Algebra Practice Questions — GRE General Test — Page 4

What value of $x$ satisfies the exponential equation $25^x - 4 \cdot 5^{x+1} = 125$ ?

Show answer & explanation

Answer: 2

Answer

The correct answer is

2

.

Rewriting

25^x

as

(5^x)^2

and

4 \cdot 5^{x+1}

as

20 \cdot 5^x

transforms the equation into

(5^x)^2 - 20(5^x) - 125 = 0

. Substituting

u = 5^x

produces

u^2 - 20u - 125 = 0

, which factors into

(u - 25)(u + 5) = 0

. Because

5^x

must be greater than zero for all real values of

x

,

u = -5

yields no valid real solution. Thus,

5^x = 25 = 5^2

, giving

x = 2

.

Step-by-Step Solution

1

Convert exponential expressions to a common base of 5.

25^x = (5^2)^x = (5^x)^2

and

4 \cdot 5^{x+1} = 4 \cdot 5 \cdot 5^x = 20 \cdot 5^x

.

Applying exponent laws

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

and

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

expresses terms in quadratic form with respect to

5^x

.

2

Formulate and factor the quadratic equation in terms of

u = 5^x

.

u^2 - 20u - 125 = 0 \implies (u - 25)(u + 5) = 0

, yielding

u = 25

or

u = -5

.

The equation reduces to standard quadratic form, which factors easily.

3

Solve for

x

while rejecting non-viable real roots.

5^x = 25 = 5^2 \implies x = 2

.

5^x = -5

has no real solution.

An exponential function with a positive base produces strictly positive output values for all real domain inputs.

Key Concept

Solving exponential equations reducible to quadratic form using exponent laws

Question 62Question →

At a charity fundraising event, standard tickets were sold for $\$45$ each and VIP tickets were sold for $\$80$ each. The number of standard tickets sold was $12$ more than twice the number of VIP tickets sold. If the total revenue generated from standard tickets exceeded the total revenue from VIP tickets by $\$1,830$ , how many VIP tickets were sold?

Show answer & explanation

Answer: 129

Answer

129 VIP tickets were sold.

Letting

v

represent the number of VIP tickets sold, the number of standard tickets sold is

2v + 12

. Expressing the revenue condition yields the linear equation

45(2v + 12) - 80v = 1830

. Distributing 45 gives

90v + 540 - 80v = 1830

. Combining like terms results in

10v + 540 = 1830

. Subtracting 540 from both sides gives

10v = 1290

, which yields

v = 129

.

Step-by-Step Solution

1

Define the unknown variable

Let

v

be the number of VIP tickets sold.

The question asks for the number of VIP tickets, making

v

a direct choice for the variable.

2

Translate the relationship between ticket quantities into an algebraic expression

Standard tickets sold

= 2v + 12

'12 more than twice the number of VIP tickets' translates directly to

2v + 12

.

3

Formulate total revenue expressions and construct the single-variable linear equation

45(2v + 12) - 80v = 1830

Total standard revenue minus total VIP revenue equals the given excess of

\$1,830

.

4

Distribute and combine like terms to solve for

v

90v + 540 - 80v = 1830 \implies 10v + 540 = 1830 \implies 10v = 1290 \implies v = 129

Applying standard algebraic operations isolates

v

on one side of the equation.

Key Concept

Formulating and solving a linear equation in one variable from a multi-step word problem context

Estimated Time:2m 0s

Question 63Question →

A reservoir initially contains $4,800$ gallons of water. A drainage pump operates at a constant rate of $120$ gallons per minute to empty the reservoir. How many minutes of continuous pumping will it take for the amount of water remaining in the reservoir to be reduced to $1,200$ gallons?

10

25

30

40

50

Show answer & explanation

Answer: $30$

Answer

30 minutes

To find the time needed for

1,200

gallons to remain, subtract

1,200

from the initial

4,800

gallons to determine that

3,600

gallons of water must be removed. Dividing the

3,600

gallons by the drainage rate of

120

gallons per minute gives exactly

30

minutes.

Step-by-Step Solution

1

Calculate the total volume of water that must be drained from the reservoir.

4,800 \text{ gallons} - 1,200 \text{ gallons} = 3,600 \text{ gallons}

The question asks for the time until

1,200

gallons remain, so the amount removed is the difference between the starting volume and ending volume.

2

Set up the linear rate equation to solve for elapsed time

t

in minutes.

\text{Rate} \times t = \text{Volume Drained} \implies 120t = 3,600

The rate of drainage is constant at

120

gallons per minute.

