integer values of

k

for which

g(x) = k

has a solution in

S

are

0, 1, 2, 3, 4, 5,

and

6

, making a total of 7 integers.

Adım Adım Çözüm

1

Solve the quadratic inequality to define the set

S

.

x^2 - 6x \leq 0 \iff x(x - 6) \leq 0 \iff 0 \leq x \leq 6

. Thus,

S = [0, 6]

.

The solution set of

x(x-6) \leq 0

lies between the roots

x = 0

and

x = 6

inclusive.

2

Analyze the inner function

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

for

x \in [0, 6]

.

Critical points of absolute values occur at

x = -2

and

x = 4

. Within

[0, 6]

, we split at

x = 4

.

The signs of

(x - 4)

and

(x + 2)

determine how the absolute value bars simplify.

3

Evaluate

g(x) = |f(x)|

on the sub-interval

[0, 4]

.

For

0 \leq x \leq 4

:

|x - 4| = 4 - x

and

|x + 2| = x + 2

. Thus,

f(x) = 2 - 2x

and

g(x) = |2 - 2x|

.

Evaluating

g(x)

at key points gives

g(0) = 2

,

g(1) = 0

, and

g(4) = 6

. By continuity,

g(x)

covers all values in

[0, 6]

on this interval.

4

Evaluate

g(x)

on the sub-interval

[4, 6]

.

For

4 \leq x \leq 6

:

|x - 4| = x - 4

and

|x + 2| = x + 2

. Thus,

f(x) = -6

and

g(x) = |-6| = 6

.

g(x)

remains constant at 6 for

x \in [4, 6]

.

5

Determine the range of

g(x)

on

S

and count the integer values of

k

.

The range of

g(x)

for

x \in [0, 6]

is

[0, 6]

. The integer values in this interval are

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

.

The equation

g(x) = k

has a solution in

S

if and only if

k

lies in the range of

g(x)

over

S

. There are

6 - 0 + 1 = 7

such integers.

Anahtar Kavram

Absolute value functions case evaluation and finding the range over a restricted domain.

Soru 18Soru →

If $|3x + 4| = 19$ , what is the positive value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The positive value of

x

is 5.

Solving the equation

|3x + 4| = 19

yields two cases:

3x + 4 = 19

, giving

x = 5

, and

3x + 4 = -19

, giving

x = -\frac{23}{3}

. The positive value among these solutions is 5.

Adım Adım Çözüm

1

Split the absolute value equation into two linear equations.

3x + 4 = 19

or

3x + 4 = -19

The equation

|a| = b

for

b \ge 0

implies

a = b

or

a = -b

.

2

Solve for

x

in both cases.

x = 5

or

x = -\frac{23}{3}

Subtract 4 from both sides and divide by 3.

3

Select the value satisfying the problem constraint.

5

The problem asks specifically for the positive value of

x

.

Anahtar Kavram

Solving Linear Absolute Value Equations

Soru 19Soru →

What is the real solution to the equation $|x - 5| = 2x + 1$ ?

x = \frac{4}{3}

x = -6

x = -6

and

x = \frac{4}{3}

x = -\frac{4}{3}

x = 6

Cevabı ve açıklamayı göster

Cevap: $x = \frac{4}{3}$

Cevap

x = \frac{4}{3}

The correct answer is

x = \frac{4}{3}

. Setting

x - 5 = -(2x + 1)

gives

x - 5 = -2x - 1

, which solves to

3x = 4

or

x = \frac{4}{3}

. Plugging

x = \frac{4}{3}

into both sides of the original equation yields

\frac{11}{3} = \frac{11}{3}

, satisfying the equation.

Adım Adım Çözüm

1

Set up the two cases for the absolute value equation

Case 1:

x - 5 = 2x + 1

; Case 2:

x - 5 = -(2x + 1)

By definition,

|A| = B

implies

A = B

or

A = -B

, provided

B \geq 0

.

2

Solve Case 1 algebraically

x - 5 = 2x + 1 \implies x = -6

Subtracting

x

and

1

from both sides isolates

x

.

3

Solve Case 2 algebraically

x - 5 = -2x - 1 \implies 3x = 4 \implies x = \frac{4}{3}

Distributing the negative sign and adding

2x

and

5

to both sides isolates

x

.

4

Check for extraneous solutions in the original equation

|x - 5| = 2x + 1

For

x = -6

:

|-6 - 5| = |-11| = 11

, but

2(-6) + 1 = -11 \neq 11

(extraneous). For

x = \frac{4}{3}

:

|\frac{4}{3} - 5| = |-\frac{11}{3}| = \frac{11}{3}

, and

2(\frac{4}{3}) + 1 = \frac{11}{3}

(valid).

An absolute value expression cannot equal a negative number.

Anahtar Kavram

Solving absolute value equations with a variable on the right-hand side requires checking for extraneous solutions.

Soru 20Soru →

For how many integer values of $k$ does the equation $|x - 2| + |x + 4| = kx + 1$ have no real solutions for $x$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

4

The equation has no real solutions when the line y = kx + 1 lies completely below the graph of f(x) = |x - 2| + |x + 4|. Analyzing the three piecewise regions of f(x) shows that f(x) = 6 on [-4, 2], with linear rays extending upward outside this interval. Setting the line to remain below f(x) forces -1.25 < k <= 2. The four integers in this range are -1, 0, 1, and 2.

Adım Adım Çözüm

1

Decompose the sum of absolute values into a piecewise linear function

f(x) = -2x - 2 for x < -4; f(x) = 6 for -4 <= x < 2; f(x) = 2x + 2 for x >= 2

The critical points x = -4 and x = 2 split the real number line into three intervals where each absolute value expression maintains a constant sign.

2

Set up conditions for zero intersections with the line g(x) = kx + 1

g(x) must remain strictly below f(x) for all x

Any intersection point between y = f(x) and y = g(x) corresponds to a real solution of the equation.

3

Evaluate boundary points and slope constraints for each interval

From the middle interval and right ray: k <= 2. From the left ray: k > -1.25.

For x >= 2, the line slope k cannot exceed the ray slope of 2. For x <= -4, g(-4) < 6 requires -4k + 1 < 6, which yields k > -1.25.

4

Determine the allowable interval for k and count integer values

-1.25 < k <= 2, giving integer values k in {-1, 0, 1, 2}

The integer values strictly inside (-1.25, 2] are -1, 0, 1, and 2, making a total of 4 integer values.

Anahtar Kavram

Piecewise analysis of absolute value functions and linear line intersection conditions

Inequalities and Absolute Value Equations

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