Algebra Practice Questions — SAT — Page 10

Question 181Question →

A drone's battery charge is at $98\%$ when it begins a mission. During the mission, the battery charge decreases by $1.5\%$ per minute when the drone is hovering, and by $2.5\%$ per minute when it is flying horizontally. The drone flies horizontally for exactly twice as many minutes as it hovers. If the battery charge is at $33\%$ at the end of the mission, for how many minutes did the drone hover?

Show answer & explanation

Answer: 10

Answer

The drone hovered for 10 minutes.

Let

t

represent the number of minutes the drone hovers. Since the drone flies horizontally for twice as long as it hovers, it flies horizontally for

2t

minutes. The total percentage drop in battery is

98 - 33 = 65

. The equation representing the total decrease in battery charge is

1.5t + 2.5(2t) = 65

. Simplifying the equation yields

1.5t + 5t = 65

, which becomes

6.5t = 65

. Dividing both sides of the equation by

6.5

gives

t = 10

.

Step-by-Step Solution

1

Define the variable for the unknown quantity.

Let

t

be the number of minutes the drone spent hovering. The time spent flying horizontally is then

2t

minutes.

The problem states that the horizontal flight time is exactly twice the hovering time.

2

Set up an equation representing the total decrease in battery charge.

The total percentage decrease is

1.5t + 2.5(2t) = 98 - 33

.

The battery decreases by

1.5\%

per minute of hovering,

2.5\%

per minute of horizontal flight, and the total change is from

98\%

to

33\%

.

3

Simplify the equation and solve for

t

.

6.5t = 65

, which gives

t = 10

.

Combine like terms and divide both sides by

6.5

to isolate the variable.

Key Concept

Setting up and solving a linear equation in one variable from a real-world scenario.

Question 182Question →

A florist is making bouquets. Each small bouquet requires $2$ roses, and each large bouquet requires $5$ roses. The florist has at most $40$ roses available for these bouquets. Additionally, the florist wants to make more than $10$ bouquets in total. If $x$ represents the number of small bouquets and $y$ represents the number of large bouquets, which of the following systems of inequalities represents this situation?

2x + 5y \geq 40

and

x + y < 10

5x + 2y \leq 40

and

x + y > 10

2x + 5y \leq 40

and

x + y > 10

2x + 5y \leq 10

and

x + y > 40

Show answer & explanation

Answer: $2x + 5y \leq 40$ and $x + y > 10$

Answer

The system of inequalities

2x + 5y \leq 40

and

x + y > 10

.

The correct system of inequalities matches the constraints given in the problem. The total number of roses used by

x

small bouquets and

y

large bouquets is

2x + 5y

. Since the florist has at most

40

roses, this is represented by

2x + 5y \leq 40

. The total number of bouquets is

x + y

. Since the florist wants to make more than

10

bouquets, this is represented by

x + y > 10

.

Step-by-Step Solution

1

Identify the inequality representing the constraint on the total number of roses.

2x + 5y \leq 40

Each small bouquet uses

2

roses (

2x

) and each large bouquet uses

5

roses (

5y

). The total roses must be 'at most'

40

, which corresponds to the inequality symbol

\leq

.

2

Identify the inequality representing the constraint on the total number of bouquets.

x + y > 10

The total number of bouquets is the sum of small bouquets (

x

) and large bouquets (

y

). The florist wants 'more than'

10

bouquets, which corresponds to the inequality symbol

>

.

3

Combine the two inequalities into a single system.

The system is

2x + 5y \leq 40

and

x + y > 10

.

Both conditions must be met simultaneously.

Key Concept

Translating real-world constraints into a system of linear inequalities in two variables.

Question 183Question →

In the system of linear equations below,

a

,

b

, and

c

are constants.

\begin{aligned} \frac{3}{2}x - \frac{2}{3}y &= 12 \\ ax + by &= c \end{aligned}

If the system has infinitely many solutions, and the point

(4, b)

lies on the line represented by the second equation, what is the value of

c

?

-162

72

108

162

Show answer & explanation

Answer: 162

Answer

The correct answer is 162.

The correct answer is 162. Since the system of linear equations has infinitely many solutions, the two equations represent the same line. Therefore, any point that lies on the second line must also lie on the first line. Substituting the coordinates of the point (4, b) into the first equation gives 3/2(4) - 2/3(b) = 12, which simplifies to 6 - 2/3(b) = 12. Solving this equation for b yields b = -9. Since the equations represent the same line, the ratio of their corresponding coefficients must be equal: c / 12 = b / (-2/3). Solving this proportion for c gives c = -18b. Substituting the value of b = -9 into this relation yields c = -18(-9) = 162.

Step-by-Step Solution

1

Determine the relationship between the two equations based on the number of solutions.

Since the system has infinitely many solutions, the two equations represent coincident lines in the coordinate plane. Therefore, any point that lies on the second line must also satisfy the first equation.

Infinitely many solutions in a system of two linear equations indicate that they represent the same line.

2

Substitute the point

(4, b)

into the first equation to solve for

b

.

\frac{3}{2}(4) - \frac{2}{3}b = 12 \implies 6 - \frac{2}{3}b = 12 \implies -\frac{2}{3}b = 6 \implies b = -9

Plugging the coordinates of the point into the first equation allows us to find the value of the unknown coordinate.

3

Set up the ratio of corresponding coefficients for the coincident lines.

\frac{a}{\frac{3}{2}} = \frac{b}{-\frac{2}{3}} = \frac{c}{12}

Equivalent equations must have proportional coefficients and constant terms.