3

Solve for

t

.

t = \frac{3,600}{120} = 30 \text{ minutes}

Dividing total gallons to drain by the rate gives the required time.

Key Concept

Linear Modeling and Distance-Rate-Time / Work-Rate Relationships

Estimated Time:1m 0s

Question 64Question →

A logistics warehouse uses two sizes of shipping crates: small crates and large crates. A shipment containing $3$ small crates and $5$ large crates has a total weight of $47$ pounds. A second shipment containing $6$ small crates and $2$ large crates has a total weight of $38$ pounds. What is the total weight, in pounds, of a shipment containing $2$ small crates and $3$ large crates?

11

22

26

29

33

Show answer & explanation

Answer: 29

Answer

29 pounds

By setting up the linear system

3s + 5l = 47

and

6s + 2l = 38

, using elimination gives

s = 4

pounds for a small crate and

l = 7

pounds for a large crate. Substituting these values into the target expression

2s + 3l

yields

2(4) + 3(7) = 29

pounds.

Step-by-Step Solution

1

Set up a system of linear equations from the given context.

Let

s

be the weight of a small crate and

l

be the weight of a large crate. The equations are: (1)

3s + 5l = 47

and (2)

6s + 2l = 38

.

Translate the physical constraints of each shipment into algebraic equations.

2

Solve for the variable

l

by eliminating

s

.

Multiply Equation (1) by

2

to get

6s + 10l = 94

. Subtract Equation (2) (

6s + 2l = 38

) from this new equation:

(6s + 10l) - (6s + 2l) = 94 - 38 \implies 8l = 56 \implies l = 7

.

Align the coefficients of

s

so that elimination via subtraction yields a single-variable linear equation for

l

.

3

Substitute

l = 7

back into Equation (2) to solve for

s

.

6s + 2(7) = 38 \implies 6s + 14 = 38 \implies 6s = 24 \implies s = 4

.

Determine the individual weight of a small crate.

4

Calculate the requested total weight for

2

small crates and

3

large crates.

2s + 3l = 2(4) + 3(7) = 8 + 21 = 29

.

Substitute the individual values of

s

and

l

into the target expression.

Key Concept

Solving 2x2 Systems of Linear Equations by Elimination and Linear Combination Evaluation

Estimated Time:1m 30s

Question 65Question →

Two water reservoirs, Reservoir A and Reservoir B, contain a combined total of $360$ liters of water. After $30$ liters of water are transferred from Reservoir A to Reservoir B, the volume of water in Reservoir B is equal to $\frac{2}{3}$ of the volume of water remaining in Reservoir A. How many liters of water were originally in Reservoir A?

114

210

216

246

228

Show answer & explanation

Answer: $246$

Answer

246 liters were originally in Reservoir A.

Let

x

represent the original volume of Reservoir A. Since the combined volume is

360

liters, Reservoir B initially contains

360 - x

liters. After transferring

30

liters from A to B, Reservoir A contains

x - 30

liters and Reservoir B contains

(360 - x) + 30 = 390 - x

liters. Setting up the equation

390 - x = \frac{2}{3}(x - 30)

and multiplying by 3 gives

1170 - 3x = 2x - 60

. Combining like terms yields

5x = 1230

, so

x = 246

liters. Thus,

246

is the correct original volume.

Step-by-Step Solution

1

Define the variable for the unknown quantity.

Let

x

be the original volume of water in Reservoir A (in liters).

Choosing a single variable simplifies setting up a linear equation.

2

Express the original volume of Reservoir B in terms of

x

.

Original volume in Reservoir B is

360 - x

.

The total volume across both reservoirs is given as

360

liters.

3

Write expressions for the volumes in each reservoir after the transfer of 30 liters.

Volume in Reservoir A after transfer:

x - 30

.
Volume in Reservoir B after transfer:

(360 - x) + 30 = 390 - x

.

Transferring 30 liters removes 30 liters from A and adds 30 liters to B.

4

Set up the linear equation based on the given relationship.

390 - x = \frac{2}{3}(x - 30)

The problem specifies that the new volume in Reservoir B is

\frac{2}{3}

of the new volume in Reservoir A.

5

Solve the equation for

x

.

3(390 - x) = 2(x - 30) \implies 1170 - 3x = 2x - 60 \implies 1230 = 5x \implies x = 246

.

Multiply both sides by 3 to eliminate the fraction, then collect like terms.

Key Concept

Formulating and solving a linear equation in one variable from a real-world conservation/transfer scenario.

