Algebra Practice Questions — GRE General Test — Page 17

Question 321Question →

A startup hired two freelance software developers, Developer A and Developer B, for a combined total of $45$ hours on a single project. Developer A charges $\$65$ per hour, and Developer B charges $\$80$ per hour. If the total amount paid to both developers was $\$3{}225$ , how many hours did Developer A work on the project?

Show answer & explanation

Answer: 25

Answer

Developer A worked on the project for 25 hours.

Let

x

be the hours Developer A worked and

y

be the hours Developer B worked. From

x + y = 45

, we get

y = 45 - x

. Substituting into

65x + 80y = 3225

yields

65x + 80(45 - x) = 3225

. Expanding gives

65x + 3600 - 80x = 3225

, which simplifies to

-15x = -375

, so

x = 25

.

Step-by-Step Solution

1

Set up a system of linear equations representing the total hours and total cost.

x + y = 45

and

65x + 80y = 3225

, where

x

is Developer A's hours and

y

is Developer B's hours.

Translating the verbal conditions into mathematical equations.

2

Express

y

in terms of

x

from the hours equation.

y = 45 - x

Prepares the linear system for substitution.

3

Substitute

y = 45 - x

into the total cost equation and solve for

x

.

65x + 80(45 - x) = 3225 \implies -15x = -375 \implies x = 25

Solves for the requested variable

x

directly.

Key Concept

Solving word problems using 2x2 systems of linear equations via substitution or elimination.

Question 322Question →

For all real numbers $x$ and $y$ such that $|x| \neq |y|$ and $x^2 + y^2 \neq 0$ , consider the algebraic expression:

P(x, y) = \left(\frac{x^3 + y^3 - x^2 y - xy^2}{x^4 - y^4}\right) \cdot \left(\frac{x^3 + y^3}{x^2 - xy + y^2}\right)

Which of the following expressions are equivalent to $P(x, y)$ for all valid values of $x$ and $y$ ? Indicate all such expressions.

Select all that apply

\frac{x^4 - y^4}{(x^2 + y^2)^2}

1 - \frac{2y^2}{x^2 + y^2}

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

\frac{x - y}{x + y}

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

Show answer & explanation

Answer: $\frac{x^4 - y^4}{(x^2 + y^2)^2}$ ; $1 - \frac{2y^2}{x^2 + y^2}$

Answer

The expressions equivalent to

P(x, y)

are

\frac{x^4 - y^4}{(x^2 + y^2)^2}

and

1 - \frac{2y^2}{x^2 + y^2}

.

The given expression simplifies to

\frac{x^2 - y^2}{x^2 + y^2}

. The option

\frac{x^4 - y^4}{(x^2 + y^2)^2}

simplifies directly to

\frac{x^2 - y^2}{x^2 + y^2}

after factoring the numerator. The option

1 - \frac{2y^2}{x^2 + y^2}

simplifies to

\frac{x^2 + y^2 - 2y^2}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2}

when combined over a common denominator.

Step-by-Step Solution

1

Factor the numerator of the first rational term by grouping terms.

x^3 + y^3 - x^2 y - xy^2 = x^2(x - y) - y^2(x - y) = (x^2 - y^2)(x - y) = (x - y)^2 (x + y)

Grouping allows rewriting four polynomial terms into product of linear/quadratic factors.

2

Factor the denominator of the first rational term as a difference of squares.

x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) = (x - y)(x + y)(x^2 + y^2)

Decomposing

x^4 - y^4

into simpler factors reveals common factors with the numerator.

3

Simplify the first rational term by canceling common factors

(x - y)(x + y)

.

\frac{(x - y)^2(x + y)}{(x - y)(x + y)(x^2 + y^2)} = \frac{x - y}{x^2 + y^2}

Canceling non-zero common factors simplifies the fraction.

4

Factor the numerator of the second rational term using the sum of cubes formula.

x^3 + y^3 = (x + y)(x^2 - xy + y^2)

Applying the sum of cubes identity exposes the irreducible quadratic factor present in the denominator.

5

Simplify the second rational term and multiply the result by the simplified first term.

P(x, y) = \left(\frac{x - y}{x^2 + y^2}\right) \cdot (x + y) = \frac{(x - y)(x + y)}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2}

Multiplying the simplified forms yields the simplest explicit representation of

P(x, y)

.

6

Verify equivalence of the options against

\frac{x^2 - y^2}{x^2 + y^2}

.

The option

\frac{x^4 - y^4}{(x^2 + y^2)^2} = \frac{(x^2 - y^2)(x^2 + y^2)}{(x^2 + y^2)^2} = \frac{x^2 - y^2}{x^2 + y^2}

, and the option

1 - \frac{2y^2}{x^2 + y^2} = \frac{x^2 + y^2 - 2y^2}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2}

. Both match

P(x, y)

.

Transforming algebraic expressions under common denominators or factoring confirms equivalence.

Key Concept

Simplifying complex algebraic expressions using factoring by grouping, difference of squares, sum of cubes, and common denominator manipulation.

Question 323Question →

A bakery packages two types of gift baskets containing gourmet croissants and blueberry muffins. Basket X contains $4$ croissants and $3$ muffins, with a total production cost of $\$19.00$ . Basket Y contains $2$ croissants and $5$ muffins, with a total production cost of $\$16.50$ . Assuming the cost per croissant and the cost per muffin are constant across all baskets, what is the production cost, in dollars, of a single croissant?

