Algebra Practice Questions — GRE General Test — Page 6

For a constant

a

, consider the following linear equation in one variable

x

:

\frac{a(x - 2)}{3} - \frac{2x + 1}{4} = \frac{(a - 3)x + 5}{12}

Which of the following statements are true? Select all such statements.

Select all that apply

If

a = 1

, the equation has no solution for

x

.

If

a = 5

, then

x = 3

.

If

a = -1

, then

x = 0

.

If

x = 4

, then

a = 5

.

If

a = 0

, the equation has a positive solution for

x

.

Show answer & explanation

Answer: If $a = 1$ , the equation has no solution for $x$ .; If $a = -1$ , then $x = 0$ .; If $x = 4$ , then $a = 5$ .

Answer

The correct statements are the ones asserting that

a = 1

leaves the equation with no solution,

a = -1

leads to

x = 0

, and

x = 4

requires

a = 5

.

Multiplying the equation by 12 and simplifying yields

3(a - 1)x = 8(a + 1)

. When

a = 1

, the left side becomes 0 while the right side becomes 16, creating an inconsistent equation

0 = 16

with no solution. When

a = -1

, the equation becomes

-6x = 0

, which gives

x = 0

. When

x = 4

, substituting into the simplified equation gives

12(a - 1) = 8(a + 1)

, which simplifies to

4a = 20 \implies a = 5

.

Step-by-Step Solution

1

Clear denominators by multiplying every term of the equation by the least common multiple, 12.

4a(x - 2) - 3(2x + 1) = (a - 3)x + 5

Eliminating fractional coefficients simplifies the distribution and collection of variable terms.

2

Expand products on both sides of the equation.

4ax - 8a - 6x - 3 = ax - 3x + 5

Applying the distributive property isolates individual algebraic terms.

3

Rearrange terms to collect all

x

-terms on the left side and constant terms on the right side.

(4a - 6 - a + 3)x = 8a + 8 \implies 3(a - 1)x = 8(a + 1)

Factoring out

x

provides the canonical linear form

A x = B

.

4

Evaluate the given conditions for

a

and

x

against the canonical form

3(a - 1)x = 8(a + 1)

.

For

a = 1

:

0 = 16

(no solution). For

a = 5

:

12x = 48 \implies x = 4

. For

a = -1

:

-6x = 0 \implies x = 0

. For

x = 4

:

12(a - 1) = 8(a + 1) \implies a = 5

. For

a = 0

:

-3x = 8 \implies x = -\frac{8}{3}

(negative).

Direct substitution verifies which algebraic relationships hold true.

Key Concept

Linear equations in one variable containing symbolic parameters can be analyzed for existence of solutions, zero-roots, and explicit values by reducing to the form

A x = B

.

Estimated Time:2m 30s

Question 102Question →

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

Select all that apply

-4

-1

1

3

-5

Show answer & explanation

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

Answer

The values of

x

that satisfy the inequality are

-4

,

-1

, and

1

.

Solving the compound inequality

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

yields the solution range

-4 \le x \le 2

. The candidate values

-4

,

-1

, and

1

all lie within this range, making them valid solutions.

Step-by-Step Solution

1

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

Since an absolute value

|2x - 1|

must be non-negative, we must have

5 - x \ge 0

, which simplifies to

x \le 5

.

An absolute value cannot be less than a negative number.

2

Rewrite the absolute value inequality

|2x - 1| \le 5 - x

as a compound linear inequality.

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

, which expands to

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

.

For any non-negative expression

B

,

|A| \le B

is logically equivalent to

-B \le A \le B

.

3

Solve the left-hand inequality

-5 + x \le 2x - 1

.

Subtracting

x

from both sides gives

-5 \le x - 1

. Adding

1

to both sides yields

x \ge -4

.

Isolating

x

establishes the lower bound of the solution set.

4

Solve the right-hand inequality

2x - 1 \le 5 - x

.

Adding

x

to both sides gives

3x - 1 \le 5

. Adding

1

yields

3x \le 6

, so

x \le 2

.

Isolating

x

establishes the upper bound of the solution set.

5

Intersect all constraints to find the valid domain for

x

.

Combining

x \ge -4

,

x \le 2

, and

x \le 5

gives the closed interval

[-4, 2]

.

A valid value of

x

must satisfy all component inequalities simultaneously.

6

Evaluate the candidate choices against the solution interval

[-4, 2]

.

The values

-4

,

-1

, and

1

fall within

[-4, 2]

, whereas

3

and

-5

fall outside this interval.

Only numbers inside

[-4, 2]

satisfy the original inequality.

Key Concept

Solving Absolute Value Inequalities with Variable Expressions

Question 103Question →

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

Select all that apply

-5

-3

3

5

7

Show answer & explanation

Answer: $3$ ; $5$

Answer

The values 3 and 5 satisfy both inequalities.

