Algebra Practice Questions — GRE General Test

Question 1Question →

If $b > 1$ , which of the following expressions are equivalent to $\frac{\sqrt{b^3 \cdot \sqrt[3]{b^2}}}{b^{-\frac{1}{6}}}$ ? Select all that apply.

Select all that apply

(\sqrt[3]{b})^6

\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}}

\sqrt{b^4 + b^2}

b^{\frac{11}{6}} \cdot b^{-\frac{1}{6}}

\left(b^{-\frac{1}{2}}\right)^{-4}

Show answer & explanation

Answer: $(\sqrt[3]{b})^6$ ; $\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}}$ ; $\left(b^{-\frac{1}{2}}\right)^{-4}$

Answer

The equivalent expressions are

(\sqrt[3]{b})^6

,

\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}}

, and

\left(b^{-\frac{1}{2}}\right)^{-4}

.

Simplifying the original expression yields

\frac{\sqrt{b^3 \cdot b^{2/3}}}{b^{-1/6}} = \frac{b^{11/6}}{b^{-1/6}} = b^{11/6 - (-1/6)} = b^{12/6} = b^2

. The expression

(\sqrt[3]{b})^6

equals

b^{6/3} = b^2

. The expression

\frac{b^{7/3}}{b^{1/3}}

equals

b^{7/3 - 1/3} = b^2

. The expression

\left(b^{-1/2}\right)^{-4}

equals

b^{(-1/2)(-4)} = b^2

. All three equal

b^2

.

Step-by-Step Solution

1

Simplify the expression inside the square root in the numerator.

b^3 \cdot \sqrt[3]{b^2} = b^3 \cdot b^{\frac{2}{3}} = b^{3 + \frac{2}{3}} = b^{\frac{11}{3}}

Convert radical to rational exponent and use product rule for exponents.

2

Apply the square root to the numerator.

\sqrt{b^{\frac{11}{3}}} = \left(b^{\frac{11}{3}}\right)^{\frac{1}{2}} = b^{\frac{11}{6}}

Taking the square root is equivalent to raising to the power of

\frac{1}{2}

.

3

Divide by the denominator.

\frac{b^{\frac{11}{6}}}{b^{-\frac{1}{6}}} = b^{\frac{11}{6} - \left(-\frac{1}{6}\right)} = b^{\frac{12}{6}} = b^2

Apply quotient rule for exponents: subtract the exponent of the denominator from the numerator.

4

Evaluate each given option to determine which simplify to

b^2

.

The expressions

(\sqrt[3]{b})^6 = b^2

,

\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}} = b^2

, and

\left(b^{-\frac{1}{2}}\right)^{-4} = b^2

are all equivalent.

Matching each simplified candidate expression to the target simplified value

b^2

.

Key Concept

Simplification of nested algebraic radicals and rational exponents using exponent rules

Estimated Time:2m 0s

Question 2Question →

If $x$ is a real number that satisfies the compound absolute value inequality $||2x - 5| - 7| \le 4$ , which of the following values could be the value of $x$ ? Select all such values.

Select all that apply

-2

0

3

6

9

Show answer & explanation

Answer: $-2$ ; $0$ ; $6$

Answer

The real numbers

-2

,

0

, and

6

satisfy the inequality.

The solution set to the compound inequality

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

is the union of two intervals:

[-3, 1]

and

[4, 8]

. Among the options provided,

-2

and

0

fall into the interval

[-3, 1]

, and

6

falls into the interval

[4, 8]

. Therefore, these three values satisfy the original inequality.

Step-by-Step Solution

1

Unfold the outer absolute value inequality.

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

For any expression

U

and constant

c \ge 0

,

|U| \le c

is equivalent to

-c \le U \le c

.

2

Isolate the inner absolute value term by adding

7

across all parts.

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

Adding a constant preserves inequality directions.

3

Break the compound inequality into two separate absolute value conditions.

Condition 1:

|2x - 5| \le 11

; Condition 2:

|2x - 5| \ge 3

Both conditions must hold simultaneously for

x

.

4

Solve Condition 1 (

|2x - 5| \le 11

).

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

Expanding the bounded absolute value inequality and solving for

x

.

5

Solve Condition 2 (

|2x - 5| \ge 3

).

2x - 5 \ge 3

or

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

or

2x \le 2 \implies x \ge 4

or

x \le 1

For

|U| \ge c

,

U \ge c

or

U \le -c

.

6

Find the intersection of the two solution sets.

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

Intersecting

[-3, 8]

with

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

yields

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

.

7

Test the provided choices against the solution set

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

.

-2 \in [-3, 1]

(valid),

0 \in [-3, 1]

(valid),

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

(invalid),

6 \in [4, 8]

(valid),

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

(invalid).