Show answer & explanation

Answer: 3.25

Answer

The production cost of a single croissant is 3.25 dollars.

Let

c

represent the cost of a croissant and

m

represent the cost of a muffin. The given situation translates to the system of equations

4c + 3m = 19.00

and

2c + 5m = 16.50

. Multiplying the second equation by

2

gives

4c + 10m = 33.00

. Subtracting

4c + 3m = 19.00

from

4c + 10m = 33.00

results in

7m = 14.00

, which gives

m = 2.00

. Substituting

m = 2.00

into

2c + 5(2.00) = 16.50

yields

2c + 10 = 16.50

, so

2c = 6.50

and

c = 3.25

. Thus, a single croissant costs $3.25 dollars.

Step-by-Step Solution

1

Set up a system of linear equations

4c + 3m = 19.00

and

2c + 5m = 16.50

Translate the contents and costs of Basket X and Basket Y into algebraic equations where

c

is the price of a croissant and

m

is the price of a muffin.

2

Eliminate variable c

4c + 10m = 33.00

, then subtracting

4c + 3m = 19.00

yields

7m = 14.00

, so

m = 2.00

Multiplying the second equation by 2 aligns the coefficients of

c

, allowing elimination by subtraction.

3

Solve for variable c

2c + 5(2.00) = 16.50 \implies 2c = 6.50 \implies c = 3.25

Substitute the value found for

m

back into one of the original linear equations to calculate the cost of a croissant.

Key Concept

Solving Systems of Linear Equations via Elimination

Alternative Method

Express

c

in terms of

m

using the second equation:

c = 8.25 - 2.5m

. Substitute this expression into the first equation:

4(8.25 - 2.5m) + 3m = 19.00 \implies 33 - 10m + 3m = 19.00 \implies -7m = -14.00 \implies m = 2.00

. Finally, calculate

c = 8.25 - 2.5(2.00) = 3.25

.

Estimated Time:1m 30s

Question 324Question →

For all real numbers

x

and

y

such that

|x| \neq |y|

, which of the following expressions is equivalent to

\frac{x^3 + x^2y + x^2 - y^3 - xy^2 - y^2}{x^2 - y^2}?

x + y + 1

x - y + 1

x + y - 1

(x + y)^2 + 1

x^2 + xy + y^2 + 1

Show answer & explanation

Answer: $x + y + 1$

Answer

x + y + 1

Factoring the numerator by grouping yields

(x - y)(x + y)(x + y + 1)

. Factoring the denominator gives

(x - y)(x + y)

. Canceling the non-zero common factor

(x - y)(x + y)

leaves the simplified expression

x + y + 1

.

Step-by-Step Solution

1

Group terms in the numerator to identify common factor pairs.

N = (x^3 - y^3) + (x^2y - xy^2) + (x^2 - y^2)

Grouping cubic terms, quadratic cross-terms, and difference of squares separately allows factoring out fundamental algebraic patterns.

2

Apply standard algebraic formulas to each grouped term.

N = (x - y)(x^2 + xy + y^2) + xy(x - y) + (x - y)(x + y)

Using difference of cubes

x^3 - y^3 = (x-y)(x^2+xy+y^2)

and difference of squares

x^2 - y^2 = (x-y)(x+y)

reveals a common

(x-y)

factor across all terms.

3

Factor out

(x - y)

from the numerator and simplify the remaining polynomial.

N = (x - y)\left[(x^2 + xy + y^2) + xy + (x + y)\right] = (x - y)\left[x^2 + 2xy + y^2 + x + y\right]

Combining like terms inside the bracket simplifies the expression.

4

Recognize the perfect square trinomial inside the expression.

N = (x - y)\left[(x + y)^2 + (x + y)\right] = (x - y)(x + y)(x + y + 1)

Rewriting

x^2 + 2xy + y^2

as

(x + y)^2

allows factoring out

(x + y)

.

5

Divide the factored numerator by the denominator.

\frac{(x - y)(x + y)(x + y + 1)}{(x - y)(x + y)} = x + y + 1

Since

|x| \neq |y|

, both

(x - y)

and

(x + y)

are non-zero and can be canceled.

Key Concept

Factoring multivariable polynomials using grouping, difference of cubes, and difference of squares formulas.

Question 325Question →

Consider the system of linear equations in two variables

x

and

y

shown below, where

k

is a constant:

\begin{aligned} 2x + ky &= 10 \\ 3x - 6y &= 15 \end{aligned}

Which of the following statements must be true? Select all such statements.

Select all that apply

If

k = -4

, the system has infinitely many solutions.

If

k = 0

, the system has a unique solution

(x, y) = (5, 0)

.

If

k = 4

, the system has no solution.

If

k = 2

, the solution to the system satisfies

x + y = 7

.

There is no real value of

k

for which the system has a solution with

y > 0

.

Show answer & explanation

Answer: If $k = -4$ , the system has infinitely many solutions.; If $k = 0$ , the system has a unique solution $(x, y) = (5, 0)$ .

Answer

The correct statements are the ones stating that if

k = -4

, the system has infinitely many solutions, and if

k = 0

, the system has a unique solution

(x, y) = (5, 0)

.

Analyzing the simplified relation

(k + 4)y = 0

demonstrates that setting

k = -4

makes the equation identity

0y = 0

, yielding infinitely many solutions. For any other value of

k

, including

k = 0

,

y

must equal

0

, which gives

x = 5

, establishing a unique solution at

(5, 0)

.