Solving the first inequality

|3x + 4| \ge 11

yields

x \le -5

or

x \ge \frac{7}{3}

. Solving the second inequality

|x - 1| < 6

yields

-5 < x < 7

. Intersecting these two regions, the interval

x \le -5

does not overlap with

-5 < x < 7

because

-5

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

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

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

Step-by-Step Solution

1

Solve the absolute value inequality

|3x + 4| \ge 11

.

x \le -5

or

x \ge \frac{7}{3}

.

An absolute value inequality of the form

|u| \ge k

(where

k > 0

) splits into two separate inequalities:

u \ge k

or

u \le -k

. Solving

3x + 4 \ge 11

gives

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

. Solving

3x + 4 \le -11

gives

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

.

2

Solve the absolute value inequality

|x - 1| < 6

.

-5 < x < 7

.

An absolute value inequality of the form

|u| < k

is equivalent to the compound inequality

-k < u < k

. Thus,

-6 < x - 1 < 6

. Adding 1 to all parts yields

-5 < x < 7

.

3

Determine the intersection of the two solution sets.

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

.

The portion

x \le -5

has no overlap with

-5 < x < 7

because

-5

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

x \ge \frac{7}{3}

overlaps with

-5 < x < 7

to yield the interval

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

.

4

Evaluate which of the given options fall inside

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

.

The numbers 3 and 5 belong to the interval, while

-5

,

-3

, and

7

do not.

Since

\frac{7}{3} \approx 2.33

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

Key Concept

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

Question 104Question →

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

x \ge 4

x \le -1

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

-1 \le x \le 4

x > \frac{5}{3}

Show answer & explanation

Answer: $x \ge 4$

Answer

The condition is satisfied by all values of

x

such that

x \ge 4

.

Solving

|3 - 2x| \ge 5

yields two separate intervals:

x \le -1

or

x \ge 4

. Solving

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

involves multiplying by

-2

and dividing by

-3

, both of which flip the inequality sign, leading to

x > \frac{5}{3}

. The values of

x

that satisfy both constraints are those in the overlap of

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

and

x > \frac{5}{3}

, which simplifies directly to

x \ge 4

.

Step-by-Step Solution

1

Solve the absolute value inequality

|3 - 2x| \ge 5

.

Splitting into two cases:

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

, or

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

. So

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

.

An absolute value inequality

|u| \ge k

(for

k > 0

) decouples into

u \ge k

or

u \le -k

.

2

Solve the linear inequality

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

.

Multiply both sides by

-2

(reversing the inequality):

7 - 3x < 2

. Subtract 7:

-3x < -5

. Divide by

-3

(reversing the inequality again):

x > \frac{5}{3}

.

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

3

Find the intersection of the two solution sets.

We require

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

. Since

(-\infty, -1]

has no overlap with

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

, the intersection is

[4, \infty)

, or

x \ge 4

.

A real number must satisfy both inequalities simultaneously.

Key Concept

Solving systems of absolute value and linear inequalities with negative multipliers

Question 105Question →

For all non-zero real numbers

x

, the function

f

satisfies the functional equation

f(x) + 2f\left(-\frac{1}{x}\right) = 3x

Which of the following statements must be true for all $x \neq 0$ ? Select all such statements.

Select all that apply

f(-1) = 3

f(x) = f\left(-\frac{1}{x}\right)

for all

x \neq 0

f(x) \leq -2\sqrt{2}

for all

x > 0

f(x) + f(-x) = 6x

for all

x \neq 0

f(2) = 3

Show answer & explanation

Answer: $f(-1) = 3$ ; $f(x) \leq -2\sqrt{2}$ for all $x > 0$

Answer

The statements stating that

f(-1) = 3

and that

f(x) \leq -2\sqrt{2}

for all

x > 0

are the correct choices.

Solving the functional equation yields

f(x) = -x - \frac{2}{x}

. Substituting

x = -1

gives

f(-1) = 3

, making the statement asserting

f(-1) = 3

correct. Furthermore, for any positive real number

x

, the sum

x + \frac{2}{x} \geq 2\sqrt{2}

by the AM-GM inequality, so

f(x) = -\left(x + \frac{2}{x}\right) \leq -2\sqrt{2}

, making the statement asserting

f(x) \leq -2\sqrt{2}

for all

x > 0

correct.

Step-by-Step Solution

1

Set up a system of functional equations by substituting

x \to -\frac{1}{x}

.

Replacing

x

with

-\frac{1}{x}

in the original equation

f(x) + 2f\left(-\frac{1}{x}\right) = 3x

gives

f\left(-\frac{1}{x}\right) + 2f(x) = 3\left(-\frac{1}{x}\right) = -\frac{3}{x}

.

This creates a system of two linear algebraic equations with two unknowns:

f(x)

and

f\left(-\frac{1}{x}\right)

.

2

Solve the system of equations for

f(x)

.

Multiply the second equation by

2

to obtain

4f(x) + 2f\left(-\frac{1}{x}\right) = -\frac{6}{x}

. Subtracting the first equation

f(x) + 2f\left(-\frac{1}{x}\right) = 3x

from this gives

3f(x) = -3x - \frac{6}{x}

, so

f(x) = -x - \frac{2}{x}

.