Values lying within the solution intervals satisfy the original inequality.

Key Concept

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

Question 3Question →

A commercial coffee roaster creates a custom blend by mixing two existing bean blends: Blend X and Blend Y. Blend X consists of 60% Arabica beans and 40% Robusta beans by weight, whereas Blend Y consists of 30% Arabica beans and 70% Robusta beans by weight. The roaster mixes a quantity of Blend X with a quantity of Blend Y to produce a total of 50 pounds of a new mixture that is 42% Arabica beans by weight. How many pounds of Blend X are in the final mixture?

Show answer & explanation

Answer: 20

Answer

20 pounds

Let

x

represent the number of pounds of Blend X. The remaining weight of the mixture,

(50 - x)

pounds, comes from Blend Y. Setting up the equation for the total weight of Arabica beans gives

0.60x + 0.30(50 - x) = 0.42(50)

. Simplifying this expression yields

0.60x + 15 - 0.30x = 21

, which reduces to

0.30x = 6

. Dividing by

0.30

gives

x = 20

. Therefore, 20 pounds of Blend X were used.

Step-by-Step Solution

1

Define variables for component weights

Let

x

be the pounds of Blend X. The weight of Blend Y used is

50 - x

pounds.

The total combined weight of the mixture is given as 50 pounds.

2

Formulate an equation for the total weight of Arabica beans

0.60x + 0.30(50 - x) = 0.42(50)

, which simplifies to

0.60x + 15 - 0.30x = 21

.

The sum of Arabica beans contributed by each blend must equal the total weight of Arabica beans in the combined mixture.

3

Solve the linear equation for

x

0.30x + 15 = 21 \implies 0.30x = 6 \implies x = 20

.

Subtract 15 from both sides to isolate the variable term, then divide by 0.30.

Key Concept

Linear Modeling and Mixture Problems

Estimated Time:1m 30s

Question 4Question →

If $x$ is a real number that satisfies the inequality $|5 - 2x| \le 7$ , what is the minimum possible value of the expression $3 - 4x$ ?

-21

-1

7

27

-27

Show answer & explanation

Answer: $-21$

Answer

-21

Solving

|5 - 2x| \le 7

yields

-7 \le 5 - 2x \le 7

. Subtracting

5

gives

-12 \le -2x \le 2

. Dividing by

-2

and flipping the inequality direction gives

-1 \le x \le 6

. To minimize

3 - 4x

, we choose the maximum value of

x

because the term

-4x

decreases as

x

increases. Substituting

x = 6

yields

3 - 4(6) = -21

.

Step-by-Step Solution

1

Unpack the absolute value inequality into a compound inequality

-7 \le 5 - 2x \le 7

For any real constant

k \ge 0

,

|u| \le k

is equivalent to

-k \le u \le k

.

2

Isolate the variable term by subtracting

5

from all parts of the inequality

-12 \le -2x \le 2

Subtracting a constant from all parts preserves inequality direction.

3

Divide all parts by

-2

and reverse the inequality signs

6 \ge x \ge -1

, which is equivalent to

-1 \le x \le 6

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

4

Find the minimum value of

3 - 4x

over the interval

[-1, 6]

The minimum value occurs at

x = 6

:

3 - 4(6) = 3 - 24 = -21

Because the linear expression

3 - 4x

has a negative coefficient for

x

, it is a decreasing function; its minimum occurs at the largest allowed value of

x

.

Key Concept

Linear Inequalities and Absolute Value Bounds

Estimated Time:1m 30s

Question 5Question →

If $x + y = 8$ and $2x - y = 7$ , what is the value of $x$ ?

5

3

1

15

11

Show answer & explanation

Answer: 5

Answer

5

Adding the two equations yields

(x + y) + (2x - y) = 8 + 7

, which simplifies to

3x = 15

. Dividing both sides by 3 gives the correct value,

x = 5

.

Step-by-Step Solution

1

Add the two equations together to eliminate the variable

y

.

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

Since the coefficients of

y

are

+1

and

-1

, adding the equations eliminates

y

directly.

2

Solve for

x

by dividing both sides of the equation by 3.

x = \frac{15}{3} = 5

Isolating

x

gives the required solution.

Key Concept

Elimination Method in Systems of Linear Equations

Question 6Question →

If $\frac{2x - 3}{4} - \frac{x - 5}{3} = \frac{x + 15}{12}$ , what is the value of $3x - 2$ ?

10

4

37

55

130

Show answer & explanation

Answer: 10

Answer

The value of

3x - 2

is

10

.

Multiplying the given linear equation by the least common denominator 12 yields

3(2x - 3) - 4(x - 5) = x + 15

. Expanding both sides gives

6x - 9 - 4x + 20 = x + 15

, which simplifies to

2x + 11 = x + 15

, leading to

x = 4

. Substituting

x = 4

into

3x - 2

gives

3(4) - 2 = 10

.