Step-by-Step Solution

1

Simplify the second equation to express

x

in terms of

y

.

3x - 6y = 15 \implies x - 2y = 5 \implies x = 2y + 5

.

Expressing

x

explicitly allows direct substitution into the first linear equation.

2

Substitute

x = 2y + 5

into the first equation

2x + ky = 10

.

2(2y + 5) + ky = 10 \implies 4y + 10 + ky = 10 \implies (k + 4)y = 0

.

This reduces the 2x2 system to a single linear equation in

y

parameterized by

k

.

3

Analyze the conditions for

y

based on the parameter

k

.

If

k = -4

, the equation becomes

0y = 0

, which is true for all real

y

(infinitely many solutions). If

k \neq -4

, then

y = 0

and

x = 5

(a unique solution).

Determines system consistency and solution multiplicity across all values of

k

.

Key Concept

Parametric Analysis of 2x2 Linear Systems

Question 326Question →

For pairwise distinct real numbers $x$ , $y$ , and $z$ , consider the algebraic expression:

E(x, y, z) = \frac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y - z)^3 + (z - x)^3}

If $x = 5$ , $y = 3$ , and $z = 1$ , what is the numerical value of $E(5, 3, 1)$ ?

Show answer & explanation

Answer: 192

Answer

192

Using the identity that

a^3 + b^3 + c^3 = 3abc

whenever

a + b + c = 0

, both the numerator and denominator can be factored directly. Factoring the difference of squares in the numerator yields

3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)

. Dividing this by the factored denominator

3(x - y)(y - z)(z - x)

simplifies the expression to

(x + y)(y + z)(z + x)

. Substituting

x = 5

,

y = 3

, and

z = 1

yields

(8)(4)(6) = 192

.

Step-by-Step Solution

1

Use the conditional cubic identity

a + b + c = 0 \implies a^3 + b^3 + c^3 = 3abc

on the denominator.

Denominator becomes

3(x - y)(y - z)(z - x)

.

The sum of the three terms

(x - y) + (y - z) + (z - x)

equals 0.

2

Apply the same identity to the numerator.

Numerator becomes

3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)

.

The sum of the squared difference terms

(x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2)

also equals 0.

3

Factor each difference of squares in the numerator.

Numerator becomes

3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)

.

Using the difference of squares identity

u^2 - v^2 = (u - v)(u + v)

on each term.

4

Simplify the fraction by dividing the common factors in the numerator and denominator.

E(x, y, z) = (x + y)(y + z)(z + x)

.

The factors

3

,

(x - y)

,

(y - z)

, and

(z - x)

cancel out completely.

5

Evaluate the simplified product for

x = 5

,

y = 3

, and

z = 1

.

(5 + 3)(3 + 1)(1 + 5) = 8 \times 4 \times 6 = 192

.

Direct evaluation after algebraic simplification.

Key Concept

Simplifying rational expressions involving sum of cubes identity

a^3 + b^3 + c^3 = 3abc

when

a + b + c = 0

and difference of squares factoring.

Question 327Question →

A quadratic function is defined by $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are real constants with $a > 0$ . If the vertex of the parabola $y = f(x)$ lies in the third quadrant of the $xy$ -plane, which of the following statements must be true? Select all that apply.

Select all that apply

b > 0

The equation

f(x) = 0

has two distinct real solutions.

c

must be negative.

The sum of the two solutions of

f(x) = 0

is positive.

The discriminant

b^2 - 4ac

is less than zero.

Show answer & explanation

Answer: $b > 0$ ; The equation $f(x) = 0$ has two distinct real solutions.

Answer

The statements '

b > 0

' and 'The equation

f(x) = 0

has two distinct real solutions' must be true.

The vertex of the parabola

y = ax^2 + bx + c

is

(h, k) = \left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)

. Because the vertex is in the third quadrant,

h < 0

and

k < 0

. With

a > 0

, the inequality

-\frac{b}{2a} < 0

directly requires

b > 0

. Furthermore, because

a > 0

(parabola opens upward) and the minimum value

k < 0

lies below the x-axis, the graph must cross the x-axis at two distinct points, establishing that

f(x) = 0

has two distinct real solutions.

Step-by-Step Solution

1

Analyze the coordinates of the vertex

(h, k)

relative to the third quadrant.

The third quadrant requires

h < 0

and

k < 0

.

Points in Quadrant III have negative x-coordinates and negative y-coordinates.

2

Evaluate the sign of

b

using the x-coordinate formula

h = -\frac{b}{2a}

.

b > 0

.

Since

h < 0

and

a > 0

,

-\frac{b}{2a} < 0 \implies \frac{b}{2a} > 0 \implies b > 0

.

3

Determine the number of real solutions using the vertex y-coordinate

k

and orientation

a > 0

.

The discriminant

b^2 - 4ac > 0

, giving two distinct real solutions.

For an upward-opening parabola (

a > 0

), having a vertex below the x-axis (

k < 0

) guarantees the curve crosses the x-axis twice.

4

Test counterexamples for

c

and evaluate the root sum

-\frac{b}{a}

.

c

is not strictly constrained in sign, and the root sum is negative.

The function

f(x) = (x+2)^2 - 1 = x^2 + 4x + 3

has vertex

(-2, -1) \in \text{QIII}

with

c = 3 > 0

. Also, since

a > 0

and

b > 0

, root sum

-\frac{b}{a} < 0

.