Question 66Question →

If $x$ is a real number satisfying the exponential equation $4^{x+1} - 4^{x-1} = 120$ , what is the value of $2^{2x + 1}$ ?

16

32

64

128

-64

Show answer & explanation

Answer: 64

Answer

The value of

2^{2x + 1}

is 64.

Factoring

4^{x-1}

from

4^{x+1} - 4^{x-1}

gives

4^{x-1}(16 - 1) = 15 \cdot 4^{x-1} = 120

. Dividing by 15 yields

4^{x-1} = 8

. Rewriting with base 2 gives

(2^2)^{x-1} = 2^{2x-2} = 2^3

, so

2x - 2 = 3

, meaning

2x = 5

. Substituting

2x = 5

into

2^{2x+1}

gives

2^{5+1} = 2^6 = 64

.

Step-by-Step Solution

1

Factor out

4^{x-1}

from the left side of the equation

4^{x+1} - 4^{x-1} = 120

.

4^{x-1}(4^2 - 1) = 120

, which simplifies to

4^{x-1}(15) = 120

.

Factoring out the lowest power of 4 allows simplification of the terms on the left side.

2

Divide both sides by 15 to isolate

4^{x-1}

.

4^{x-1} = \frac{120}{15} = 8

.

Isolating the exponential expression is necessary to solve for

x

.

3

Express both sides with a common base of 2.

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

.

Converting to a common prime base allows equating the exponents.

4

Equate exponents to solve for

2x

.

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

.

Since the bases are equal and positive, their exponents must be equal.

5

Substitute

2x = 5

into the target expression

2^{2x + 1}

.

2^{5 + 1} = 2^6 = 64

.

Evaluating the exact expression requested in the stem.

Key Concept

Factoring exponential expressions with variable exponents and converting bases

Estimated Time:1m 30s

Question 67Question →

A manufacturing company's monthly total production cost

C

, in dollars, is related to the number of units produced,

x

, by the linear equation:

\frac{3(C - 1,200)}{5} = 2x + 180

If the total production cost in a given month was between

\$4,000

and

\$5,000

, inclusive, which of the following could be the number of units produced in that month? Select all such numbers.

Select all that apply

700

800

950

1,000

1,100

Show answer & explanation

Answer: 800; 950; 1,000

Answer

The numbers of units that could be produced are 800, 950, and 1,000.

Solving the linear equation for

x

gives

x = \frac{3C - 4,500}{10}

. Substituting the boundary cost values

C = 4,000

and

C = 5,000

yields

x = 750

and

x = 1,050

, respectively. Therefore, any valid number of units must lie in the range

[750, 1,050]

. The values 800, 950, and 1,000 fall strictly within this interval.

Step-by-Step Solution

1

Express x in terms of C by isolating the variable x in the linear equation.

3(C - 1,200) = 5(2x + 180) \implies 3C - 3,600 = 10x + 900 \implies 10x = 3C - 4,500 \implies x = \frac{3C - 4,500}{10}

Isolating

x

allows direct evaluation of the unit limits based on the cost boundary values.

2

Calculate the minimum value of x corresponding to the minimum cost C = 4,000.

x_{\min} = \frac{3(4,000) - 4,500}{10} = \frac{12,000 - 4,500}{10} = \frac{7,500}{10} = 750

Determines the lower bound of units produced.

3

Calculate the maximum value of x corresponding to the maximum cost C = 5,000.

x_{\max} = \frac{3(5,000) - 4,500}{10} = \frac{15,000 - 4,500}{10} = \frac{10,500}{10} = 1,050

Determines the upper bound of units produced.

4

Filter the given choices to find all values within the interval [750, 1,050].

The valid values are 800, 950, and 1,000.

Only values between 750 and 1,050 inclusive satisfy the cost constraint.

Key Concept

Linear Equations in One Variable

Question 68Question →

If $x$ satisfies the equation $\frac{x + 5}{3} - 2 = \frac{x - 1}{4}$ , what is the value of $x$ ?

-21

-7

1

7

17

Show answer & explanation

Answer: 1

Answer

1

Multiplying the equation

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

by the common denominator

12

eliminates fractions to produce

4(x + 5) - 24 = 3(x - 1)

. Expanding both sides yields

4x + 20 - 24 = 3x - 3

, which simplifies to

4x - 4 = 3x - 3

. Subtracting

3x

from both sides and adding

4

to both sides gives

x = 1

.