Eliminating the composite term

f\left(-\frac{1}{x}\right)

yields an explicit expression for

f(x)

.

3

Evaluate

f(-1)

.

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

.

This confirms that the statement asserting

f(-1) = 3

is correct.

4

Analyze the maximum value of

f(x)

for

x > 0

using the AM-GM inequality.

For

x > 0

,

x + \frac{2}{x} \geq 2\sqrt{x \cdot \frac{2}{x}} = 2\sqrt{2}

. Therefore,

f(x) = -\left(x + \frac{2}{x}\right) \leq -2\sqrt{2}

.

This confirms that the statement asserting

f(x) \leq -2\sqrt{2}

for all

x > 0

is correct.

Key Concept

Solving functional equations via variable substitution and analyzing function bounds via the AM-GM inequality.

Estimated Time:2m 30s

Question 106Question →

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

-7

0

7

9

14

Show answer & explanation

Answer: 9

Answer

9

Solving the absolute value inequality

|4x - 5| \le 13

yields

-13 \le 4x - 5 \le 13

, which simplifies to

-2 \le x \le 4.5

. Next, solving

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

requires multiplying both sides by

-3

and reversing the inequality sign, giving

2x - 1 \ge 3

, which simplifies to

x \ge 2

. Combining

-2 \le x \le 4.5

and

x \ge 2

for integer

x

gives the set

\{2, 3, 4\}

. The sum of these values is

2 + 3 + 4 = 9

.

Step-by-Step Solution

1

Solve the absolute value inequality

|4x - 5| \le 13

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

An absolute value inequality

|u| \le k

unfolds into the compound inequality

-k \le u \le k

.

2

Solve the linear inequality

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

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

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

3

Find the intersection of the two solution sets for integer

x

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

The integers satisfying both

-2 \le x \le 4.5

and

x \ge 2

are 2, 3, and 4.

4

Calculate the sum of the possible integer values

2 + 3 + 4 = 9

Summing the valid integer solutions yields the final requested answer.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:2m 0s

Question 107Question →

If $x$ and $y$ are real numbers greater than $1$ such that $x^{\sqrt{y}} = y^{\sqrt{x}}$ and $x^3 = y^2$ , what is the value of $x$ ?

\frac{81}{16}

\frac{9}{4}

\frac{27}{8}

\frac{729}{64}

\frac{243}{32}

Show answer & explanation

Answer: $\frac{81}{16}$

Answer

The correct answer is

\frac{81}{16}

.

Expressing

y

as

x^{3/2}

and substituting it into

x^{\sqrt{y}} = y^{\sqrt{x}}

transforms the equation into

x^{x^{3/4}} = x^{\frac{3}{2}x^{1/2}}

. Equating exponents yields

x^{3/4} = \frac{3}{2}x^{1/2}

, which simplifies to

x^{1/4} = \frac{3}{2}

. Raising both sides to the fourth power gives

x = \frac{81}{16}

.

Step-by-Step Solution

1

Express

y

in terms of

x

using the second given equation.

Since

x > 1

and

y > 1

, taking the square root of both sides of

y^2 = x^3

gives

y = x^{3/2}

.

Converting

y

to an exponential expression of

x

allows single-variable substitution into the first equation.

2

Substitute

y = x^{3/2}

into the first equation

x^{\sqrt{y}} = y^{\sqrt{x}}

.

Note that

\sqrt{y} = \sqrt{x^{3/2}} = (x^{3/2})^{1/2} = x^{3/4}

. Thus, the left side becomes

x^{x^{3/4}}

, and the right side becomes

(x^{3/2})^{\sqrt{x}} = x^{\frac{3}{2}x^{1/2}}

.

Applying exponent rules

(a^m)^n = a^{mn}

simplifies both sides to base

x

expressions.

3

Equate the exponents since the bases are equal and greater than 1.

x^{3/4} = \frac{3}{2}x^{1/2}

.

If

x^a = x^b

for

x > 1

, then

a = b

.

4

Divide both sides by

x^{1/2}

to isolate the power of

x

.

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

.

Using the quotient rule for exponents,

\frac{x^a}{x^b} = x^{a-b}

where

3/4 - 1/2 = 1/4

.

5

Raise both sides to the 4th power to solve for

x

.

x = \left(\frac{3}{2}\right)^4 = \frac{3^4}{2^4} = \frac{81}{16}

.

Raising

(x^{1/4})^4

eliminates the fractional exponent to give

x

.

Key Concept

Solving systems of exponential equations using fractional exponent rules and base equality properties.

Question 108Question →

A container originally holds a liquid mixture consisting of substance A and substance B, where substance A constitutes $\frac{2}{5}$ of the total volume. After $15$ liters of substance B are added to the container and $3$ liters of substance A evaporate, the volume of substance A in the container becomes $\frac{1}{4}$ of the new total liquid volume. What was the original total volume, in liters, of the liquid mixture in the container?