4

Solve for

c

using the ratio containing

b

and

c

.

\frac{c}{12} = \frac{b}{-\frac{2}{3}} \implies c = -18b \implies c = -18(-9) = 162

Substituting the solved value of

b = -9

into the proportion yields the value of

c

.

Key Concept

Systems of Linear Equations with Infinitely Many Solutions

Question 184Question →

A school club is planning a fundraiser by selling two types of customized keychains: acrylic keychains and metal keychains. Acrylic keychains cost $3 to make and sell for$ 5, while metal keychains cost $5 to make and sell for$ 8. The club has a budget of at most $150 to produce the keychains, and they want to make at least 40 keychains in total. Let$ x $represent the number of acrylic keychains and$ y$ represent the number of metal keychains. Which of the following systems of inequalities represents this situation?

3x + 5y \leq 150

and

x + y \geq 40

3x + 5y \geq 150

and

x + y \leq 40

5x + 8y \leq 150

and

x + y \geq 40

5x + 3y \leq 150

and

x + y \geq 40

Show answer & explanation

Answer: $3x + 5y \leq 150$ and $x + y \geq 40$

Answer

3x + 5y \leq 150

and

x + y \geq 40

The correct system of inequalities must reflect the constraints on the budget and the total number of keychains. The budget is at most

150, which means the total production cost must be less than or equal to 150. Since each acrylic keychain costs

3 to make and each metal keychain costs

5, the cost constraint is represented by

3x + 5y \leq 150

. The club also wants to make at least 40 keychains, which means the total count must be greater than or equal to 40. This constraint is represented by

x + y \geq 40

. Together, these form the system

3x + 5y \leq 150

and

x + y \geq 40$.

Step-by-Step Solution

1

Identify the variables and their meanings from the problem statement.

x

represents the number of acrylic keychains, and

y

represents the number of metal keychains.

Understanding the variable definitions is necessary to construct the constraints correctly.

2

Set up the budget inequality using the production costs.

3x + 5y \leq 150

Each acrylic keychain costs

3 to make, each metal keychain costs

5, and the total cost must be at most (less than or equal to) $150.

3

Set up the quantity inequality.

x + y \geq 40

The club wants to make a total of at least (greater than or equal to) 40 keychains.

Key Concept

Systems of linear inequalities in context

Question 185Question →

A linear relationship exists between the variables $p$ and $q$ . When the value of $p$ increases by $6$ , the value of $q$ decreases by $4$ . If $q = 15$ when $p = 2$ , which of the following equations correctly expresses $p$ in terms of $q$ ?

p = -\frac{3}{2}q + \frac{49}{2}

p = -\frac{2}{3}q + \frac{49}{3}

p = \frac{3}{2}q - \frac{41}{2}

p = -\frac{2}{3}q + 12

Show answer & explanation

Answer: $p = -\frac{3}{2}q + \frac{49}{2}$

Answer

p = -\frac{3}{2}q + \frac{49}{2}

The correct equation is found by first calculating the rate of change of

q

with respect to

p

, which is

\frac{-4}{6} = -\frac{2}{3}

. Using the point-slope form with

(p, q) = (2, 15)

, we set up the equation

q - 15 = -\frac{2}{3}(p - 2)

. To express

p

in terms of

q

, we isolate

p

by multiplying both sides by

-\frac{3}{2}

to get

-\frac{3}{2}q + \frac{45}{2} = p - 2

, and then adding

2

to both sides, which yields

p = -\frac{3}{2}q + \frac{49}{2}

.

Step-by-Step Solution

1

Calculate the rate of change of

q

relative to

p

using the given changes.

The rate of change is

-\frac{4}{6} = -\frac{2}{3}

.

Since

q

decreases by

4

when

p

increases by

6

, the slope

m

in terms of

\Delta q / \Delta p

is

-\frac{2}{3}

.

2

Set up the point-slope form of the linear equation using the point

(p, q) = (2, 15)

.

q - 15 = -\frac{2}{3}(p - 2)

The point-slope form is

q - q_1 = m(p - p_1)

, where

(p_1, q_1) = (2, 15)

and

m = -\frac{2}{3}

.

3

Solve the equation for

p

by first clearing the coefficient of the

p

-term.

-\frac{3}{2}(q - 15) = p - 2 \implies -\frac{3}{2}q + \frac{45}{2} = p - 2

Multiplying both sides by

-\frac{3}{2}

simplifies isolation of the variable

p

.

4

Complete the isolation of

p

by adding

2

to both sides.

p = -\frac{3}{2}q + \frac{45}{2} + 2 \implies p = -\frac{3}{2}q + \frac{49}{2}

Adding

2

(which is

\frac{4}{2}

) to

\frac{45}{2}

isolates

p

on one side of the equation.

Key Concept

Linear relationships and isolating variables in two-variable linear equations.

Question 186Question →

In the $xy$ -plane, the graph of a linear function $f$ passes through the point $(1, 3)$ and has a slope of $m$ , where $m \neq 0$ . The graph of a second linear function, $g$ , is obtained by reflecting the graph of $f$ across the $y$ -axis and then translating it up by $4$ units. If the graphs of $f$ and $g$ intersect at a point on the $x$ -axis, what is the value of $m$ ?

1

-2

5

3

Show answer & explanation

Answer: 5

Answer

5

The correct value is

5

. By using the point-slope form, the first function is

f(x) = mx - m + 3

. Reflecting it across the

y

-axis gives

f(-x) = -mx - m + 3

, and translating it up by

4

units yields

g(x) = -mx - m + 7

. Since they intersect on the

x

-axis, the

y

-value at their intersection point is

0

. This gives the system of equations

mx_0 = m - 3

and

-mx_0 = m - 7

. Adding these equations eliminates the

mx_0

term, leaving

2m - 10 = 0

, which simplifies to

m = 5

.