Step-by-Step Solution

1

Multiply every term on both sides of the equation by the least common denominator of 3 and 4, which is 12.

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

, which simplifies to

4(x + 5) - 24 = 3(x - 1)

.

Clearing fractional denominators simplifies the equation into integer-coefficient linear form.

2

Expand both sides by distributing the numeric multipliers.

4x + 20 - 24 = 3x - 3

, which combines like terms to

4x - 4 = 3x - 3

.

Distributing coefficients removes grouping symbols so variable and constant terms can be combined.

3

Isolate the variable term

x

on one side.

Subtract

3x

from both sides to get

x - 4 = -3

, then add

4

to both sides to get

x = 1

.

Standard algebraic reduction requires grouping all terms containing the unknown variable on one side and numerical constants on the other.

Key Concept

Solving single-variable linear equations containing fractional terms by clearing denominators

Estimated Time:1m 0s

Question 69Question →

A chemist mixes $x$ ounces of a 40% acid solution with $y$ ounces of a 70% acid solution to produce a 20-ounce mixture that is 52% acid. What is the value of $x - y$ ?

-4

4

8

12

20

Show answer & explanation

Answer: 4

Answer

4

Setting up the linear system

x + y = 20

and

0.40x + 0.70y = 10.4

leads to

x = 12

and

y = 8

. Subtracting

y

from

x

yields

12 - 8 = 4

.

Step-by-Step Solution

1

Set up the system of linear equations based on total solution volume and pure acid content.

Total volume equation:

x + y = 20

. Pure acid equation:

0.40x + 0.70y = 0.52(20) = 10.4

.

The sum of the component volumes equals the total volume, and the sum of the pure acid contents equals the total acid content.

2

Multiply the acid equation by 10 to eliminate decimals.

4x + 7y = 104

.

Clearing decimals simplifies the subsequent elimination calculation.

3

Solve for

y

using the elimination method.

Multiply

x + y = 20

by 4 to get

4x + 4y = 80

. Subtract this from

4x + 7y = 104

:

(4x + 7y) - (4x + 4y) = 104 - 80 \implies 3y = 24 \implies y = 8

.

Eliminating the variable

x

isolates

y

.

4

Solve for

x

and compute

x - y

.

x = 20 - 8 = 12

. Therefore,

x - y = 12 - 8 = 4

.

Substitute

y = 8

back into the first equation and calculate the requested expression.

Key Concept

Systems of Linear Equations in Mixture Problems

Question 70Question →

For all real numbers $x$ such that $x \neq -3$ , which of the following expressions is equivalent to $\frac{2x^2 - 18}{4x + 12}$ ?

\frac{x - 3}{2}

\frac{x + 3}{2}

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

\frac{2x - 3}{4}

\frac{x - 9}{2}

Show answer & explanation

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

Answer

The expression

\frac{x - 3}{2}

is equivalent to the given rational expression.

Factoring the numerator gives

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

and factoring the denominator gives

4(x + 3)

. Canceling the non-zero common terms

2(x + 3)

leaves

\frac{x - 3}{2}

.

Step-by-Step Solution

1

Factor out the greatest common factor from the numerator and denominator.

Numerator:

2x^2 - 18 = 2(x^2 - 9)

; Denominator:

4x + 12 = 4(x + 3)

.

Factoring out common numerical coefficients simplifies the expression and reveals algebraic patterns.

2

Apply the difference of squares formula to factor

x^2 - 9

.

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

, so the numerator becomes

2(x - 3)(x + 3)

.

The algebraic identity

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

allows complete factoring of the numerator.

3

Cancel common factors shared by the numerator and denominator.

\frac{2(x - 3)(x + 3)}{4(x + 3)} = \frac{2(x - 3)}{4} = \frac{x - 3}{2}.

Since

x \neq -3

, the factor

(x + 3)

is non-zero and can be safely canceled along with reducing the constant ratio

\frac{2}{4}

to

\frac{1}{2}

.

Key Concept

Simplifying rational algebraic expressions by factoring common numerical factors and applying the difference of squares identity.

Question 71Question →

If $x$ satisfies the linear equation $\frac{x + 2}{2} + \frac{x - 1}{3} = 4$ , what is the value of $3x - 2$ ?

10

4

-2

26.5

7.6

Show answer & explanation

Answer: $10$

Answer

The value of

3x - 2

is

10

.