32

40

45

50

52

Show answer & explanation

Answer: 40

Answer

40 liters

Letting

V

represent the original volume of the liquid mixture in liters, the initial amount of substance A is

\frac{2}{5}V

. After adding 15 liters of substance B and losing 3 liters of substance A to evaporation, the updated volume of substance A is

\frac{2}{5}V - 3

, and the updated total volume is

V + 15 - 3 = V + 12

. Setting up the relationship

\frac{2}{5}V - 3 = \frac{1}{4}(V + 12)

and expanding the right side gives

\frac{2}{5}V - 3 = \frac{1}{4}V + 3

. Subtracting

\frac{1}{4}V

from both sides yields

\frac{3}{20}V = 6

, which solves to

V = 40

liters.

Step-by-Step Solution

1

Define the unknown variable and express initial quantities algebraically.

Let

V

be the original total volume of the liquid mixture in liters. The original volume of substance A is

\frac{2}{5}V

.

Establishing a variable for the initial total volume allows all changes to be modeled in terms of one variable.

2

Express the modified quantities after additions and evaporation.

New volume of substance A

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

. New total volume

= V + 15 - 3 = V + 12

.

Adding 15 liters of substance B increases the total volume by 15, and losing 3 liters of substance A decreases both substance A and the total volume by 3.

3

Set up the linear equation based on the given ratio condition.

\frac{2}{5}V - 3 = \frac{1}{4}(V + 12)

Substance A forms one-fourth of the updated total liquid volume.

4

Expand and solve the linear equation for

V

.

\frac{2}{5}V - 3 = \frac{1}{4}V + 3 \implies \frac{2}{5}V - \frac{1}{4}V = 6 \implies \frac{8 - 5}{20}V = 6 \implies \frac{3}{20}V = 6 \implies V = 40.

Clearing terms and subtracting

\frac{1}{4}V

from

\frac{2}{5}V

gives

\frac{3}{20}V = 6

, which yields

V = 40

.

Key Concept

Linear Equations in One Variable

Question 109Question →

If $x$ and $y$ are positive real numbers such that $\left(\frac{x^{-1}y^{\frac{2}{3}}}{\sqrt{x^3 y^{-2}}}\right)^{-\frac{3}{4}} = x^a y^b$ , what is the value of $a + b$ ?

\frac{5}{8}

-\frac{5}{8}

\frac{17}{8}

-\frac{1}{8}

-\frac{13}{8}

Show answer & explanation

Answer: $\frac{5}{8}$

Answer

The value of

a + b

is

\frac{5}{8}

.

Simplifying the fraction inside the parentheses gives

x^{-\frac{5}{2}} y^{\frac{5}{3}}

. Raising this to the power of

-\frac{3}{4}

produces

x^{\frac{15}{8}} y^{-\frac{5}{4}}

. Equating exponents yields

a = \frac{15}{8}

and

b = -\frac{5}{4} = -\frac{10}{8}

. Adding these together gives

a + b = \frac{5}{8}

.

Step-by-Step Solution

1

Rewrite the radical expression in the denominator using fractional exponents.

\sqrt{x^3 y^{-2}} = (x^3 y^{-2})^{\frac{1}{2}} = x^{\frac{3}{2}} y^{-1}

Applying the power rule

(u^v)^w = u^{v \cdot w}

to radical expressions converts square roots to fractional exponents of

\frac{1}{2}

.

2

Simplify the expression inside the main parentheses by combining like base exponents.

\frac{x^{-1}y^{\frac{2}{3}}}{x^{\frac{3}{2}} y^{-1}} = x^{-1 - \frac{3}{2}} y^{\frac{2}{3} - (-1)} = x^{-\frac{5}{2}} y^{\frac{5}{3}}

When dividing exponential expressions with the same base, subtract the denominator exponent from the numerator exponent.

3

Apply the outer exponent

-\frac{3}{4}

to each factor inside the parentheses.

(x^{-\frac{5}{2}} y^{\frac{5}{3}})^{-\frac{3}{4}} = x^{(-\frac{5}{2})(-\frac{3}{4})} y^{(\frac{5}{3})(-\frac{3}{4})} = x^{\frac{15}{8}} y^{-\frac{5}{4}}

According to exponent rules,

(u \cdot v)^p = u^p \cdot v^p

and

(u^p)^q = u^{p \cdot q}

.

4

Identify

a

and

b

and calculate their sum

a + b

.

a = \frac{15}{8}, \quad b = -\frac{5}{4} = -\frac{10}{8} \implies a + b = \frac{15}{8} - \frac{10}{8} = \frac{5}{8}

Match the simplified expression with

x^a y^b

and add the resulting fractional exponents using a common denominator.

Key Concept

Simplifying nested algebraic expressions with negative and rational exponents using fundamental laws of exponents.

Estimated Time:2m 0s

Question 110Question →

What is the least integer value of $x$ that satisfies the inequality $|4x - 5| \le 11$ ?

Show answer & explanation

Answer: -1

Answer

The least integer value of

x

that satisfies the inequality is

-1

.