Step-by-Step Solution

1

Write the equation of the linear function

f(x)

using the point-slope form.

f(x) = m(x - 1) + 3 = mx - m + 3

The function passes through the point

(1, 3)

and has a slope of

m

.

2

Determine the equation of the linear function

g(x)

by applying the reflection and translation transformations.

g(x) = f(-x) + 4 = -mx - m + 7

Reflecting across the

y

-axis replaces

x

with

-x

, and translating up by

4

units adds

4

to the function.

3

Set the

y

-coordinates of the intersection point

(x_0, 0)

on the

x

-axis to

0

for both functions.

mx_0 - m + 3 = 0 \implies mx_0 = m - 3

and

-mx_0 - m + 7 = 0 \implies -mx_0 = m - 7

An intersection on the

x

-axis means that the

y

-coordinate of the intersection point is

0

.

4

Solve the system of equations to find the value of

m

.

0 = (m - 3) + (m - 7) \implies 2m = 10 \implies m = 5

Adding the two equations eliminates the term

mx_0

, allowing us to solve directly for

m

.

Key Concept

Applying transformations (reflections and translations) to linear functions and finding their intercepts in the coordinate plane.

Question 187Question →

A chef is preparing portions of roasted vegetables and mashed potatoes for a catered event. Each portion of roasted vegetables requires $3$ ounces of potatoes, and each portion of mashed potatoes requires $6$ ounces of potatoes. The chef has a total of at most $90$ ounces of potatoes available. The chef must prepare at least $8$ portions of roasted vegetables and at least $5$ portions of mashed potatoes. If $v$ represents the number of portions of roasted vegetables that the chef prepares, what is the maximum possible value of $v$ ?

Show answer & explanation

Answer: 20

Answer

The maximum possible value of

v

is 20.

The maximum number of portions of roasted vegetables the chef can prepare is 20 because minimizing the number of mashed potato portions to its boundary constraint of 5 maximizes the remaining resources, yielding

3v \leq 90 - 6(5) \implies 3v \leq 60 \implies v \leq 20

.

Step-by-Step Solution

1

Formulate the inequality representing the potato weight constraint.

3v + 6p \leq 90

, where

v

is the number of portions of roasted vegetables and

p

is the number of portions of mashed potatoes.

Each portion of roasted vegetables uses

3

ounces of potatoes, each portion of mashed potatoes uses

6

ounces of potatoes, and the total amount used cannot exceed

90

ounces.

2

Formulate the inequalities representing the minimum quantity constraints.

v \geq 8

and

p \geq 5

The chef must make at least

8

portions of roasted vegetables and at least

5

portions of mashed potatoes.

3

Isolate the variable

v

in the potato limit inequality.

v \leq 30 - 2p

Subtracting

6p

from both sides of

3v + 6p \leq 90

yields

3v \leq 90 - 6p

, and dividing the entire inequality by

3

gives

v \leq 30 - 2p

.

4

Determine the maximum value of

v

by substituting the minimum possible value of

p

.

v \leq 30 - 2(5) \implies v \leq 20

To maximize

v

, we must minimize the subtracted term

2p

. The minimum allowed value of

p

is

5

.

Key Concept

Optimization in systems of linear inequalities

Question 188Question →

For the linear function $f$ , the graph of $y = f(x)$ has a slope of $3$ and contains the point $(2, 7)$ . The graph of another linear function, $g$ , is parallel to the graph of $f$ . If $g(0) = -4$ , what is the value of $g(5)$ ?

Show answer & explanation

Answer: 11

Answer

11

Since the graph of

g

is parallel to the graph of

f

, it has the same slope of

3

. The value

g(0) = -4

indicates that the

y

-intercept of the graph of

g

is

-4

. Thus, the equation for

g(x)

is

g(x) = 3x - 4

. Evaluating this function at

x = 5

gives

g(5) = 3(5) - 4 = 11

.

Step-by-Step Solution

1

Determine the slope of the function

g

.

The slope of

g

is

3

.

Parallel lines in the coordinate plane have equal slopes, and the slope of the graph of

f

is given as

3

.

2

Write the equation for

g(x)

.

g(x) = 3x - 4

Using the slope-intercept form

g(x) = mx + b

with slope

m = 3

and

y

-intercept

b = g(0) = -4

.

3

Evaluate

g(5)

.

11

Substitute

x = 5

into the equation for

g(x)

to get

g(5) = 3(5) - 4 = 11

.

Key Concept

Parallel lines have the same slope. A linear function can be written in slope-intercept form,

y = mx + b

, where

m

is the slope and

b

is the

y

-intercept.

Question 189Question →

A local library has two types of study rooms: small rooms, which can accommodate up to $3$ people, and large rooms, which can accommodate up to $8$ people. The library has a total of $15$ study rooms. If the maximum capacity of all the study rooms combined is $80$ people, how many large study rooms does the library have?

Show answer & explanation

Answer: 7

Answer

The correct answer is 7.

By representing the number of small rooms as

s

and large rooms as

l

, we set up the system of linear equations

s + l = 15

and

3s + 8l = 80

. Solving the first equation for

s

gives

s = 15 - l

. Substituting this into the second equation yields

3(15 - l) + 8l = 80

. Distributing and combining like terms results in

45 + 5l = 80

. Subtracting 45 from both sides gives

5l = 35

, which simplifies to

l = 7

. Thus, there are 7 large study rooms.