To solve

\frac{x + 2}{2} + \frac{x - 1}{3} = 4

, multiply the entire equation by the common denominator 6 to clear fractions, yielding

3(x + 2) + 2(x - 1) = 24

. Distributing gives

3x + 6 + 2x - 2 = 24

, which simplifies to

5x + 4 = 24 \implies 5x = 20 \implies x = 4

. Substituting

x = 4

into

3x - 2

gives

3(4) - 2 = 10

.

Step-by-Step Solution

1

Find a common denominator to clear the fractions from the equation.

The least common multiple of 2 and 3 is 6. Multiplying both sides of the equation by 6 gives

6 \cdot \left(\frac{x + 2}{2}\right) + 6 \cdot \left(\frac{x - 1}{3}\right) = 6 \cdot 4

, which simplifies to

3(x + 2) + 2(x - 1) = 24

.

Clearing denominators simplifies multi-step fractional equations into standard linear form.

2

Expand terms and solve for

x

.

3x + 6 + 2x - 2 = 24 \implies 5x + 4 = 24 \implies 5x = 20 \implies x = 4

.

Combining like terms isolates the variable

x

.

3

Substitute

x = 4

into the target expression

3x - 2

.

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

.

The question asks for the value of the algebraic expression

3x - 2

, not

x

itself.

Key Concept

Solving linear equations with fractional coefficients by clearing denominators and evaluating target algebraic expressions.

Estimated Time:1m 15s

Question 72Question →

Consider the following system of linear equations:

\begin{aligned} 2x + 3y &= 12 \\ 4x + 6y &= 24 \end{aligned}

Which of the following statements about this system must be true? Select all that apply.

Select all that apply

The system has infinitely many solutions.

The graphs of the two equations represent the exact same line in the

xy

-plane.

The ordered pair

(3, 2)

is a solution to the system.

The system is inconsistent and has no solution.

The

y

-intercept of the line is

(6, 0)

.

Show answer & explanation

Answer: The system has infinitely many solutions.; The graphs of the two equations represent the exact same line in the $xy$ -plane.; The ordered pair $(3, 2)$ is a solution to the system.

Answer

The correct statements are that the system has infinitely many solutions, the graphs represent the exact same line, and

(3, 2)

is a solution to the system.

Dividing

4x + 6y = 24

by

2

produces

2x + 3y = 12

, showing that both equations represent the exact same line. Therefore, the system has infinitely many solutions. Substituting

x = 3

and

y = 2

yields

2(3) + 3(2) = 12

, confirming that

(3, 2)

is one of the infinitely many valid solutions.

Step-by-Step Solution

1

Analyze the relationship between the two linear equations

Dividing the second equation

4x + 6y = 24

by

2

gives

2x + 3y = 12

, which is identical to the first equation.

Comparing coefficients and constants determines whether equations in a system are dependent, independent, or inconsistent.

2

Determine the number of solutions and geometric structure

Because the equations are mathematically equivalent, they describe the same line in the coordinate plane and have infinitely many intersection points.

Identical linear equations form a dependent system with infinitely many solutions.

3

Test the ordered pair

(3, 2)

Evaluating

2(3) + 3(2) = 6 + 6 = 12

confirms that

(3, 2)

lies on the line.

Any point satisfying one equation satisfies the entire system of equivalent equations.

4

Determine the axis intercepts

Setting

y=0

gives

2x=12 \Rightarrow x=6

(the

x

-intercept is

(6,0)

); setting

x=0

gives

3y=12 \Rightarrow y=4

(the

y

-intercept is

(0,4)

).

Checking coordinates of axis intersections prevents mislabeling

x

- and

y

-intercepts.

Key Concept

Dependent Systems of Linear Equations

Question 73Question →

For all non-zero real numbers $x$ and $y$ , which of the following expressions is equivalent to $\frac{12x^4y^2 - 8x^2y^3}{4x^2y}$ ?

3x^2y - 2y^2

3x^2y - 8x^2y^3

3x^6y^3 - 2x^4y^4

3x^2y + 2y^2

3x^2 - 2y

Show answer & explanation

Answer: $3x^2y - 2y^2$

Answer

3x^2y - 2y^2

Dividing each term in the numerator by

4x^2y

yields

\frac{12x^4y^2}{4x^2y} - \frac{8x^2y^3}{4x^2y} = 3x^2y - 2y^2

, which matches the correct expression.

Step-by-Step Solution

1

Split the rational algebraic expression into two separate fractions by distributing the denominator.

\frac{12x^4y^2}{4x^2y} - \frac{8x^2y^3}{4x^2y}

When dividing a polynomial by a monomial, each term of the numerator must be divided independently by the denominator.