Step-by-Step Solution

1

Clear the denominators by multiplying both sides of the equation by the least common multiple of 4, 3, and 12, which is 12.

12 \cdot \left(\frac{2x - 3}{4}\right) - 12 \cdot \left(\frac{x - 5}{3}\right) = 12 \cdot \left(\frac{x + 15}{12}\right) \implies 3(2x - 3) - 4(x - 5) = x + 15

Eliminating fractions simplifies the algebraic manipulation.

2

Distribute the coefficients through the terms in parentheses, paying careful attention to the negative sign on the second term.

6x - 9 - 4x + 20 = x + 15

Distributing

-4

across

(x - 5)

yields

-4x + 20

.

3

Combine like terms on the left side of the equation.

2x + 11 = x + 15

Grouping variable terms

(6x - 4x = 2x)

and constant terms

(-9 + 20 = 11)

.

4

Isolate the variable

x

by subtracting

x

and

11

from both sides.

x = 4

Solves the linear equation for

x

.

5

Substitute

x = 4

into the target expression

3x - 2

.

3(4) - 2 = 12 - 2 = 10

Calculates the final answer requested in the stem.

Key Concept

Linear Equations with Fractional Coefficients

Question 7Question →

For all real numbers

x

such that

x \neq 3

and

x \neq -5

, the algebraic expression

\frac{(x^2 - 9)^2 - 4(x - 3)^2}{(x - 3)(x + 5)}

can be simplified to the equivalent polynomial expression

x^2 + ax + b

, where

a

and

b

are constants. What is the value of

a + b

?

Show answer & explanation

Answer: -5

Answer

The value of

a + b

is

-5

.

Factoring the numerator yields

(x - 3)^2(x + 1)(x + 5)

. Canceling the common factors

(x - 3)

and

(x + 5)

with the denominator simplifies the expression to

(x - 3)(x + 1) = x^2 - 2x - 3

. Comparing this to

x^2 + ax + b

identifies

a = -2

and

b = -3

, giving a sum of

a + b = -5

.

Step-by-Step Solution

1

Factor the numerator by recognizing

(x^2 - 9) = (x - 3)(x + 3)

(x^2 - 9)^2 - 4(x - 3)^2 = [(x - 3)(x + 3)]^2 - 4(x - 3)^2 = (x - 3)^2 (x + 3)^2 - 4(x - 3)^2

Applying the difference of squares identity inside the squared term allows factoring out common factors.

2

Factor out

(x - 3)^2

from the numerator

(x - 3)^2 \left[ (x + 3)^2 - 4 \right]

Extracting the greatest common algebraic factor simplifies the remaining expression.

3

Apply difference of squares to

(x + 3)^2 - 4

(x + 3)^2 - 2^2 = ((x + 3) - 2)((x + 3) + 2) = (x + 1)(x + 5)

Recognizing

(x + 3)^2 - 2^2

as

A^2 - B^2

yields factored linear terms directly.

4

Substitute the fully factored numerator back into the rational expression and simplify

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

Canceling non-zero common factors

(x - 3)

and

(x + 5)

simplifies the rational function.

5

Expand

(x - 3)(x + 1)

and determine

a + b

(x - 3)(x + 1) = x^2 - 2x - 3

, so

a = -2

and

b = -3

. Therefore,

a + b = -2 + (-3) = -5

.

Matching coefficients with

x^2 + ax + b

gives

a = -2

and

b = -3

.

Key Concept

Simplifying complex rational expressions through nested difference of squares factoring.

Question 8Question →

Consider the system of linear equations in two variables

x

and

y

:

\begin{aligned} 2x - 3y &= a \\ 4x - 6y &= b \end{aligned}

where

a

and

b

are real constants. Which of the following statements MUST be true? Select all that apply.

Select all that apply

If

b = 2a

, the system has infinitely many solutions.

If

b \neq 2a

, the system has no solution.

If

a = 0

and

b = 0

, the system has a unique solution at

(0, 0)

.

There exist real values of

a

and

b

for which the system has exactly one solution.

If

b = 2a + 1

, the lines represented by the equations in the

xy

-plane are parallel and distinct.

Show answer & explanation

Answer: If $b = 2a$ , the system has infinitely many solutions.; If $b \neq 2a$ , the system has no solution.; If $b = 2a + 1$ , the lines represented by the equations in the $xy$ -plane are parallel and distinct.

Answer

The correct statements are: 'If

b = 2a

, the system has infinitely many solutions.', 'If

b \neq 2a

, the system has no solution.', and 'If

b = 2a + 1

, the lines represented by the equations in the

xy

-plane are parallel and distinct.'