Step-by-Step Solution

1

Set up the system of linear equations representing the total number of rooms and their total capacity.

s + l = 15

and

3s + 8l = 80

We define

s

as the number of small study rooms and

l

as the number of large study rooms. The total number of rooms is 15, and the total capacity is 80.

2

Express the number of small rooms,

s

, in terms of the number of large rooms,

l

, using the first equation.

s = 15 - l

This allows for substitution into the capacity equation so that we can solve directly for the number of large study rooms.

3

Substitute the expression for

s

into the capacity equation and solve the resulting single-variable equation for

l

.

3(15 - l) + 8l = 80 \implies 45 - 3l + 8l = 80 \implies 45 + 5l = 80 \implies 5l = 35 \implies l = 7

Substituting

s

eliminates one variable, leaving a linear equation in terms of

l

that can be solved using standard algebraic steps.

Key Concept

Solving systems of linear equations in a real-world context using substitution.

Question 190Question →

For the constants $p$ and $q$ , the given system of linear equations in $x$ and $y$ has infinitely many solutions:

\begin{aligned} px - 3y &= q \\ 4x + (p - 7)y &= 12 \end{aligned}

Which of the following is a possible value of $q$ ?

3

6

9

16

Show answer & explanation

Answer: 9

Answer

9

For the system of linear equations to have infinitely many solutions, the equations must be equivalent. This means the ratio of the coefficients of

x

, the coefficients of

y

, and the constant terms must all be equal:

\frac{p}{4} = \frac{-3}{p-7} = \frac{q}{12}

. Cross-multiplying the first two ratios gives

p(p-7) = -12

, which simplifies to

p^2 - 7p + 12 = 0

. Factoring this quadratic equation gives

(p-3)(p-4) = 0

, which yields

p = 3

or

p = 4

. Using the relation between the first and third ratios, we have

\frac{q}{12} = \frac{p}{4}

, which simplifies to

q = 3p

. Substituting the possible values of

p

gives

q = 3(3) = 9

or

q = 3(4) = 12

. Since 9 is one of the options, it is a possible value of

q

.

Step-by-Step Solution

1

Set up the condition for infinitely many solutions by equating the ratios of the coefficients and the constants.

The coefficients and constants must satisfy the proportion:

\frac{p}{4} = \frac{-3}{p - 7} = \frac{q}{12}

For a system of two linear equations to have infinitely many solutions, the two equations must represent the exact same line, meaning all corresponding coefficients and constant terms must be proportional.

2

Solve the equation formed by the first two ratios to find the possible values of the parameter

p

.

\frac{p}{4} = \frac{-3}{p - 7} \Rightarrow p(p - 7) = -12 \Rightarrow p^2 - 7p + 12 = 0

. Factoring the quadratic yields

(p - 3)(p - 4) = 0

, which gives

p = 3

or

p = 4

.

This step determines the values of

p

for which the two lines are parallel (i.e., they have equal slopes).

3

Find the corresponding values of

q

using the relation between the first and third ratios.

\frac{p}{4} = \frac{q}{12} \Rightarrow q = 3p

. Substituting the values of

p

:
- If

p = 3

, then

q = 3(3) = 9

.
- If

p = 4

, then

q = 3(4) = 12

.

This step ensures that the parallel lines are coincident by equating their y-intercepts.

4

Compare the possible values of

q

with the given options.

The value 9 is a possible value of

q

and matches one of the choices.

To identify which of the two mathematically valid solutions for

q

is listed in the multiple-choice options.

Key Concept

Conditions for infinitely many solutions in a system of linear equations

Alternative Method

Instead of using ratios directly, we can write both equations in slope-intercept form:

y = mx + b

. For the first equation,

px - 3y = q \Rightarrow y = \frac{p}{3}x - \frac{q}{3}

. For the second equation,

4x + (p-7)y = 12 \Rightarrow y = -\frac{4}{p-7}x + \frac{12}{p-7}

(assuming

p \neq 7

). For the system to have infinitely many solutions, the two lines must have the same slope and the same y-intercept. Equating the slopes gives

\frac{p}{3} = -\frac{4}{p-7}

, which simplifies to

p^2 - 7p + 12 = 0

, yielding

p = 3

or

p = 4

. Equating the y-intercepts gives

-\frac{q}{3} = \frac{12}{p-7} \Rightarrow q = -\frac{36}{p-7}

. Substituting

p=3

gives

q = -\frac{36}{-4} = 9

, and substituting

p=4

gives

q = -\frac{36}{-3} = 12

.

Estimated Time:3m 0s

Question 191Question →

A small factory manufactures two types of toys: wood blocks and toy cars. Let $x$ represent the number of wood blocks produced daily, and let $y$ represent the number of toy cars produced daily. The daily production must satisfy the following constraints:

* The total number of toys produced daily cannot exceed 40: $x + y \le 40$
* Each wood block requires 2 minutes of painting, and each toy car requires 1 minute of painting. The total daily painting time is at most 60 minutes: $2x + y \le 60$
* The number of wood blocks produced cannot exceed the number of toy cars produced by more than 15: $y \ge x - 15$

If the factory must produce a non-negative number of both types of toys, what is the maximum possible number of wood blocks the factory can produce daily?

20

25

30

40

Show answer & explanation

Answer: 25

Answer

The maximum possible number of wood blocks the factory can produce daily is 25.