Expanding the absolute value inequality

|4x - 5| \le 11

produces

-11 \le 4x - 5 \le 11

. Adding 5 across all sections yields

-6 \le 4x \le 16

, and dividing by 4 results in

-1.5 \le x \le 4

. The set of integers within this range is

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

, making

-1

the least integer value.

Step-by-Step Solution

1

Rewrite the absolute value inequality as a double inequality.

-11 \le 4x - 5 \le 11

An absolute value inequality of the form

|u| \le a

for

a \ge 0

is equivalent to

-a \le u \le a

.

2

Add 5 to all parts of the compound inequality.

-6 \le 4x \le 16

To isolate

x

, first add 5 to eliminate the constant term

-5

.

3

Divide all parts by 4.

-1.5 \le x \le 4

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

4

Select the minimum integer contained within the solution interval

[-1.5, 4]

.

-1

The integer values satisfying

-1.5 \le x \le 4

are

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

. The smallest among these is

-1

.

Key Concept

Linear inequalities involving absolute value

Question 111Question →

What is the sum of all real solutions to the equation $\sqrt{3x + 10} - x = 2$ ?

2

-1

-3

5

3

Show answer & explanation

Answer: $2$

Answer

The sum of all real solutions is

2

.

To solve

\sqrt{3x + 10} - x = 2

, isolate the radical to get

\sqrt{3x + 10} = x + 2

. Squaring both sides yields

3x + 10 = x^2 + 4x + 4

. Rearranging into standard quadratic form gives

x^2 + x - 6 = 0

, which factors as

(x + 3)(x - 2) = 0

, yielding candidates

x = 2

and

x = -3

. Substituting

x = 2

into the original equation yields

\sqrt{16} - 2 = 2

, which is true. Substituting

x = -3

yields

\sqrt{1} - (-3) = 4 \neq 2

, so

x = -3

is extraneous. The only valid solution is

x = 2

, so the sum of all valid solutions is

2

.

Step-by-Step Solution

1

Isolate the radical expression on one side of the equation

\sqrt{3x + 10} = x + 2

Isolating the radical allows squaring both sides cleanly to eliminate the radical sign.

2

Square both sides of the equation

3x + 10 = (x + 2)^2 = x^2 + 4x + 4

Squaring eliminates the square root, converting the equation into a polynomial form.

3

Rearrange terms into standard quadratic form

ax^2 + bx + c = 0

x^2 + x - 6 = 0

Grouping all terms on one side sets up the equation for factoring.

4

Factor the quadratic equation to find candidate solutions

(x + 3)(x - 2) = 0 \implies x = -3 \text{ or } x = 2

Factoring determines the values of

x

that solve the algebraic polynomial.

5

Test candidate solutions in the original radical equation to filter extraneous roots

For

x = 2

:

\sqrt{3(2) + 10} - 2 = \sqrt{16} - 2 = 4 - 2 = 2

(Valid). For

x = -3

:

\sqrt{3(-3) + 10} - (-3) = \sqrt{1} + 3 = 4 \neq 2

(Extraneous). Thus,

x = 2

is the only valid solution.

Squaring an equation can introduce extraneous roots that do not satisfy the original principal root definition.

Key Concept

Solving Radical Equations and Filtering Extraneous Roots

Estimated Time:1m 30s

Question 112Question →

A coffee shop owner creates a $30$ -pound blend of coffee by mixing Bean X, which costs $\$8$ per pound, with Bean Y, which costs $\$14$ per pound. If the final mixture costs $\$10$ per pound, how many pounds of Bean X are in the mixture?

Show answer & explanation

Answer: 20

Answer

The mixture contains 20 pounds of Bean X.

The quantity of Bean X required is 20 pounds. Setting up the cost equation

8x + 14(30 - x) = 300

simplifies to

-6x = -120

, giving

x = 20

.

Step-by-Step Solution

1

Define the unknown variables representing the quantities of each component.

Let

x

be the weight of Bean X in pounds. The weight of Bean Y is

(30 - x)

pounds.

Since the total weight of the mixture is 30 pounds, expressing Bean Y in terms of

x

creates a single-variable system.

2

Set up a linear equation based on total financial value.

8x + 14(30 - x) = 10 \times 30

The total cost of Bean X plus the total cost of Bean Y equals the total cost of the combined 30-pound mixture.

3

Solve the algebraic equation for

x

.

8x + 420 - 14x = 300 \implies -6x = -120 \implies x = 20

Combining like terms isolates the variable

x

to yield the required quantity of Bean X.

Key Concept

Linear mixture modeling and single-variable algebraic modeling

Question 113Question →

If $x$ is a real number that satisfies both $|5 - 2x| \le 11$ and $-3x + 7 < 1$ , what is the least possible integer value of $x$ ?

-3

2

3

-8

8

Show answer & explanation

Answer: 3

Answer

The least possible integer value of

x

is 3.