Statements asserting that the system has infinitely many solutions when

b = 2a

, no solution when

b \neq 2a

, and parallel distinct lines when

b = 2a + 1

are all mathematically sound. Multiplying the first equation by 2 reveals that the left sides are identical (

4x - 6y

). Equality of the right sides (

b = 2a

) makes the lines identical, whereas inequality (

b \neq 2a

) makes them parallel and distinct.

Step-by-Step Solution

1

Analyze the coefficients of the system of equations.

The coefficients of

x

and

y

in the second equation (

4

and

-6

) are exactly twice the coefficients of

x

and

y

in the first equation (

2

and

-3

).

Comparing coefficient ratios determines line relationships (slopes).

2

Multiply the first equation by 2.

2(2x - 3y) = 2(a) \implies 4x - 6y = 2a

.

This puts the left side of the first equation in exact alignment with the second equation (

4x - 6y = b

).

3

Evaluate the condition for infinitely many solutions.

If

b = 2a

, the two equations become

4x - 6y = 2a

and

4x - 6y = 2a

, which describe the exact same line, giving infinitely many solutions.

Coincident lines intersect at every point along the line.

4

Evaluate the condition for no solution.

If

b \neq 2a

, subtracting the equations gives

0 = b - 2a \neq 0

, a contradiction. Hence, the lines are parallel and distinct, meaning no solution exists.

Distinct parallel lines never intersect.

5

Examine specific cases such as

b = 2a + 1

and

a=0, b=0

.

For

b = 2a + 1

, since

2a + 1 \neq 2a

,

b \neq 2a

, confirming parallel distinct lines. For

a=0, b=0

,

b = 2(0) = 0

, which yields infinitely many solutions rather than a unique solution.

Verifies specific option claims.

Key Concept

Systems of Linear Equations (Solvability and Geometric Interpretation)

Question 9Question →

If the pair $(x, y)$ satisfies the system of linear equations below, what is the value of $x - y$ ?

\begin{aligned} 3x - 5y &= 14 \\ 2x + 7y &= -1 \end{aligned}

-1

2

3

4

5

Show answer & explanation

Answer: 4

Answer

The value of

x - y

is

4

.

Solving the system of linear equations by elimination gives

x = 3

and

y = -1

. Substituting these values into the target expression

x - y

yields

3 - (-1) = 3 + 1 = 4

.

Step-by-Step Solution

1

Eliminate one variable using the elimination method.

Multiply the first equation by

2

and the second equation by

3

:

6x - 10y = 28

6x + 21y = -3

Creating matching coefficients for

x

allows elimination of

x

by subtraction.

2

Subtract the first modified equation from the second modified equation to solve for

y

.

(6x + 21y) - (6x - 10y) = -3 - 28

31y = -31 \implies y = -1

Subtracting eliminates

x

, isolating

y

.

3

Substitute

y = -1

back into one of the original equations to solve for

x

.

3x - 5(-1) = 14 \implies 3x + 5 = 14 \implies 3x = 9 \implies x = 3

Plugging in

y

allows direct solution for

x

.

4

Evaluate the requested expression

x - y

.

x - y = 3 - (-1) = 3 + 1 = 4

Subtracting a negative value is equivalent to adding its positive value.

Key Concept

Solving systems of linear equations using elimination and evaluating algebraic expressions.

Estimated Time:1m 30s

Question 10Question →

The quadratic equation $x^2 - kx + 36 = 0$ , where $k$ is a positive constant, has two distinct real roots $r_1$ and $r_2$ such that $r_2 - r_1 = 5$ . What is the value of $k$ ?

Show answer & explanation

Answer: 13

Answer

The value of

k

is 13.

According to Vieta's formulas, for the quadratic equation

x^2 - kx + 36 = 0

, the sum of the roots is

r_1 + r_2 = k

and the product of the roots is

r_1 r_2 = 36

. Using the identity

(r_2 - r_1)^2 = (r_1 + r_2)^2 - 4r_1 r_2

, we substitute the known values

r_2 - r_1 = 5

,

r_1 + r_2 = k

, and

r_1 r_2 = 36

. This gives

5^2 = k^2 - 4(36)

, which simplifies to

25 = k^2 - 144

. Solving for

k^2

gives

k^2 = 169

. Since

k

is specified as a positive constant,

k = 13

.

Step-by-Step Solution

1

Apply Vieta's formulas to the given quadratic equation

The sum of the roots is

r_1 + r_2 = k

and the product of the roots is

r_1 r_2 = 36

.

For any quadratic equation

ax^2 + bx + c = 0

, the sum of roots is

-b/a

and the product of roots is

c/a

.