The correct answer is the value that represents the maximum

x

-coordinate within the feasible region. By graphing the inequalities, we find the vertices of the shaded region are

(0, 0)

,

(15, 0)

,

(25, 10)

,

(20, 20)

, and

(0, 40)

. The maximum

x

-value among these vertices is 25, which occurs at the intersection of the painting time constraint and the demand constraint.

Step-by-Step Solution

1

Identify the system of inequalities representing the constraints.

The system of inequalities is:
1)

x + y \le 40

2)

2x + y \le 60

3)

y \ge x - 15

4)

x \ge 0

5)

y \ge 0

This establishes the boundaries of the feasible region on the coordinate plane.

2

Find the boundary intersection points (vertices of the feasible region) that could maximize

x

.

The relevant intersections are:
- The intersection of (1) and (2):

x + y = 40

and

2x + y = 60 \implies x = 20, y = 20

.
- The intersection of (2) and (3):

2x + y = 60

and

y = x - 15 \implies 2x + (x - 15) = 60 \implies 3x = 75 \implies x = 25, y = 10

.
- The intersection of (3) and the x-axis (

y = 0

):

0 = x - 15 \implies x = 15, y = 0

.

The maximum value of a variable in a bounded linear feasibility region must occur at one of the vertices of the region.

3

Test the vertices to verify they satisfy all inequalities in the system.

- For

(20, 20)

:

20 + 20 \le 40

(True),

2(20) + 20 \le 60

(True),

20 \ge 20 - 15

(True).
- For

(25, 10)

:

25 + 10 = 35 \le 40

(True),

2(25) + 10 = 60 \le 60

(True),

10 \ge 25 - 15

(True).
- For

(15, 0)

:

15 + 0 \le 40

(True),

2(15) + 0 \le 60

(True),

0 \ge 15 - 15

(True).

This ensures that the intersection points actually lie within the feasible region defined by all constraints.

4

Compare the

x

-values of the valid vertices to find the maximum.

The candidate

x

-values are 20, 25, and 15. The maximum value is 25.

This determines the absolute maximum coordinate value within the bounded region.

Key Concept

Solving systems of linear inequalities by identifying the vertices of the feasible region to optimize a coordinate value.

Question 192Question →

If $\frac{1}{4}(8x - 12) - \frac{2}{3}(3 - 6x) = 13$ , what is the value of $3x - 2$ ?

2

3

7

-29

Show answer & explanation

Answer: 7

Answer

7

Distributing the coefficients yields

\frac{1}{4}(8x - 12) = 2x - 3

and

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

. Combining these terms gives the simplified equation

6x - 5 = 13

. Adding

5

to both sides yields

6x = 18

, and dividing by

6

gives

x = 3

. Substituting

x = 3

into the expression

3x - 2

results in

3(3) - 2 = 7

.

Step-by-Step Solution

1

Distribute the fraction coefficients to the terms inside the parentheses.

\frac{1}{4}(8x - 12) = 2x - 3

and

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

.

This simplifies the equation by removing the parentheses.

2

Combine like terms on the left side of the equation.

(2x - 3) + (-2 + 4x) = 6x - 5

, transforming the equation to

6x - 5 = 13

.

Grouping the variable terms and constant terms is necessary to isolate the variable.

3

Isolate the variable by adding 5 to both sides and dividing by 6.

6x = 18 \Rightarrow x = 3

.

This determines the value of the variable.

4

Substitute the value of

x

into the expression

3x - 2

.

3(3) - 2 = 7

.

The question asks for the value of the expression

3x - 2

, not the value of

x

itself.

Key Concept

Linear Equations in One Variable

Alternative Method

An alternative method is to multiply both sides of the equation by the least common multiple of the denominators

4

and

3

, which is

12

. Multiplying the entire equation by

12

yields

3(8x - 12) - 8(3 - 6x) = 156

. Distributing gives

24x - 36 - 24 + 48x = 156

, which simplifies to

72x - 60 = 156

. Adding

60

gives

72x = 216

, so

x = 3

. Substituting

x = 3

into

3x - 2

gives

7

.

Estimated Time:1m 30s

Question 193Question →

Which of the following ordered pairs $(x, y)$ is a solution to the system of inequalities below?

\begin{aligned} y &\geq 3x + 1 \\ y &< -2x + 5 \end{aligned}

(0, 2)

(1, 2)

(0, 6)

(3, 10)

Show answer & explanation

Answer: $(0, 2)$

Answer

(0, 2)

The ordered pair

(0, 2)

is correct because substituting these values into both inequalities results in true statements. Specifically,

2 \geq 3(0) + 1

becomes

2 \geq 1

, which is true, and

2 < -2(0) + 5

becomes

2 < 5

, which is also true.

Step-by-Step Solution

1

Substitute the coordinates of the correct ordered pair into the first inequality,

y \geq 3x + 1

.

2 \geq 3(0) + 1 \Rightarrow 2 \geq 1

, which is true.

To be a solution to the system, an ordered pair must satisfy both inequalities in the system.

2

Substitute the coordinates of the same ordered pair into the second inequality,

y < -2x + 5

.

2 < -2(0) + 5 \Rightarrow 2 < 5

, which is true.

To verify the solution, we check if it satisfies the second constraint.

Key Concept

Testing coordinate solutions in a system of linear inequalities

Question 194Question →

An artist sells small prints for $4$ dollars each and large prints for $9$ dollars each. The artist wants to sell at least $15$ prints in total and earn at least $80$ dollars from these sales. If the artist sells $5$ small prints, what is the minimum number of large prints the artist must sell to meet both conditions?

Show answer & explanation

Answer: 10

Answer

The minimum number of large prints the artist must sell is 10.