2

Simplify the first term

\frac{12x^4y^2}{4x^2y}

.

\frac{12}{4} \cdot x^{4-2} \cdot y^{2-1} = 3x^2y

Divide the numerical coefficients and apply the exponent quotient rule

a^m / a^n = a^{m-n}

for each variable.

3

Simplify the second term

\frac{8x^2y^3}{4x^2y}

.

\frac{8}{4} \cdot x^{2-2} \cdot y^{3-1} = 2(1)y^2 = 2y^2

Divide the coefficients and apply the exponent quotient rule, noting that

x^0 = 1

for non-zero

x

.

4

Combine the simplified terms with the original subtraction operator.

3x^2y - 2y^2

Combine terms to form the final simplified expression.

Key Concept

Simplifying algebraic fractions by distributing a monomial denominator across numerator terms and applying exponent rules.

Estimated Time:45s

Question 74Question →

A shipping company calculates the total cost $C$ , in dollars, to deliver a package of weight $w$ pounds using the linear relationship $C = kw + b$ , where $k$ and $b$ are positive constants. The delivery cost for a $4$ -pound package is $\$19$ , and the delivery cost for a $9$ -pound package is $\$39$ . If a customer pays a total delivery cost of $\$71$ for a single package, what is the weight of the package, in pounds?

13

15

17

18

21

Show answer & explanation

Answer: $17$

Answer

The weight of the package is

17

pounds.

Using the two points

(4, 19)

and

(9, 39)

, the rate of change

k

is calculated as

\frac{39 - 19}{9 - 4} = \frac{20}{5} = 4

dollars per pound. Substituting

k = 4

into

4(4) + b = 19

yields

b = 3

. Setting the linear equation

4w + 3 = 71

and solving for

w

gives

4w = 68

, so

w = 17

pounds.

Step-by-Step Solution

1

Set up a system of linear equations using the given data points

(4, 19)

and

(9, 39)

.

4k + b = 19

and

9k + b = 39

.

The cost model follows

C = kw + b

for weight

w

.

2

Subtract the first equation from the second equation to solve for

k

.

(9k + b) - (4k + b) = 39 - 19 \implies 5k = 20 \implies k = 4

.

Subtracting eliminates the constant

b

to determine the unit rate per pound.

3

Substitute

k = 4

into

4k + b = 19

to solve for

b

.

4(4) + b = 19 \implies 16 + b = 19 \implies b = 3

.

Finding

b

establishes the full linear equation model:

C = 4w + 3

.

4

Substitute

C = 71

into the linear equation

4w + 3 = 71

and solve for

w

.

4w = 71 - 3 \implies 4w = 68 \implies w = 17

.

Solving for

w

yields the required weight corresponding to a

\$71

delivery cost.

Key Concept

Linear Modeling and Single-Variable Linear Equations

Estimated Time:1m 30s

Question 75Question →

Consider the following system of linear equations:

\begin{aligned} x + 2y &= 10 \\ 3x - y &= 9 \end{aligned}

Which of the following statements are true? Select all that apply.

Select all that apply

The value of

x

is

4

.

The value of

x + y

is

7

.

The value of

y

is

4

.

The value of

2x + y

is

11

.

The value of

x - y

is

5

.

Show answer & explanation

Answer: The value of $x$ is $4$ .; The value of $x + y$ is $7$ .; The value of $2x + y$ is $11$ .

Answer

The true statements are those asserting that

x = 4

, that

x + y = 7

, and that

2x + y = 11

.

Solving the system of linear equations yields the unique solution

x = 4

and

y = 3

. Substituting these values into the given choices demonstrates that the statements claiming

x = 4

,

x + y = 7

, and

2x + y = 11

are all mathematically correct.

Step-by-Step Solution

1

Express

y

in terms of

x

using the second equation.

y = 3x - 9

Isolating

y

allows for simple substitution into the first equation.

2

Substitute

y = 3x - 9

into the first equation

x + 2y = 10

.

x + 2(3x - 9) = 10 \implies x + 6x - 18 = 10 \implies 7x = 28 \implies x = 4

This yields a single linear equation in terms of

x

.

3

Calculate

y

using

x = 4

.

y = 3(4) - 9 = 3

Substituting

x = 4

back gives the unique solution for

y

.

4

Evaluate the given statements with

(x, y) = (4, 3)

.

x = 4

is true;

x + y = 4 + 3 = 7

is true;

y = 4

is false (

y = 3

);

2x + y = 2(4) + 3 = 11

is true;

x - y = 4 - 3 = 1

is false.