Key Concept

Parabola Vertex and Quadratic Discriminant Analysis

Question 328Question →

For all real numbers

x

and

y

such that

x \neq y

,

x \neq -y

, and

x^2 + y^2 \neq 0

, which of the following expressions is equivalent to

\frac{x^4 - y^4}{x^3 - x^2 y + x y^2 - y^3} - \frac{2 x y}{x + y}

\frac{x^2 + y^2}{x + y}

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

\frac{x + y - 2xy}{1 + x + y}

\frac{x^2 - y^2}{x + y}

\frac{x + y}{x - y}

Show answer & explanation

Answer: $\frac{x^2 + y^2}{x + y}$

Answer

\frac{x^2 + y^2}{x + y}

Factoring

x^4 - y^4

as

(x - y)(x + y)(x^2 + y^2)

and

x^3 - x^2 y + x y^2 - y^3

as

(x - y)(x^2 + y^2)

simplifies the first term to

x + y

. Combining

x + y

with

-\frac{2xy}{x+y}

over the common denominator

(x + y)

gives

\frac{(x+y)^2 - 2xy}{x+y} = \frac{x^2 + y^2}{x+y}

.

Step-by-Step Solution

1

Factor the numerator and denominator of the first rational expression

Numerator:

x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) = (x - y)(x + y)(x^2 + y^2)

. Denominator:

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

.

Factoring allows for cancellation of common factors in rational expressions.

2

Simplify the first rational expression by canceling common terms

\frac{(x - y)(x + y)(x^2 + y^2)}{(x - y)(x^2 + y^2)} = x + y

.

Since

x \neq y

and

x^2 + y^2 \neq 0

, the common terms

(x - y)

and

(x^2 + y^2)

are non-zero and can be divided out.

3

Subtract the second expression using a common denominator

(x + y) - \frac{2xy}{x + y} = \frac{(x + y)^2}{x + y} - \frac{2xy}{x + y} = \frac{(x + y)^2 - 2xy}{x + y}

.

Combining terms under the common denominator

(x + y)

enables algebraic reduction.

4

Expand the squared binomial in the numerator and simplify like terms

\frac{x^2 + 2xy + y^2 - 2xy}{x + y} = \frac{x^2 + y^2}{x + y}.

Expanding

(x + y)^2 = x^2 + 2xy + y^2

allows the

+2xy

and

-2xy

terms to cancel out.

Key Concept

Simplifying rational expressions using polynomial factoring (difference of squares and grouping) and common denominators.

Estimated Time:2m 0s

Question 329Question →

A research laboratory operates two types of automated centrifuges, Model X and Model Y. A single operating cycle of Model X processes $40$ biological samples and consumes $10$ kilowatt-hours (kWh) of electricity. A single operating cycle of Model Y processes $25$ biological samples and consumes $15$ kWh of electricity. On a given day, the laboratory processed a total of $775$ biological samples and consumed $325$ kWh of electricity using only these two models. What is the total number of operating cycles completed by Model X and Model Y combined on that day?

10

15

25

30

35

Show answer & explanation

Answer: 25

Answer

The total number of combined operating cycles completed by Model X and Model Y is 25.

The correct answer is 25. Setting up the linear system

40x + 25y = 775

and

10x + 15y = 325

and solving via elimination gives

x = 10

cycles for Model X and

y = 15

cycles for Model Y. Adding these together gives

10 + 15 = 25

total cycles.

Step-by-Step Solution

1

Define variables and set up the system of linear equations based on total samples and total electricity consumption.

Let

x

be the number of cycles for Model X and

y

be the number of cycles for Model Y.
System of equations:
1)

40x + 25y = 775

(Sample constraint)
2)

10x + 15y = 325

(Energy constraint)

Modeling the word problem as two linear equations in two variables allows for algebraic elimination.

2

Simplify both equations by dividing by their greatest common factors.

Divide Equation 1 by 5:

8x + 5y = 155

Divide Equation 2 by 5:

2x + 3y = 65

Simplifying coefficients reduces computational complexity and minimizes arithmetic errors.

3

Eliminate variable

x

by multiplying the simplified second equation by 4 and subtracting the simplified first equation.

4 \times (2x + 3y = 65) \Rightarrow 8x + 12y = 260

Subtract

(8x + 5y = 155)

:

(8x - 8x) + (12y - 5y) = 260 - 155

7y = 105 \Rightarrow y = 15

Equalizing the coefficients of

x

enables solving directly for

y

.

4

Substitute

y = 15

back into

2x + 3y = 65

to find

x

.

2x + 3(15) = 65 \Rightarrow 2x + 45 = 65 \Rightarrow 2x = 20 \Rightarrow x = 10

Evaluating

x

completes the solution for individual cycle counts.

5

Calculate the requested combined total

x + y

.

x + y = 10 + 15 = 25

The question specifically requests the sum of operating cycles of both models combined.

Key Concept

Systems of Linear Equations in Two Variables

Estimated Time:1m 45s

Question 330Question →

For all non-zero real numbers $x$ and $y$ such that $|x| \neq |y|$ and $x^2 + 2xy + 2y^2 \neq 0$ , consider the algebraic expression:

Q(x, y) = \frac{x^4 + 4y^4}{x^2 + 2xy + 2y^2} + \frac{2x^3 y - 2xy^3}{x^2 - y^2}

Which of the following expressions are equivalent to $Q(x, y)$ ? Indicate all such expressions.