Substituting the value of small prints (

x = 5

) into the total count inequality

x + y \geq 15

gives

5 + y \geq 15

, which simplifies to

y \geq 10

. Substituting it into the earnings inequality

4x + 9y \geq 80

yields

20 + 9y \geq 80

, which simplifies to

y \geq 6.67

. To satisfy both conditions, the value of

y

must be at least 10.

Step-by-Step Solution

1

Set up the inequalities for the system.

Let

x

represent the number of small prints and

y

represent the number of large prints. The constraint for the minimum number of prints is

x + y \geq 15

. The constraint for the minimum earnings is

4x + 9y \geq 80

.

To represent the given conditions as a system of linear inequalities in two variables.

2

Substitute the known value of small prints (

x = 5

) into the inequalities.

Substituting

x = 5

into the first inequality gives

5 + y \geq 15

, which simplifies to

y \geq 10

. Substituting

x = 5

into the second inequality gives

4(5) + 9y \geq 80

, which simplifies to

20 + 9y \geq 80

, then

9y \geq 60

, resulting in

y \geq \frac{20}{3} \approx 6.67

.

To determine the range of values for the number of large prints (

y

) under both constraints.

3

Determine the minimum integer value for

y

that satisfies both inequalities.

The first constraint requires

y \geq 10

, and the second constraint requires

y \geq 6.67

. Since the number of prints must be an integer and both conditions must be satisfied, the minimum value is

10

.

To satisfy both inequalities simultaneously with the smallest possible integer value.

Key Concept

Systems of Linear Inequalities in Two Variables

Estimated Time:1m 0s

Question 195Question →

The system of linear equations shown below contains constants

a

and

b

.

\begin{aligned} \frac{1}{2}(ax - 4y) &= x - \frac{1}{2}y + 6 \\ 2x - y &= b \end{aligned}

If the system has infinitely many solutions, what is the value of

a + b

?

Show answer & explanation

Answer: 12

Answer

The correct answer is 12.

To find the value of

a + b

that yields infinitely many solutions, we rewrite both equations in the standard form

Ax + By = C

. Simplifying the first equation gives

(a - 2)x - 3y = 12

. Multiplying the second equation

2x - y = b

by

3

gives

6x - 3y = 3b

. For the system to have infinitely many solutions, the two equations must be equivalent, meaning

a - 2 = 6

(which gives

a = 8

) and

3b = 12

(which gives

b = 4

). The sum of these values is

8 + 4 = 12

.

Step-by-Step Solution

1

Distribute the fraction

\frac{1}{2}

on the left side of the first equation.

\frac{1}{2}ax - 2y = x - \frac{1}{2}y + 6

To expand the expression and prepare it for simplification.

2

Group the

x

and

y

terms on the left side of the equation and the constants on the right side.

\left(\frac{1}{2}a - 1\right)x - \frac{3}{2}y = 6

To write the equation in standard linear form.

3

Multiply the entire equation by

2

to eliminate the fractional coefficients.

(a - 2)x - 3y = 12

To simplify comparison with the second equation by working with integer coefficients.

4

Multiply the second equation,

2x - y = b

, by

3

to align the

y

-coefficients with the first equation.

6x - 3y = 3b

Two linear equations have infinitely many solutions if they represent the same line, which requires matching coefficients and constants.

5

Equate the corresponding

x

-coefficients and constant terms from

(a - 2)x - 3y = 12

and

6x - 3y = 3b

.

a - 2 = 6 \implies a = 8

and

3b = 12 \implies b = 4

To solve for the values of the constants

a

and

b

.

6

Calculate the sum of

a

and

b

.

8 + 4 = 12

To find the final requested value of

a + b

.

Key Concept

Determining parameters for infinitely many solutions in a system of linear equations

Question 196Question →

A company plans to install standard-charging ports and fast-charging ports at its office building. The company must install at least $15$ charging ports in total. Each fast-charging port requires $12$ kilowatts of power, and each standard-charging port requires $4$ kilowatts of power. The electrical grid can supply a maximum of $120$ kilowatts of power for these ports. Additionally, the number of standard-charging ports must be at least twice the number of fast-charging ports. What is the maximum number of fast-charging ports the company can install?

Show answer & explanation

Answer: 6

Answer

The maximum number of fast-charging ports the company can install is 6.

Representing the number of fast-charging ports by

x

and standard-charging ports by

y

yields the system of inequalities:

x + y \geq 15

,

12x + 4y \leq 120

, and

y \geq 2x

. The power constraint simplifies to

3x + y \leq 30

, or

y \leq 30 - 3x

. Combining this with

y \geq 2x

gives

2x \leq 30 - 3x

, which simplifies to

5x \leq 30

, or

x \leq 6

. Substituting

x = 6

into the constraints gives

y = 12

, which satisfies the total port constraint because

6 + 12 = 18 \geq 15

. Therefore, the maximum number of fast-charging ports is 6.

Step-by-Step Solution

1

Define variables and write the system of inequalities representing the constraints.

Let

x

be the number of fast-charging ports and

y

be the number of standard-charging ports. The constraints are:
1)

x + y \geq 15

2)

12x + 4y \leq 120

3)

y \geq 2x

4)

x \geq 0, y \geq 0

This translates the word problem constraints into a system of linear inequalities.

2

Simplify the power capacity inequality.

3x + y \leq 30

, which can be rewritten as

y \leq 30 - 3x

.

Dividing the terms by 4 simplifies the coefficients, making algebraic manipulation easier.

3

Combine the simplified inequality with the charging port ratio constraint to find the upper limit for

x

.