2

Relate the difference of the roots to their sum and product

(r_2 - r_1)^2 = (r_1 + r_2)^2 - 4r_1 r_2

Expanding both sides shows that

r_2^2 - 2r_1 r_2 + r_1^2 = r_1^2 + 2r_1 r_2 + r_2^2 - 4r_1 r_2

, which is an algebraic identity.

3

Substitute the known values into the identity

5^2 = k^2 - 4(36) \implies 25 = k^2 - 144

We are given that

r_2 - r_1 = 5

,

r_1 r_2 = 36

, and

r_1 + r_2 = k

.

4

Solve for the positive constant

k

k^2 = 169 \implies k = 13

Adding 144 to both sides gives

k^2 = 169

. Taking the positive square root because

k > 0

yields

k = 13

.

Key Concept

Vieta's Formulas and Root Difference Identity

Question 11Question →

If $x$ and $y$ are positive integers such that $2^{x+1} + 2^x = 3^{y+2} - 3^y$ , what is the value of $x + y$ ?

Show answer & explanation

Answer: 4

Answer

The value of

x + y

is

4

.

Factoring the left side gives

2^x(2 + 1) = 3 \cdot 2^x

, while factoring the right side gives

3^y(9 - 1) = 8 \cdot 3^y = 2^3 \cdot 3^y

. Equating the two expressions gives

3 \cdot 2^x = 2^3 \cdot 3^y

. Rearranging terms to separate bases yields

2^{x-3} = 3^{y-1}

. Because 2 and 3 share no common prime factors, this equality holds for integers if and only if both exponents are equal to 0. Solving

x - 3 = 0

gives

x = 3

, and solving

y - 1 = 0

gives

y = 1

. Both are positive integers. Thus,

x + y = 3 + 1 = 4

.

Step-by-Step Solution

1

Factor out common terms on both sides of the equation.

2^x(2 + 1) = 3^y(9 - 1) \implies 3 \cdot 2^x = 8 \cdot 3^y

Factoring simplifies sums of powers with identical bases.

2

Rewrite integers using prime factorizations and re-group bases.

3 \cdot 2^x = 2^3 \cdot 3^y \implies 2^{x-3} = 3^{y-1}

Dividing both sides by

2^3 \cdot 3

separates the base-2 and base-3 exponential terms.

3

Set each exponent to zero using prime independence.

x - 3 = 0 \implies x = 3

and

y - 1 = 0 \implies y = 1

Powers of distinct prime numbers 2 and 3 can only be equal if both powers equal

1

(

2^0 = 3^0 = 1

).

4

Calculate the required sum

x + y

.

3 + 1 = 4

Evaluates the requested combined value of the variables.

Key Concept

Solving exponential equations involving distinct prime bases through factoring and exponent properties.

Question 12Question →

If

x

is a real number satisfying the equation

\frac{\left(2^{3x+2} + 2^{3x}\right)^3}{\left(4^{2x+1} - 4^{2x}\right)^2} = \frac{2000}{9}

what is the value of

x

?

Show answer & explanation

Answer: 4

Answer

4

Factoring out

2^{3x}

in the numerator gives

2^{3x}(2^2 + 1) = 5 \cdot 2^{3x}

. Cubing this yields

125 \cdot 2^{9x}

. In the denominator, rewriting

4^{2x}

as

2^{4x}

and factoring gives

2^{4x}(4 - 1) = 3 \cdot 2^{4x}

. Squaring this yields

9 \cdot 2^{8x}

. Taking the quotient gives

\frac{125}{9} \cdot 2^x

. Setting this equal to

\frac{2000}{9}

leads directly to

125 \cdot 2^x = 2000 \implies 2^x = 16 \implies x = 4

.

Step-by-Step Solution

1

Factor out common exponential terms inside the parentheses

Numerator inside becomes

5 \cdot 2^{3x}

and denominator inside becomes

3 \cdot 2^{4x}

Factoring out

2^{3x}

from

2^{3x+2} + 2^{3x}

isolates the constant multiplier

(4+1)

, and expressing

4^{2x}

as

2^{4x}

allows base unification.

2

Raise the simplified terms to their respective outer powers

Numerator becomes

125 \cdot 2^{9x}

and denominator becomes

9 \cdot 2^{8x}

Using power rules

(a \cdot b)^n = a^n b^n

and

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

.

3

Simplify the fraction by subtracting exponents of like bases

The left side simplifies to

\frac{125}{9} \cdot 2^x

\ me{2^{9x}}{2^{8x}} = 2^{9x-8x} = 2^x

using the quotient rule for exponents.

4

Solve the resulting single-variable exponential equation

2^x = 16

, which yields

x = 4

Multiplying both sides by

9

yields

125 \cdot 2^x = 2000

, so

2^x = 16 = 2^4

.