Select all that apply

(x - y)^2 + 2xy + y^2

\frac{x^3 + 2xy^2}{x}

\frac{(x^2 + 2y^2)^2 - 4x^2 y^2}{x^2 - 2xy + 2y^2}

\frac{x^4 - 4y^4}{x^2 + 2y^2}

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

Show answer & explanation

Answer: $(x - y)^2 + 2xy + y^2$ ; \frac{x^3 + 2xy^2}{x}

Answer

The expressions equivalent to

Q(x, y)

are

(x - y)^2 + 2xy + y^2

and

\frac{x^3 + 2xy^2}{x}

.

First, simplify

Q(x, y)

by factoring each term. Using Sophie Germain's identity on the numerator of the first term gives

x^4 + 4y^4 = (x^2 + 2y^2)^2 - (2xy)^2 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)

. Dividing this by

(x^2 + 2xy + 2y^2)

results in

x^2 - 2xy + 2y^2

. For the second term, factoring out

2xy

gives

\frac{2xy(x^2 - y^2)}{x^2 - y^2} = 2xy

. Adding the two simplified terms yields

Q(x, y) = (x^2 - 2xy + 2y^2) + 2xy = x^2 + 2y^2

.

Evaluating the options for equivalence:
- The expression

(x - y)^2 + 2xy + y^2

expands to

x^2 - 2xy + y^2 + 2xy + y^2 = x^2 + 2y^2

.
- The expression

\frac{x^3 + 2xy^2}{x}

factors as

\frac{x(x^2 + 2y^2)}{x} = x^2 + 2y^2

.
Therefore, both of these expressions are equivalent to

Q(x, y)

.

Step-by-Step Solution

1

Simplify the first algebraic fraction using Sophie Germain's identity.

\frac{x^4 + 4y^4}{x^2 + 2xy + 2y^2} = \frac{(x^2 + 2y^2)^2 - (2xy)^2}{x^2 + 2xy + 2y^2} = \frac{(x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)}{x^2 + 2xy + 2y^2} = x^2 - 2xy + 2y^2

Completing the square on

x^4 + 4y^4

allows it to be factored into the product of two quadratic expressions.

2

Simplify the second algebraic fraction by factoring out common factors.

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

The term

(x^2 - y^2)

cancels out since

x \neq \pm y

.

3

Combine the simplified terms to find the closed-form expression for

Q(x, y)

.

Q(x, y) = (x^2 - 2xy + 2y^2) + 2xy = x^2 + 2y^2

The

-2xy

and

+2xy

terms sum to zero.

4

Test each option for equivalence to

x^2 + 2y^2

.

The option

(x - y)^2 + 2xy + y^2

expands to

x^2 - 2xy + y^2 + 2xy + y^2 = x^2 + 2y^2

. The option

\frac{x^3 + 2xy^2}{x}

simplifies directly to

x^2 + 2y^2

.

Both expressions reduce identically to

x^2 + 2y^2

for all non-zero

x

and

y

.

Key Concept

Simplifying complex algebraic expressions using polynomial factoring identities (Sophie Germain identity and difference of squares) and algebraic reduction.

Estimated Time:2m 0s

Question 331Question →

A community theater sold a total of $250$ tickets for a performance, consisting of adult tickets for $\$15$ each and student tickets for $\$10$ each. The total revenue collected from ticket sales was $\$3,100$ . Let $a$ represent the number of adult tickets sold and $s$ represent the number of student tickets sold. Which of the following statements must be true? Select all such statements.

Select all that apply

The number of student tickets sold was

130

.

The number of adult tickets sold was

130

.

The total revenue generated from adult ticket sales was

\$1,800

.

The ratio of adult tickets sold to student tickets sold was

13

to

12

.

More adult tickets were sold than student tickets.

Show answer & explanation

Answer: The number of student tickets sold was $130$ .; The total revenue generated from adult ticket sales was $\$1,800$ .

Answer

The correct statements are that the number of student tickets sold was 130, and the total revenue generated from adult ticket sales was $1,800.

The system of equations

a + s = 250

and

15a + 10s = 3100

uniquely yields

a = 120

adult tickets and

s = 130

student tickets. Therefore, the statement that 130 student tickets were sold is true. Furthermore, multiplying 120 adult tickets by their price of

15 per ticket gives a total adult ticket revenue of

1,800, making that statement true as well.

Step-by-Step Solution

1

Set up a system of two linear equations in terms of

a

and

s

.

Equation 1 (total tickets):

a + s = 250

; Equation 2 (total revenue):

15a + 10s = 3100

.

The sum of the ticket quantities equals total tickets, and the sum of monetary contributions equals total revenue.

2

Express

s

in terms of

a

using Equation 1 and substitute into Equation 2.

s = 250 - a \implies 15a + 10(250 - a) = 3100 \implies 15a + 2500 - 10a = 3100

.

Using substitution eliminates variable

s

to solve for

a

.

3

Solve for

a

and then find

s

.

5a = 600 \implies a = 120

;

s = 250 - 120 = 130

.

Dividing

600

by

5

gives

a = 120

, and subtracting

120

from

250

gives

s = 130

.

4

Evaluate the statements against the values

a = 120

and

s = 130

.

s = 130

is true. Revenue from adult tickets is

120 \times 15 = \$1,800

, which is also true.