Since

2x \leq y

and

y \leq 30 - 3x

, we have

2x \leq 30 - 3x

.
Adding

3x

to both sides yields

5x \leq 30

.
Dividing by 5 gives

x \leq 6

.

This determines the maximum possible value for the number of fast-charging ports.

4

Verify that the upper limit

x = 6

satisfies all system constraints with integer values.

If

x = 6

, then

y \geq 2(6) = 12

and

y \leq 30 - 3(6) = 12

, which means

y = 12

.
Checking the total port constraint:

x + y = 6 + 12 = 18

. Since

18 \geq 15

, the point

(6, 12)

satisfies all constraints.

Since the number of ports must be integers, we must confirm that

x = 6

yields a valid integer coordinate

(6, 12)

that lies within the feasible region.

Key Concept

Maximizing a variable under a system of linear inequalities in two variables.

Question 197Question →

In the $xy$ -plane, the graph of the linear function $f$ passes through the points $(-2, 5)$ and $(4, 1)$ . If the linear function $g$ is defined by $g(x) = f(x) + 3$ , what is the $x$ -intercept of the graph of $g$ ?

6

\frac{11}{2}

10

\frac{20}{3}

Show answer & explanation

Answer: $10$

Answer

10

The correct answer is

10

. First, find the slope of

f

using the formula

m = \frac{y_2 - y_1}{x_2 - x_1}

, which gives

m = \frac{1 - 5}{4 - (-2)} = -\frac{2}{3}

. Using the point

(4, 1)

in the point-slope formula, the equation of

f

is

f(x) = -\frac{2}{3}x + \frac{11}{3}

. Since

g(x) = f(x) + 3

, the equation of

g

is

g(x) = -\frac{2}{3}x + \frac{20}{3}

. Setting

g(x) = 0

to find the

x

-intercept gives

x = 10

.

Step-by-Step Solution

1

Calculate the slope of the linear function

f

m = -\frac{2}{3}

The slope

m

of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is given by

\frac{y_2 - y_1}{x_2 - x_1}

. Substituting

(-2, 5)

and

(4, 1)

yields

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

.

2

Find the equation of function

f

f(x) = -\frac{2}{3}x + \frac{11}{3}

Using the point-slope form

y - y_1 = m(x - x_1)

with the point

(4, 1)

, we get

y - 1 = -\frac{2}{3}(x - 4) \Rightarrow y = -\frac{2}{3}x + \frac{8}{3} + 1 = -\frac{2}{3}x + \frac{11}{3}

.

3

Determine the equation of function

g

g(x) = -\frac{2}{3}x + \frac{20}{3}

Since

g(x) = f(x) + 3

, we add

3

to the expression for

f(x)

:

g(x) = -\frac{2}{3}x + \frac{11}{3} + 3 = -\frac{2}{3}x + \frac{20}{3}

.

4

Find the

x

-intercept of the graph of

g

x = 10

The

x

-intercept is the value of

x

when

g(x) = 0

. Setting

g(x) = 0

gives

0 = -\frac{2}{3}x + \frac{20}{3} \Rightarrow \frac{2}{3}x = \frac{20}{3} \Rightarrow x = 10

.

Key Concept

Finding the equation of a linear function from two points and applying vertical translations to find key features such as intercepts.

Estimated Time:1m 30s

Question 198Question →

In the $xy$ -plane, the graph of the equation $ax + by = 24$ , where $a$ and $b$ are constants, is a line. The line has an $x$ -intercept of $(d, 0)$ and a $y$ -intercept of $(0, d - 6)$ , where $d > 6$ is a constant. If the line passes through the point $(2, 5)$ , what is the value of $a + b$ ?

Show answer & explanation

Answer: 6

Answer

6

Substituting the intercepts

(d, 0)

and

(0, d-6)

into

ax + by = 24

gives

a = 24/d

and

b = 24/(d-6)

. Using the point

(2, 5)

, we get

2(24/d) + 5(24/(d-6)) = 24

, which simplifies to

2/d + 5/(d-6) = 1

. Solving for

d

yields

d^2 - 13d + 12 = 0

, giving

d = 12

or

d = 1

. Given

d > 6

, we have

d = 12

. Substituting

d = 12

back gives

a = 2

and

b = 4

, so

a + b = 6

.

Step-by-Step Solution

1

Find expressions for

a

and

b

in terms of

d

by substituting the given intercepts.

a = \frac{24}{d}

and

b = \frac{24}{d-6}

The

x

-intercept

(d, 0)

and

y

-intercept

(0, d-6)

lie on the line

ax + by = 24

.

2

Substitute the point

(2, 5)

into the line's equation.

2a + 5b = 24

The point

(2, 5)

lies on the line.

3

Substitute the expressions for

a

and

b

into

2a + 5b = 24

and simplify.

\frac{2}{d} + \frac{5}{d-6} = 1

To create a single equation in terms of the variable

d

.

4

Solve the rational equation for

d

.

d = 12

Multiplying by the common denominator

d(d-6)

leads to the quadratic equation

d^2 - 13d + 12 = 0

, which factors as

(d-12)(d-1) = 0

. Since the problem states

d > 6

, the only valid solution is

d = 12

.

5

Calculate

a

and

b

using

d = 12

, and find their sum.

a = 2

,

b = 4

, and

a + b = 6

Substituting

d = 12

into the expressions for

a

and

b

yields the constant coefficients, and summing them provides the final requested value.

Key Concept

Using intercepts and a given point on a line in the coordinate plane to solve for the parameters of its linear equation.