Testing each condition determines which options are valid.

Key Concept

Solving a 2x2 system of linear equations using substitution to evaluate linear expressions.

Estimated Time:1m 0s

Question 76Question →

In the $xy$ -plane, line $k$ has a slope of $\frac{2}{5}$ and passes through the point $(5, 8)$ . Line $m$ is perpendicular to line $k$ and has the same $y$ -intercept as line $k$ . What is the $x$ -intercept of line $m$ ?

Show answer & explanation

Answer: 2.4

Answer

2.4

To find the

x

-intercept of line

m

, first determine the equation of line

k

. Using slope-intercept form

y = mx + b

with

m = \frac{2}{5}

and point

(5, 8)

, we get

8 = \frac{2}{5}(5) + b \implies b = 6

. Thus, the

y

-intercept of line

k

(and line

m

) is

(0, 6)

. Next, line

m

is perpendicular to line

k

, so its slope is the negative reciprocal of

\frac{2}{5}

, which is

-\frac{5}{2}

. The equation for line

m

is

y = -\frac{5}{2}x + 6

. Setting

y = 0

to solve for the

x

-intercept yields

0 = -\frac{5}{2}x + 6 \implies \frac{5}{2}x = 6 \implies x = \frac{12}{5} = 2.4

.

Step-by-Step Solution

1

Find the equation and

y

-intercept of line

k

Line

k

has equation

y = \frac{2}{5}x + 6

, with

y

-intercept at

(0, 6)

.

Using point-slope form

y - y_1 = m(x - x_1)

with

m = \frac{2}{5}

and

(x_1, y_1) = (5, 8)

.

2

Calculate the slope of line

m

The slope of line

m

is

-\frac{5}{2}

.

Perpendicular lines have negative reciprocal slopes:

-\frac{1}{2/5} = -\frac{5}{2}

.

3

Construct the equation of line

m

Line

m

has equation

y = -\frac{5}{2}x + 6

.

Line

m

shares the

y

-intercept

(0, 6)

with line

k

.

4

Solve for the

x

-intercept of line

m

x = 2.4

Set

y = 0

in

y = -\frac{5}{2}x + 6

to get

0 = -\frac{5}{2}x + 6 \implies \frac{5}{2}x = 6 \implies x = 2.4

.

Key Concept

Perpendicular Slopes and Line Intercepts

Question 77Question →

If $x$ is a solution to the quadratic equation $x^2 - 11x + 28 = 0$ such that $x > 5$ , what is the value of $x$ ?

Show answer & explanation

Answer: 7

Answer

The value of

x

is 7.

Factoring

x^2 - 11x + 28 = 0

yields

(x - 7)(x - 4) = 0

, giving solutions

x = 7

and

x = 4

. Since the problem specifies that

x > 5

, the only valid value for

x

is 7.

Step-by-Step Solution

1

Factor the quadratic expression

(x - 7)(x - 4) = 0

Find two numbers that multiply to 28 and sum to -11, which are -7 and -4.

2

Find the roots of the equation

x = 7

or

x = 4

Set each linear factor equal to 0 and solve for

x

.

3

Apply the given constraint

x > 5

x = 7

Out of the two roots 4 and 7, only 7 is strictly greater than 5.

Key Concept

Factoring quadratic equations of the form

x^2 + bx + c = 0

Question 78Question →

If $x$ is a negative real number such that $x^2 + 5x - 14 = 0$ , what is the value of $x + 3$ ?

-10

-7

-4

2

5

Show answer & explanation

Answer: $-4$

Answer

The value of

x + 3

is

-4

.

Factoring

x^2 + 5x - 14 = 0

yields

(x + 7)(x - 2) = 0

, giving solutions

x = -7

and

x = 2

. Given that

x

must be negative,

x = -7

. Substituting this into the target expression gives

-7 + 3 = -4

.

Step-by-Step Solution

1

Factor the quadratic equation

x^2 + 5x - 14 = 0

.

(x + 7)(x - 2) = 0

Find two numbers that multiply to

-14

and add to

5

, which are

7

and

-2

.

2

Solve for the possible values of

x

.

x = -7

or

x = 2

Set each linear factor equal to zero.

3

Apply the condition that

x

is a negative real number.

x = -7

The value

x = 2

is positive and must be discarded.

4

Evaluate the target expression

x + 3

.

-7 + 3 = -4

Substitute

x = -7

into

x + 3

.