Key Concept

Factoring exponential expressions and applying power of a power and quotient rules

Question 13Question →

If $x$ is a positive real number, which of the following is equivalent to the expression $\frac{(x^3)^2 \cdot x^{-4}}{\sqrt{x^8}}$ ?

x^{-2}

x^2

x^{-3}

x^{-4}

x^0

Show answer & explanation

Answer: $x^{-2}$

Answer

The simplified expression is equivalent to

x^{-2}

.

Applying the rules of exponents systematically yields

(x^3)^2 = x^6

in the numerator, which combines with

x^{-4}

to give

x^2

. The denominator

\sqrt{x^8}

simplifies to

x^{8/2} = x^4

. Dividing

x^2

by

x^4

gives

x^{2-4} = x^{-2}

.

Step-by-Step Solution

1

Simplify the power raised to a power in the numerator.

(x^3)^2 = x^{3 \cdot 2} = x^6

When raising a power to a power, multiply the exponents:

(a^m)^n = a^{m \cdot n}

.

2

Multiply the terms in the numerator.

x^6 \cdot x^{-4} = x^{6 + (-4)} = x^2

When multiplying exponential expressions with the same base, add the exponents:

a^m \cdot a^n = a^{m+n}

.

3

Simplify the radical in the denominator.

\sqrt{x^8} = (x^8)^{\frac{1}{2}} = x^{\frac{8}{2}} = x^4

Taking the square root of a non-negative term is equivalent to raising it to the power of

\frac{1}{2}

.

4

Divide the numerator by the denominator.

\frac{x^2}{x^4} = x^{2 - 4} = x^{-2}

When dividing exponential expressions with the same base, subtract the denominator exponent from the numerator exponent: \frac{a^m}{a^n} = a^{m-n}.

Key Concept

Laws of Exponents and Radical Simplification

Estimated Time:45s

Question 14Question →

A manufacturing company produces two models of office chairs, Model P and Model Q. Producing each Model P chair requires $3$ hours of assembly and $1$ hour of finishing. Producing each Model Q chair requires $2$ hours of assembly and $2$ hours of finishing. On a given day, the assembly department was scheduled for $96$ total hours of work and the finishing department was scheduled for $48$ total hours of work. If both departments operated at full capacity and used all scheduled hours, how many Model P chairs were produced on that day?

Show answer & explanation

Answer: 24

Answer

24

Letting

p

be the number of Model P chairs and

q

be the number of Model Q chairs, the hours used by the assembly department give

3p + 2q = 96

and the hours used by the finishing department give

p + 2q = 48

. Subtracting the finishing equation from the assembly equation eliminates

2q

, resulting in

2p = 48

, or

p = 24

.

Step-by-Step Solution

1

Define variables for the unknown quantities.

Let

p

equal the number of Model P chairs produced and

q

equal the number of Model Q chairs produced.

Assigning variables to the unknown quantities enables the construction of algebraic equations.

2

Formulate a system of linear equations representing total department hours.

Assembly department equation:

3p + 2q = 96

; Finishing department equation:

p + 2q = 48

.

The sum of hours required for both models in each department must equal that department's total scheduled hours.

3

Solve the system of equations for

p

using the elimination method.

Subtracting

p + 2q = 48

from

3p + 2q = 96

yields

2p = 48

, which gives

p = 24

.

Eliminating

q

directly provides the value for

p

, which corresponds to the target quantity requested in the problem.

Key Concept

Modeling real-world resource allocation using systems of linear equations

Question 15Question →

For a real constant

k \neq -1

, consider the linear equation in one variable

x

:

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

Which of the following values of

k

result in a solution

x

that is a positive integer? Select all such values.

Select all that apply

-2

0

1

2

3

Show answer & explanation

Answer: -2; 2; 3

Answer

The values of

k

that yield a positive integer solution for

x

are

-2

,

2

, and

3

.

Solving the linear equation for

x

in terms of

k

gives

x = \frac{4k + 22}{5 - k}

. Substituting

k = -2

yields

x = 2

, substituting

k = 2

yields

x = 10

, and substituting

k = 3

yields

x = 17

. All three resulting values of

x

are positive integers.

Step-by-Step Solution

1

Clear denominators by multiplying the entire equation by

3(k + 1)

.

6(x - 3) - (x + 1)(k + 1) = 3(k + 1)

Eliminating fractional terms simplifies isolation of the variable

x

.

2

Expand all terms on both sides of the equation.

6x - 18 - (kx + x + k + 1) = 3k + 3 \implies 5x - kx - 19 - k = 3k + 3

Carefully distribute negative signs and combine like terms.

3

Group terms containing

x

on the left side and constant/k terms on the right side.