Comparing calculated values directly confirms which options state accurate quantitative properties.

Key Concept

Solving 2x2 Systems of Linear Equations using Substitution or Elimination in Word Problems

Question 332Question →

Consider the system of linear equations in two variables

x

and

y

shown below:

\begin{aligned} 3x + 2y &= 16 \\ x - 4y &= -4 \end{aligned}

If

(x, y)

is the unique solution to the system, what is the value of

\frac{y}{x}

?

\frac{1}{2}

2

4

\frac{2}{5}

6

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

The value of

\frac{y}{x}

is

\frac{1}{2}

.

Solving the system by substitution gives

x = 4

and

y = 2

. Dividing

y

by

x

gives

\frac{2}{4} = \frac{1}{2}

, which is the target ratio.

Step-by-Step Solution

1

Isolate

x

in the second equation.

x = 4y - 4

Expressing

x

in terms of

y

allows for substitution into the first equation.

2

Substitute

x = 4y - 4

into the first equation.

3(4y - 4) + 2y = 16 \implies 12y - 12 + 2y = 16 \implies 14y = 28 \implies y = 2

Solving the single-variable equation determines the value of

y

.

3

Substitute

y = 2

back into the isolated expression for

x

.

x = 4(2) - 4 = 4

Determines the corresponding value of

x

.

4

Compute the required ratio

\frac{y}{x}

.

\frac{y}{x} = \frac{2}{4} = \frac{1}{2}

Evaluates the final expression requested by the prompt.

Key Concept

Solving systems of linear equations using substitution or elimination to evaluate a combined expression.

Question 333Question →

A commercial print shop operates two types of high-speed printers, Printer M and Printer N. Operating simultaneously for $5$ hours, $2$ units of Printer M and $3$ units of Printer N print a total of $5,500$ pages. Operating simultaneously for $4$ hours, $5$ units of Printer M and $2$ units of Printer N print a total of $6,600$ pages. What is the hourly page output of a single Printer M?

Show answer & explanation

Answer: 250

Answer

The hourly page output of a single Printer M is 250 pages per hour.

Dividing each total page output by the corresponding number of hours produces the simplified linear system:

2m + 3n = 1100

and

5m + 2n = 1650

, where

m

and

n

are the hourly rates of Printer M and Printer N. Multiplying the first equation by

2

gives

4m + 6n = 2200

, and multiplying the second equation by

3

gives

15m + 6n = 4950

. Subtracting the two equations eliminates

n

, giving

11m = 2750

, which simplifies to

m = 250

pages per hour.

Step-by-Step Solution

1

Define variables and write initial algebraic equations based on time and rate.

Let

m

represent the hourly page output of Printer M and

n

represent the hourly page output of Printer N. The total outputs give

5(2m + 3n) = 5500

and

4(5m + 2n) = 6600

.

Total page output equals total operating time multiplied by the combined hourly output rate.

2

Simplify the system by dividing each equation by its respective number of hours.

First equation:

2m + 3n = 1100

. Second equation:

5m + 2n = 1650

.

Simplifying yields a standard system of linear equations representing the combined hourly rate.

3

Eliminate variable

n

to solve for

m

.

Multiply

2m + 3n = 1100

by

2

to get

4m + 6n = 2200

. Multiply

5m + 2n = 1650

by

3

to get

15m + 6n = 4950

. Subtracting the first modified equation from the second yields

11m = 2750

, so

m = 250

.

Equating the coefficients of

n

allows direct elimination of

n

when subtracting the equations.

Key Concept

Solving a 2x2 system of linear equations using the method of elimination

Estimated Time:1m 30s

Question 334Question →

For all real numbers

x \neq -5

, the algebraic expression

\frac{x^3 + 125}{x^2 - 5x + 25} - \frac{x^2 - 25}{x + 5}

simplifies to a single constant value. What is the value of this constant?

Show answer & explanation

Answer: 10

Answer

The simplified value of the expression for all valid real numbers x is 10.

Factoring the numerators reveals that the first term reduces to

x + 5

and the second term reduces to

x - 5

. Subtracting

(x - 5)

from

(x + 5)

yields

(x + 5) - (x - 5) = 10

, which is constant for all valid values of

x

.

Step-by-Step Solution

1

Factor the numerator of the first rational term using the sum of cubes identity.

x^3 + 125 = (x + 5)(x^2 - 5x + 25)

The sum of cubes formula

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

applies with

a = x

and

b = 5

.

2

Cancel the non-zero factor

(x^2 - 5x + 25)

from the first fraction.

\frac{(x + 5)(x^2 - 5x + 25)}{x^2 - 5x + 25} = x + 5

The quadratic factor

x^2 - 5x + 25

has a negative discriminant (

(-5)^2 - 4(1)(25) = -75 < 0

), so it is never zero for any real number

x

.

3

Factor the numerator of the second rational term using the difference of squares identity.

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

The difference of squares formula

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

applies with

a = x

and

b = 5

.

4

Cancel the common factor

(x + 5)

from the second fraction.

\frac{(x - 5)(x + 5)}{x + 5} = x - 5

Given

x \neq -5

, the factor

x + 5

is non-zero and can be canceled.

5

Subtract the two simplified terms.

(x + 5) - (x - 5) = x + 5 - x + 5 = 10

Distribute the negative sign to both terms in

(x - 5)

and combine like terms.