Solving

|5 - 2x| \le 11

leads to

-11 \le 5 - 2x \le 11

. Subtracting 5 gives

-16 \le -2x \le 6

, and dividing by

-2

(flipping the inequalities) yields

-3 \le x \le 8

. Next, solving

-3x + 7 < 1

gives

-3x < -6

, which upon dividing by

-3

(and flipping the inequality sign) gives

x > 2

. Combining these two requirements yields

2 < x \le 8

. The integer values satisfying this inequality are 3, 4, 5, 6, 7, and 8. The least possible integer value among these is 3.

Step-by-Step Solution

1

Solve the absolute value inequality

|5 - 2x| \le 11

.

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

.

An absolute value inequality

|u| \le k

expands to

-k \le u \le k

. Dividing by

-2

flips the inequality direction.

2

Solve the linear inequality

-3x + 7 < 1

.

-3x < -6 \implies x > 2

.

Subtract 7 from both sides, then divide by

-3

, remembering to reverse the inequality sign.

3

Determine the intersection of both solution sets.

2 < x \le 8

.

x

must be strictly greater than 2 and less than or equal to 8.

4

Identify the smallest integer within the range

2 < x \le 8

.

3

Since

x > 2

is strict, 2 is excluded, making 3 the smallest integer in the range.

Key Concept

Solving systems of linear inequalities involving absolute values and correctly applying sign-flipping rules when multiplying or dividing by negative quantities.

Alternative Method

Test integer candidates directly: for

x = 2

,

-3(2) + 7 = 1

, which is not strictly less than 1. For

x = 3

,

-3(3) + 7 = -2 < 1

(valid) and

|5 - 2(3)| = |-1| = 1 \le 11

(valid), confirming 3 is the smallest integer solution.

Estimated Time:1m 30s

Question 114Question →

If $-5x + 3 \le 18$ , which of the following inequalities represents all possible real values of $x$ ?

x \ge -3

x \le -3

x \ge 3

x \le 3

x \ge -\frac{21}{5}

Show answer & explanation

Answer: $x \ge -3$

Answer

x \ge -3

Subtracting 3 from both sides gives

-5x \le 15

. Dividing both sides by

-5

requires flipping the inequality sign from

\le

to

\ge

, resulting in

x \ge -3

.

Step-by-Step Solution

1

Subtract 3 from both sides of the inequality

-5x + 3 \le 18

.

-5x \le 15

Isolate the variable term on the left side.

2

Divide both sides by

-5

and reverse the inequality sign from

\le

to

\ge

.

x \ge -3

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

Key Concept

Linear Inequalities and Sign Reversal

Question 115Question →

If $x$ and $y$ are real numbers that satisfy the inequalities $|3 - 2x| \le 9$ and $|4y + 3| \le 13$ , what is the maximum possible value of the expression $|3x - 4y|$ ?

Show answer & explanation

Answer: 34

Answer

The maximum possible value of

|3x - 4y|

is

34

.

To find the maximum possible value of

|3x - 4y|

, we first solve for the ranges of

x

and

y

from their respective absolute value inequalities. From

|3 - 2x| \le 9

, we get

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

. From

|4y + 3| \le 13

, we get

-13 \le 4y + 3 \le 13 \implies -4 \le y \le 2.5

. To maximize

3x - 4y

, we take the maximum value of

3x

(

3 \times 6 = 18

) and the maximum value of

-4y

(

-4 \times -4 = 16

), giving

18 + 16 = 34

. To minimize

3x - 4y

, we take the minimum value of

3x

(

3 \times -3 = -9

) and the minimum value of

-4y

(

-4 \times 2.5 = -10

), giving

-9 - 10 = -19

. The expression

3x - 4y

ranges from

-19

to

34

, so the maximum possible magnitude

|3x - 4y|

is

\max(|-19|, |34|) = 34

.

Step-by-Step Solution

1

Solve the inequality

|3 - 2x| \le 9

for

x

.

-3 \le x \le 6

Unpacking the absolute value yields

-9 \le 3 - 2x \le 9

. Subtracting

3

gives

-12 \le -2x \le 6

. Dividing by

-2

and reversing the inequality signs produces

-3 \le x \le 6

.

2

Solve the inequality

|4y + 3| \le 13

for

y

.

-4 \le y \le 2.5

Unpacking the absolute value yields

-13 \le 4y + 3 \le 13

. Subtracting

3

gives

-16 \le 4y \le 10

. Dividing by

4

gives

-4 \le y \le 2.5

.

3

Find the range of possible values for

3x

and

-4y

.

-9 \le 3x \le 18

and

-10 \le -4y \le 16

Multiplying

-3 \le x \le 6

by

3

gives

-9 \le 3x \le 18

. Multiplying

-4 \le y \le 2.5

by

-4

and flipping the signs gives

-10 \le -4y \le 16

.

4

Combine the bounds for

3x

and

-4y

to find the range for

3x - 4y

.

-19 \le 3x - 4y \le 34

The minimum value of

3x - 4y

is

(-9) + (-10) = -19

. The maximum value of

3x - 4y

is

18 + 16 = 34

.

5

Determine the maximum absolute value

|3x - 4y|

over the interval

[-19, 34]

.