Key Concept

Factoring Quadratic Equations and Evaluating Expressions

Question 79Question →

For all non-zero real numbers

a

and

b

such that

a \neq b

and

a \neq -b

, which of the following expressions is equivalent to

\frac{a^{-2} - b^{-2}}{a^{-1} - b^{-1}} \div \frac{a^2 + ab + b^2}{a^3 - b^3}?

\frac{a^2 - b^2}{ab}

\frac{b^2 - a^2}{ab}

\frac{a - b}{a + b}

\frac{a + b}{ab(a - b)}

\frac{(a - b)^2}{ab}

Show answer & explanation

Answer: $\frac{a^2 - b^2}{ab}$

Answer

\frac{a^2 - b^2}{ab}

Simplifying

\frac{a^{-2} - b^{-2}}{a^{-1} - b^{-1}}

yields

\frac{a + b}{ab}

after canceling

(b - a)

from both numerator and denominator. Simplifying

\frac{a^2 + ab + b^2}{a^3 - b^3}

yields

\frac{1}{a - b}

using the difference of cubes identity. Dividing

\frac{a + b}{ab}

by

\frac{1}{a - b}

gives

\frac{a + b}{ab} \cdot (a - b) = \frac{a^2 - b^2}{ab}

.

Step-by-Step Solution

1

Simplify the first rational expression

\frac{a^{-2} - b^{-2}}{a^{-1} - b^{-1}}

\frac{\frac{1}{a^2} - \frac{1}{b^2}}{\frac{1}{a} - \frac{1}{b}} = \frac{\frac{b^2 - a^2}{a^2 b^2}}{\frac{b - a}{ab}} = \frac{(b - a)(b + a)}{a^2 b^2} \cdot \frac{ab}{b - a} = \frac{a + b}{ab}

Convert negative exponents to fractions, find common denominators, and cancel the common non-zero factor

(b - a)

.

2

Simplify the second rational expression

\frac{a^2 + ab + b^2}{a^3 - b^3}

\frac{a^2 + ab + b^2}{(a - b)(a^2 + ab + b^2)} = \frac{1}{a - b}

Factor the denominator using the difference of cubes identity

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

.

3

Divide the simplified first expression by the simplified second expression

\frac{a + b}{ab} \div \frac{1}{a - b} = \frac{a + b}{ab} \cdot (a - b) = \frac{(a + b)(a - b)}{ab} = \frac{a^2 - b^2}{ab}

Multiply by the reciprocal of the second expression and apply the difference of squares identity.

Key Concept

Simplifying complex fractions and factoring using difference of squares and difference of cubes identities.

Question 80Question →

An investor divides a total of $\$15,000$ between two accounts, Account P and Account Q. Account P earns an annual simple interest rate of $6\%$ , while Account Q earns an annual simple interest rate of $8\%$ . If the total interest earned from both accounts combined after one year is $\$1,020$ , how much more money was invested in Account P than in Account Q?

\$1,500

\$3,000

\$4,500

\$6,000

\$9,000

Show answer & explanation

Answer: $\$3,000$

Answer

\$3,000

Setting up the system of linear equations

p + q = 15,000

and

0.06p + 0.08q = 1,020

yields

p = 9,000

and

q = 6,000

. Subtracting the amount in Account Q from Account P gives

9,000 - 6,000 = 3,000

. Thus,

\$3,000

is the correct answer.

Step-by-Step Solution

1

Define variables and set up the system of linear equations

Let

p

be the amount invested in Account P and

q

be the amount invested in Account Q. The equations are

p + q = 15,000

and

0.06p + 0.08q = 1,020

.

The first equation represents the total investment amount, and the second equation represents the total annual interest earned.

2

Simplify the interest equation and solve for one variable using elimination

Multiply

0.06p + 0.08q = 1,020

by 100 to obtain

6p + 8q = 102,000

, which simplifies to

3p + 4q = 51,000

. Multiplying the first equation by 3 yields

3p + 3q = 45,000

. Subtracting this from

3p + 4q = 51,000

gives

q = 6,000

.

Eliminating

p

allows for direct calculation of the amount invested in Account Q.

3

Calculate the amount invested in Account P

p = 15,000 - 6,000 = 9,000

.

Substitute the value of

q

back into the total investment equation.

4

Determine the requested difference

p - q = 9,000 - 6,000 = 3,000

.

The question specifically asks how much more money was invested in Account P than in Account Q.

Key Concept

Solving Systems of 2x2 Linear Equations for Word Problems

Algebra

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