Key Concept

Factoring sum of cubes and difference of squares to simplify rational expressions

Question 335Question →

For all real numbers $x \neq 2$ , which of the following expressions is equivalent to $\frac{x^4 - 16}{x^3 - 2x^2 + 4x - 8}$ ?

x + 2

x - 2

x^2 + 4

\frac{x + 2}{x - 2}

\pm (x + 2)

Show answer & explanation

Answer: $x + 2$

Answer

x + 2

Factoring the numerator

x^4 - 16

as a difference of squares yields

(x - 2)(x + 2)(x^2 + 4)

. Factoring the denominator

x^3 - 2x^2 + 4x - 8

by grouping yields

(x - 2)(x^2 + 4)

. Dividing the numerator by the denominator allows the shared factor

(x - 2)(x^2 + 4)

to cancel, leaving

x + 2

.

Step-by-Step Solution

1

Factor the numerator

x^4 - 16

x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)

Apply the difference of squares factorization identity twice.

2

Factor the denominator

x^3 - 2x^2 + 4x - 8

by grouping

x^3 - 2x^2 + 4x - 8 = x^2(x - 2) + 4(x - 2) = (x - 2)(x^2 + 4)

Group the terms in pairs and factor out the common binomial factor

(x - 2)

.

3

Simplify the rational expression by canceling common factors

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

Cancel the non-zero common factor

(x - 2)(x^2 + 4)

present in both numerator and denominator.

Key Concept

Simplifying rational expressions using difference of squares and factoring by grouping

Question 336Question →

For all real numbers

x

, the algebraic expression

\frac{x^6 - 27}{x^4 + 3x^2 + 9}

can be simplified to the polynomial form

ax^2 + bx + c

, where

a

,

b

, and

c

are real constants. What is the value of

a + b + c

?

Show answer & explanation

Answer: -2

Answer

The simplified expression is

x^2 - 3

, which corresponds to polynomial coefficients

a = 1

,

b = 0

, and

c = -3

. The sum

a + b + c

equals

-2

.

Factoring the numerator

x^6 - 27

as a difference of cubes

(x^2)^3 - 3^3

produces

(x^2 - 3)(x^4 + 3x^2 + 9)

. Canceling the non-zero factor

(x^4 + 3x^2 + 9)

from the numerator and denominator simplifies the expression to

x^2 - 3

. In standard form

ax^2 + bx + c

,

a = 1

,

b = 0

, and

c = -3

. Adding these coefficients gives

1 + 0 + (-3) = -2

.

Step-by-Step Solution

1

Factor the numerator using the difference of cubes formula

x^6 - 27 = (x^2)^3 - 3^3 = (x^2 - 3)(x^4 + 3x^2 + 9)

The expression

x^6 - 27

matches the pattern

u^3 - v^3

with

u = x^2

and

v = 3

.

2

Simplify the rational expression by canceling the common quadratic-biquadratic factor

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

Since

x^4 + 3x^2 + 9 > 0

for all real numbers

x

, the denominator is never zero, allowing direct cancellation of the common factor.

3

Match coefficients with

ax^2 + bx + c

and calculate

a + b + c

a = 1, b = 0, c = -3 \implies a + b + c = 1 + 0 + (-3) = -2

Comparing

x^2 - 3 = 1x^2 + 0x - 3

to

ax^2 + bx + c

determines the values of constants

a

,

b

, and

c

.

Key Concept

Difference of Cubes Factoring Identity

Estimated Time:1m 30s

Question 337Question →

For all real numbers

x

and

y

such that

x \neq -y

and

x \neq -2y

, which of the following expressions is equivalent to

\frac{x^3 + 2x^2y - xy^2 - 2y^3}{x^2 + 3xy + 2y^2}?

x - y

x + y

x^2 - y^2

x^2 + y^2

x - 2y

Show answer & explanation

Answer: $x - y$

Answer

x - y

Factoring the numerator by grouping gives

x^2(x + 2y) - y^2(x + 2y) = (x^2 - y^2)(x + 2y) = (x - y)(x + y)(x + 2y)

. Factoring the denominator yields

(x + y)(x + 2y)

. Dividing the numerator by the denominator cancels the common factors

(x + y)

and

(x + 2y)

, leaving

x - y

.

Step-by-Step Solution

1

Factor the numerator by grouping terms

x^3 + 2x^2y - xy^2 - 2y^3 = x^2(x + 2y) - y^2(x + 2y) = (x^2 - y^2)(x + 2y) = (x - y)(x + y)(x + 2y)

Grouping pairs of terms allows factoring out common binomial factors.

2

Factor the quadratic denominator

x^2 + 3xy + 2y^2 = (x + y)(x + 2y)

Finding two terms whose sum is

3y

and product is

2y^2

factors the quadratic in

x

.

3

Simplify the rational expression by canceling non-zero common factors

\frac{(x - y)(x + y)(x + 2y)}{(x + y)(x + 2y)} = x - y

Since

x \neq -y

and

x \neq -2y

, the factors

(x + y)

and

(x + 2y)

are non-zero and can be canceled.

Key Concept

Polynomial factoring by grouping, difference of squares, quadratic trinomial factoring, and simplifying rational algebraic expressions.

Estimated Time:1m 30s

Question 338Question →

The length of a rectangular plot of land is $3$ meters less than twice its width. If the area of the plot is $90$ square meters, what is the perimeter of the plot, in meters?