34

The absolute value of any number in the interval

[-19, 34]

ranges from

0

to

\max(|-19|, |34|) = 34

.

Key Concept

Absolute Value Inequalities and Expression Bounding

Question 116Question →

If $x$ is a real number that satisfies both $|4 - 3x| > 7$ and $\frac{1 - 2x}{3} \ge -3$ , which of the following represents the complete set of all possible values of $x$ ?

x < -1

or

\frac{11}{3} < x \le 5

-1 < x \le 5

x \ge 5

-1 < x < \frac{11}{3}

x < -1

or

x > \frac{11}{3}

Show answer & explanation

Answer: $x < -1$ or $\frac{11}{3} < x \le 5$

Answer

x < -1

or

\frac{11}{3} < x \le 5

The correct option correctly solves

|4 - 3x| > 7

to yield

x < -1

or

x > \frac{11}{3}

, solves

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

to yield

x \le 5

, and takes their intersection to produce

x < -1

or

\frac{11}{3} < x \le 5

.

Step-by-Step Solution

1

Solve the absolute value inequality

|4 - 3x| > 7

.

4 - 3x > 7

or

4 - 3x < -7

. Solving

4 - 3x > 7

gives

-3x > 3 \implies x < -1

. Solving

4 - 3x < -7

gives

-3x < -11 \implies x > \frac{11}{3}

. Thus,

x \in (-\infty, -1) \cup (\frac{11}{3}, \infty)

.

An absolute value inequality of the form

|u| > c

splits into

u > c

or

u < -c

. Dividing by a negative number reverses the inequality direction.

2

Solve the linear inequality

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

.

Multiply both sides by

3

:

1 - 2x \ge -9

. Subtract

1

:

-2x \ge -10

. Divide by

-2

and flip the inequality sign:

x \le 5

.

Isolating the variable

x

requires reversing the inequality sign when dividing by the negative constant

-2

.

3

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

We require

x

to satisfy

(x < -1 \text{ or } x > \frac{11}{3})

AND

x \le 5

. Case 1:

x < -1

automatically satisfies

x \le 5

, giving

x < -1

. Case 2:

x > \frac{11}{3}

combined with

x \le 5

gives

\frac{11}{3} < x \le 5

. Combining both cases yields

x < -1 \text{ or } \frac{11}{3} < x \le 5

.

The word 'both' in the stem indicates a logical AND (intersection) between the two conditions.

Key Concept

Solving systems involving absolute value inequalities and linear inequalities requires handling disjunctions (OR) for absolute values greater than a positive constant, reversing inequality signs when multiplying or dividing by negative numbers, and taking the intersection (AND) of all valid regions.

Estimated Time:2m 15s

Question 117Question →

If $x$ is a real number that satisfies the inequality $-2|3x - 4| + 5 \ge -7$ , which of the following inequalities represents all possible values of $x$ ?

-\frac{2}{3} \le x \le \frac{10}{3}

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

or

x \ge \frac{10}{3}

-\frac{10}{3} \le x \le \frac{2}{3}

x \le \frac{10}{3}

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

Show answer & explanation

Answer: $-\frac{2}{3} \le x \le \frac{10}{3}$

Answer

-\frac{2}{3} \le x \le \frac{10}{3}

Isolating

|3x - 4|

requires dividing

-2|3x - 4| \ge -12

by

-2

, which reverses the inequality to

|3x - 4| \le 6

. Expanding this into the compound inequality

-6 \le 3x - 4 \le 6

and solving for

x

yields the interval

-\frac{2}{3} \le x \le \frac{10}{3}

.

Step-by-Step Solution

1

Isolate the absolute value expression by subtracting 5 from both sides of the inequality.

-2|3x - 4| \ge -12

Before removing absolute value bars, the term

|3x - 4|

must be isolated on one side.

2

Divide both sides by

-2

and reverse the inequality sign.

|3x - 4| \le 6

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

3

Express the absolute value inequality

|u| \le k

as a double inequality

-k \le u \le k

.

-6 \le 3x - 4 \le 6

The distance of

3x - 4

from 0 on the number line must be at most 6 units.

4

Add 4 to all three parts of the inequality and divide by 3.

-2 \le 3x \le 10 \implies -\frac{2}{3} \le x \le \frac{10}{3}

Solving for

x

isolates the variable in the center of the double inequality.

Key Concept

Solving linear absolute value inequalities requires reversing the inequality sign when multiplying or dividing by negative numbers, and expressing

|A| \le B

as

-B \le A \le B

.

Question 118Question →

If $|x - 3| < 5$ , which of the following values could be the value of $x$ ? Select all such values.

Select all that apply

-3

-1

0

4

8

Show answer & explanation

Answer: $-1$ ; $0$ ; $4$

Answer

The possible values of

x

are

-1

,

0

, and

4

.

The absolute value inequality

|x - 3| < 5

represents all real numbers

x

whose distance from

3

on the number line is strictly less than

5

. Expressed as a compound inequality, this means

-5 < x - 3 < 5

. Adding

3

across the entire inequality yields

-2 < x < 8

. The values

-1

,

0

, and

4

are the only options that fall strictly within the interval

(-2, 8)

.