(5 - k)x = 4k + 22 \implies x = \frac{4k + 22}{5 - k}

Express

x

explicitly as a rational function of the parameter

k

.

4

Evaluate

x

for each given option to determine which produce positive integers.

For

k = -2

:

x = 14/7 = 2

(integer);
For

k = 0

:

x = 22/5 = 4.4

(not integer);
For

k = 1

:

x = 26/4 = 6.5

(not integer);
For

k = 2

:

x = 30/3 = 10

(integer);
For

k = 3

:

x = 34/2 = 17

(integer).

Test each candidate value of

k

against the condition that

x

must be a positive integer.

Key Concept

Solving parametric linear equations in one variable and evaluating integer solutions

Estimated Time:2m 30s

Question 16Question →

Line $L$ is defined by the equation $2x + y = 6$ in the $xy$ -plane. Which of the following statements about line $L$ must be true? Select all that apply.

Select all that apply

The slope of line

L

is

-2

.

The

x

-intercept of line

L

is

(3, 0)

.

The slope of line

L

is

-\frac{1}{2}

.

A line perpendicular to line

L

has a slope of

-2

.

The

y

-intercept of line

L

is

(0, -6)

.

Show answer & explanation

Answer: The slope of line $L$ is $-2$ .; The $x$ -intercept of line $L$ is $(3, 0)$ .

Answer

The slope of line

L

is

-2

, and the

x

-intercept of line

L

is

(3, 0)

.

The given line equation

2x + y = 6

can be rewritten in slope-intercept form

y = -2x + 6

. This directly shows that the slope is

-2

. Setting

y = 0

gives

2x = 6 \implies x = 3

, so the

x

-intercept is

(3, 0)

. Therefore, both the statement that the slope is

-2

and the statement that the

x

-intercept is

(3, 0)

are correct.

Step-by-Step Solution

1

Convert the equation to slope-intercept form (

y = mx + b

).

y = -2x + 6

This isolates

y

to clearly reveal the slope

m = -2

and the

y

-intercept

(0, 6)

.

2

Find the

x

-intercept by setting

y = 0

.

2x + 0 = 6 \implies x = 3

, giving coordinate

(3, 0)

The

x

-intercept is the point where the line crosses the

x

-axis.

3

Determine perpendicular slope rules.

Perpendicular slope

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

Perpendicular lines have negative reciprocal slopes, not identical slopes.

Key Concept

Linear equations, slope-intercept form, intercepts, and perpendicular slopes in coordinate geometry.

Estimated Time:1m 0s

Question 17Question →

If $(2x - 3)^2 = 25$ and $x < 0$ , what is the value of $x$ ?

-1

4

-2

-4

1

Show answer & explanation

Answer: $-1$

Answer

-1

Taking the square root of both sides of

(2x - 3)^2 = 25

yields two equations:

2x - 3 = 5

(which gives

x = 4

) and

2x - 3 = -5

(which gives

x = -1

). Because the problem specifies that

x < 0

, the correct value must be

-1

.

Step-by-Step Solution

1

Take the square root of both sides of the equation

(2x - 3)^2 = 25

2x - 3 = \pm 5

Applying the square root property to a squared binomial requires considering both positive and negative principal roots.

2

Set up two separate linear equations corresponding to the two cases

Case 1:

2x - 3 = 5

or Case 2:

2x - 3 = -5

To find all possible solutions for

x

, evaluate both root possibilities.

3

Solve each linear equation for

x

Case 1 yields

2x = 8 \implies x = 4

. Case 2 yields

2x = -2 \implies x = -1

.

Isolate

x

using standard algebraic operations.

4

Apply the given constraint

x < 0

x = -1

Since

4 > 0

, the positive root is eliminated, leaving

x = -1

as the sole valid solution.

Key Concept

Quadratic Equations and Factoring

Question 18Question →

If $x = 5$ is a solution to the quadratic equation $x^2 - (k + 3)x + 3k + 1 = 0$ , where $k$ is a constant, what is the value of the other solution?

\frac{7}{2}

\frac{11}{2}

-\frac{7}{2}

-\frac{11}{2}

\frac{5}{2}

Show answer & explanation

Answer: $\frac{7}{2}$

Answer

\frac{7}{2}

Substituting

x = 5

into

x^2 - (k + 3)x + 3k + 1 = 0

yields

25 - 5(k + 3) + 3k + 1 = 0

, which simplifies to

11 - 2k = 0

, so

k = \frac{11}{2}

. Substituting

k = \frac{11}{2}

back into the product of roots formula

x_1 \cdot x_2 = 3k + 1

gives

5 \cdot x_2 = 3\left(\frac{11}{2}\right) + 1 = \frac{35}{2}

. Dividing by

5

yields the second root

x_2 = \frac{7}{2}

.