Show answer & explanation

Answer: 39

Answer

The perimeter of the plot of land is 39 meters.

Setting the length to

2w - 3

gives an area equation of

w(2w - 3) = 90

, which expands and rearranges to

2w^2 - 3w - 90 = 0

. Factoring this quadratic equation yields

(2w - 15)(w + 6) = 0

. Since width must be positive,

w = 7.5

meters, which means the length is

12

meters. The perimeter is

2(12 + 7.5) = 39

meters.

Step-by-Step Solution

1

Express the length in terms of width and set up the area equation.

Let

w

be the width of the rectangle. Length

l = 2w - 3

. Area equation:

w(2w - 3) = 90

.

The area of a rectangle is equal to length multiplied by width.

2

Rearrange into standard quadratic form

ax^2 + bx + c = 0

.

2w^2 - 3w - 90 = 0

Distributing

w

and subtracting

90

from both sides puts the equation in standard quadratic form.

3

Factor the quadratic equation.

(2w - 15)(w + 6) = 0

Finding two numbers with a product of

2 \times (-90) = -180

and a sum of

-3

gives

-15

and

12

.

4

Determine the valid physical dimensions.

w = 7.5

meters and

l = 12

meters.

The root

w = -6

is discarded because physical length cannot be negative. Thus

w = \frac{15}{2} = 7.5

meters.

5

Calculate the perimeter.

Perimeter

= 2(l + w) = 2(12 + 7.5) = 39

meters.

The perimeter of a rectangle is given by

2 \times (\text{length} + \text{width})

.

Key Concept

Solving quadratic word problems via factoring

Estimated Time:1m 30s

Question 339Question →

For all real numbers $x$ such that $x \neq -2$ , $x \neq 2$ , and $x \neq 4$ , which of the following expressions are equivalent to $\frac{x^4 - 16x^2}{x^3 - 4x^2 - 4x + 16}$ ? Select all such expressions.

Select all that apply

\frac{x^3 + 4x^2}{x^2 - 4}

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

\frac{x^2(x - 4)}{x^2 - 4}

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

\frac{x^4 - 4x^2}{x^3 - 4}

Show answer & explanation

Answer: $\frac{x^3 + 4x^2}{x^2 - 4}$ ; $\frac{x^2(x + 4)}{(x - 2)(x + 2)}$

Answer

The equivalent expressions are

\frac{x^3 + 4x^2}{x^2 - 4}

and

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

.

The given expression factors as

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

. Canceling

(x - 4)

leaves

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

. Expanding the numerator gives

\frac{x^3 + 4x^2}{x^2 - 4}

, and factoring the denominator further gives

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

. Both represent valid equivalent forms of the expression.

Step-by-Step Solution

1

Factor the numerator of the given algebraic expression.

x^4 - 16x^2 = x^2(x^2 - 16) = x^2(x - 4)(x + 4)

Factor out the greatest common factor

x^2

, then apply the difference of squares identity

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

.

2

Factor the denominator by grouping terms.

x^3 - 4x^2 - 4x + 16 = x^2(x - 4) - 4(x - 4) = (x^2 - 4)(x - 4) = (x - 2)(x + 2)(x - 4)

Group the first two terms and last two terms, factor out common binomials, and expand the remaining difference of squares.

3

Simplify the full rational expression by canceling common factors.

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

Since

x \neq 4

, cancel the common factor

(x - 4)

from both numerator and denominator.

4

Compare the simplified form to the given choices to identify all equivalent expressions.

Both

\frac{x^3 + 4x^2}{x^2 - 4}

and

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

match the simplified algebraic forms.

Expanding the numerator or factoring the denominator yields these two equivalent representations.

Key Concept

Simplifying rational algebraic expressions by polynomial factoring and term grouping.

Question 340Question →

If $r_1$ and $r_2$ are the two distinct real solutions to the quadratic equation $x^2 - 4x - 21 = 0$ , where $r_1 > r_2$ , what is the value of $r_1 + 2r_2$ ?

1

13

-11

-5

17

Show answer & explanation

Answer: $1$

Answer

The correct value of

r_1 + 2r_2

is

1

.

Factoring the quadratic polynomial

x^2 - 4x - 21

gives

(x - 7)(x + 3) = 0

, which yields solutions

x = 7

and

x = -3

. Applying the constraint

r_1 > r_2

identifies

r_1 = 7

and

r_2 = -3

. Substituting these values into

r_1 + 2r_2

yields

7 + 2(-3) = 7 - 6 = 1

.

Step-by-Step Solution

1

Factor the quadratic equation

(x - 7)(x + 3) = 0

Find two numbers that multiply to

-21

and add up to

-4

, which are

-7

and

3

.

2

Solve for the roots of the equation

x = 7

or

x = -3

Set each linear factor equal to zero:

x - 7 = 0 \implies x = 7

and

x + 3 = 0 \implies x = -3

.

3

Assign values to

r_1

and

r_2

based on the condition

r_1 > r_2

r_1 = 7

and

r_2 = -3

Since

7 > -3

,

r_1

must be

7

and

r_2

must be

-3

.

4

Evaluate the expression

r_1 + 2r_2

7 + 2(-3) = 7 - 6 = 1

Substitute

r_1 = 7

and

r_2 = -3

into the given expression.

Key Concept

Solving quadratic equations by factoring and evaluating expressions involving roots.

Estimated Time:1m 30s

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