Step-by-Step Solution

1

Rewrite the absolute value inequality as a compound inequality.

-5 < x - 3 < 5

An absolute value inequality of the form

|u| < k

(where

k > 0

) is equivalent to

-k < u < k

.

2

Solve for

x

by adding

3

to all parts of the inequality.

-5 + 3 < x < 5 + 3 \implies -2 < x < 8

Adding a positive constant to all parts of an inequality isolates

x

while preserving inequality directions.

3

Test each given option against the range

-2 < x < 8

.

The values

-1

,

0

, and

4

lie within

(-2, 8)

, whereas

-3

and

8

lie outside.

-3

is less than or equal to

-2

, and

8

is not strictly less than

8

.

Key Concept

Solving linear absolute value inequalities

Estimated Time:45s

Question 119Question →

If

x

satisfies the linear equation

\frac{3(x - 2)}{4} - \frac{x - 5}{3} = \frac{2x + 1}{6} + 2

what is the value of

2x - 5

?

43

24

123

-1

51

Show answer & explanation

Answer: 43

Answer

43

The correct answer is obtained by clearing denominators with the least common denominator 12, yielding

9(x - 2) - 4(x - 5) = 2(2x + 1) + 24

. Expanding and simplifying gives

5x + 2 = 4x + 26

, which isolates

x = 24

. Substituting

x = 24

into

2x - 5

results in

2(24) - 5 = 43

.

Step-by-Step Solution

1

Clear the denominators by multiplying every term in the equation by the least common multiple of 4, 3, and 6, which is 12.

12 \cdot \frac{3(x - 2)}{4} - 12 \cdot \frac{x - 5}{3} = 12 \cdot \frac{2x + 1}{6} + 12 \cdot 2 \implies 9(x - 2) - 4(x - 5) = 2(2x + 1) + 24

Eliminating fractions simplifies the linear equation into standard polynomial form.

2

Distribute the constants through the parentheses, taking careful note of negative signs.

9x - 18 - 4x + 20 = 4x + 2 + 24

Parentheses must be removed to collect like variable and constant terms.

3

Combine like terms on both sides of the equation.

5x + 2 = 4x + 26

Consolidating terms on each side allows for isolating the variable.

4

Isolate

x

on one side of the equation.

5x - 4x = 26 - 2 \implies x = 24

Subtracting

4x

and 2 from both sides solves directly for

x

.

5

Evaluate the target expression

2x - 5

using

x = 24

.

2(24) - 5 = 48 - 5 = 43

The question requests the value of

2x - 5

, not

x

alone.

Key Concept

Solving multi-step linear equations in one variable with fractional terms and evaluating algebraic expressions.

Estimated Time:2m 0s

Question 120Question →

Working alone at a constant rate, Pump A can drain a full water reservoir in $6$ hours, while Pump B working alone at a constant rate can drain the same full reservoir in $4$ hours. Which of the following statements must be true? Select all such statements.

Select all that apply

Working together, Pump A and Pump B can drain the entire reservoir in

2.4

hours.

In

1

hour, Pump A and Pump B working together drain

\frac{5}{12}

of the total reservoir volume.

The combined rate of Pump A and Pump B working together is

\frac{1}{10}

of the reservoir per hour.

Pump A's individual work rate accounts for

\frac{3}{5}

of the combined hourly work rate of both pumps.

Pump B's hourly rate is

33.\overline{3}\%

greater than Pump A's hourly rate.

Show answer & explanation

Answer: Working together, Pump A and Pump B can drain the entire reservoir in $2.4$ hours.; In $1$ hour, Pump A and Pump B working together drain $\frac{5}{12}$ of the total reservoir volume.

Answer

The statements confirming that the two pumps together drain the reservoir in 2.4 hours and that they drain 5/12 of the reservoir in 1 hour are correct.

The rate for Pump A is

\frac{1}{6}

job/hr and for Pump B is

\frac{1}{4}

job/hr. Their combined rate is

\frac{1}{6} + \frac{1}{4} = \frac{5}{12}

of the reservoir per hour, which directly validates the statement regarding 1 hour of combined work. Dividing

1

full reservoir by

\frac{5}{12}

yields

2.4

hours, which validates the total elapsed time statement.

Step-by-Step Solution

1

Calculate individual work rates

Pump A rate =

\frac{1}{6}

reservoir/hr, Pump B rate =

\frac{1}{4}

reservoir/hr

Work rate is the reciprocal of time required to complete the job.

2

Calculate combined work rate

Combined rate =

\frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12}

reservoir/hr

Rates add when workers/pumps perform simultaneously.

3

Calculate time needed for full task

Total time =

\frac{1}{\text{Combined Rate}} = \frac{12}{5} = 2.4

hours

Total work (

1

) divided by combined rate gives total elapsed time.

Key Concept

Combined Work Rates and Inverse Time Relationships

Estimated Time:1m 0s

Algebra

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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