Step-by-Step Solution

1

Substitute the known solution

x = 5

into the quadratic equation to solve for

k

.

5^2 - (k + 3)(5) + 3k + 1 = 0 \implies 25 - 5k - 15 + 3k + 1 = 0 \implies 11 - 2k = 0 \implies k = \frac{11}{2}

.

Since

x = 5

is a solution, it must satisfy the equation.

2

Use Vieta's formulas to find the other solution

x_2

.

Product of roots

x_1 \cdot x_2 = 3k + 1 = 3\left(\frac{11}{2}\right) + 1 = \frac{35}{2}

. Since

x_1 = 5

,

5 \cdot x_2 = \frac{35}{2} \implies x_2 = \frac{7}{2}

.

By Vieta's formulas for a standard quadratic

x^2 + bx + c = 0

, the product of the roots equals

c

.

Key Concept

Quadratic equations, Vieta's formulas, and parameter evaluation

Question 19Question →

A hotel renovated its guest rooms and purchased a total of 100 lighting fixtures, consisting of wall sconces costing $\$45$ each and ceiling pendants costing $\$70$ each. If the average (arithmetic mean) cost per fixture for the entire purchase was $\$52$ , how many wall sconces were purchased?

Show answer & explanation

Answer: 72

Answer

72

Setting up the weighted total cost equation

45x + 70(100 - x) = 5200

simplifies to

-25x = -1800

, giving

x = 72

wall sconces.

Step-by-Step Solution

1

Define the variable for the unknown quantity

Let

x

represent the number of wall sconces purchased. The number of ceiling pendants is

100 - x

.

Defining the target quantity as

x

allows the problem to be modeled using a single-variable linear equation.

2

Set up the algebraic equation for total cost

45x + 70(100 - x) = 52 \times 100

The total cost of all fixtures is the sum of the total sconce cost and total pendant cost, which equals the overall average cost multiplied by the total number of fixtures.

3

Expand and solve the linear equation

45x + 7000 - 70x = 5200 \implies -25x = -1800 \implies x = 72

Simplifying algebraic terms isolates

x

to find the exact number of wall sconces.

Key Concept

Linear Algebraic Modeling and Weighted Averages

Question 20Question →

A chemistry laboratory prepares a $100\text{ mL}$ mixture using three solutions: Solution $X$ ( $10\%$ acid by volume), Solution $Y$ ( $30\%$ acid by volume), and Solution $Z$ ( $50\%$ acid by volume). The resulting mixture is $37\%$ acid by volume. If the volume of Solution $Y$ used is $5\text{ mL}$ more than twice the volume of Solution $X$ , what is the volume, in $\text{mL}$ , of Solution $Z$ used in the mixture?

15

35

40

50

55

Show answer & explanation

Answer: $50$

Answer

50\text{ mL}

Setting up the system of equations gives

x + y + z = 100

,

x + 3y + 5z = 370

, and

y = 2x + 5

. Substituting

y = 2x + 5

into the first two equations yields

3x + z = 95

and

7x + 5z = 355

. Solving for

x

gives

x = 15

, which leads to

z = 95 - 3(15) = 50

. Therefore,

50\text{ mL}

of Solution Z was used.

Step-by-Step Solution

1

Define variables and set up the system of linear equations

Let

x

,

y

, and

z

be the volumes in

\text{mL}

of Solutions

X

,

Y

, and

Z

, respectively.
1) Total volume:

x + y + z = 100

2) Total acid volume:

0.10x + 0.30y + 0.50z = 0.37(100) \implies x + 3y + 5z = 370

3) Relationship between

Y

and

X

:

y = 2x + 5

Translate the word problem statements into mathematical equations.

2

Substitute

y = 2x + 5

into equations (1) and (2) to reduce to a two-variable system

From equation (1):

x + (2x + 5) + z = 100 \implies 3x + z = 95 \implies z = 95 - 3x

From equation (2):

x + 3(2x + 5) + 5z = 370 \implies 7x + 15 + 5z = 370 \implies 7x + 5z = 355

Eliminating

y

simplifies the system to two equations in

x

and

z

.

3

Substitute

z = 95 - 3x

into

7x + 5z = 355

and solve for

x

7x + 5(95 - 3x) = 355 \implies 7x + 475 - 15x = 355 \implies -8x = -120 \implies x = 15

Solves for the unknown volume of Solution X.

4

Calculate

z

using

z = 95 - 3x

z = 95 - 3(15) = 95 - 45 = 50

Finds the requested volume of Solution Z.

Key Concept

Setting up and solving 3x3 systems of linear equations using substitution or elimination

Estimated Time:2m 0s

Algebra

Answer

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