Question 199Question →

\begin{aligned} x - 2y &= 5 \\ 3x - y &= 10 \end{aligned}

If $(x, y)$ is the solution to the system of equations above, what is the value of $x + y$ ?

2

3

6

8

Show answer & explanation

Answer: 2

Answer

2

The correct answer is 2. Isolating

x

in the first equation gives

x = 2y + 5

. Substituting this expression into the second equation yields

3(2y + 5) - y = 10

. Distributing the 3 gives

6y + 15 - y = 10

, which simplifies to

5y + 15 = 10

. Subtracting 15 from both sides gives

5y = -5

, so

y = -1

. Substituting

y = -1

back into the equation for

x

gives

x = 2(-1) + 5 = 3

. The sum of the solutions is

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

.

Step-by-Step Solution

1

Isolate

x

in the first equation.

x = 2y + 5

Isolating one variable makes it straightforward to substitute it into the other equation.

2

Substitute the expression for

x

into the second equation and solve for

y

.

y = -1

Substituting

x = 2y + 5

gives

3(2y + 5) - y = 10

. Distributing the 3 yields

6y + 15 - y = 10

. Combining like terms gives

5y + 15 = 10

. Subtracting 15 from both sides gives

5y = -5

, so

y = -1

.

3

Substitute

y = -1

back into the expression for

x

to find its value.

x = 3

Using the isolated expression

x = 2y + 5

, substituting

y = -1

gives

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

.

4

Calculate the sum

x + y

.

2

The question asks for the value of

x + y

, which is

3 + (-1) = 2

.

Key Concept

Solving systems of linear equations in two variables using substitution or elimination, and evaluating algebraic combinations of the variables.

Alternative Method

The system can also be solved by elimination. Multiplying the second equation by 2 gives

6x - 2y = 20

. Subtracting the first equation (

x - 2y = 5

) from this new equation eliminates

y

, resulting in

5x = 15

, or

x = 3

. Substituting

x = 3

into either original equation yields

y = -1

. Finally, the sum is

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

.

Estimated Time:1m 30s

Question 200Question →

A researcher is preparing a growth medium for a bacterial culture by mixing Nutrient A and Nutrient B. Let $x$ represent the number of grams of Nutrient A and $y$ represent the number of grams of Nutrient B in the mixture. The mixture must satisfy the following conditions:

* The total mass of the nutrients in the mixture is at most $15$ grams.
* The mass of Nutrient B is at most twice the difference of the mass of Nutrient A and $2$ grams.
* The mass of Nutrient B is at least half the mass of Nutrient A.

What is the maximum possible integer number of grams of Nutrient B that can be used in the mixture?

5

6

8

9

Show answer & explanation

Answer: 8

Answer

The maximum possible integer number of grams of Nutrient B that can be used is 8.

The system of inequalities bounding the feasible region is

x + y \leq 15

,

y \leq 2x - 4

, and

y \geq \frac{1}{2}x

. The maximum value of

y

(representing Nutrient B) occurs at the intersection of the boundary lines

x + y = 15

and

y = 2x - 4

. Solving this system gives

x = 19/3

and

y = 26/3 \approx 8.67

. Since the researcher needs an integer number of grams, the largest integer value of

y

that falls within the region is

8

. Setting

y = 8

yields the valid range

6 \leq x \leq 7

, confirming that

y = 8

is a possible value.

Step-by-Step Solution

1

Translate the verbal constraints into a system of linear inequalities.

The system of inequalities is:
1)

x + y \leq 15

2)

y \leq 2(x - 2) \implies y \leq 2x - 4

3)

y \geq \frac{1}{2}x

Establishing the mathematical model is necessary to define the feasible region in the coordinate plane.

2

Find the intersection points of the boundary lines to locate the vertices of the feasible region.

Vertex 1 (Intersection of lines 1 and 2):

x + (2x - 4) = 15 \implies 3x = 19 \implies x = 19/3

, so

y = 26/3

.

Vertex 2 (Intersection of lines 1 and 3):

x + \frac{1}{2}x = 15 \implies \frac{3}{2}x = 15 \implies x = 10

, so

y = 5

.

Vertex 3 (Intersection of lines 2 and 3):

2x - 4 = \frac{1}{2}x \implies \frac{3}{2}x = 4 \implies x = 8/3

, so

y = 4/3

.

The maximum value of a coordinate in a bounded linear inequality system occurs at one of the vertices of the feasible region.

3

Identify the vertex that maximizes the

y

-coordinate.

Comparing the

y

-coordinates of the vertices:
-

y = 4/3 \approx 1.33

at

(8/3, 4/3)

-

y = 5

at

(10, 5)

-

y = 26/3 \approx 8.67

at

(19/3, 26/3)

The maximum value of

y

in the feasible region is

26/3 \approx 8.67

.

Comparing the coordinates determines the upper limit of the variable

y

in the shaded region.

4

Find the maximum possible integer value for

y

within the feasible region.

The maximum value is

8.67

. The largest integer less than or equal to

8.67

is

8

. Verify that

y = 8

is achievable by substituting it into the inequalities:

x + 8 \leq 15 \implies x \leq 7

8 \leq 2x - 4 \implies 2x \geq 12 \implies x \geq 6

8 \geq \frac{1}{2}x \implies x \leq 16

This yields the interval

6 \leq x \leq 7

, which contains valid solutions (such as the integer points

(6, 8)

and

(7, 8)

).

Ensures that the integer value chosen actually lies within the set of valid coordinates for the system.

Key Concept

Maximizing a variable under a system of linear constraints

Estimated Time:3m 0s

Algebra

Answer

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Alternative Method