Systems of Linear Inequalities in Two Variables Practice Questions — SAT — Page 2

Question 21Question →

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

Question 22Question →

In the $xy$ -plane, a point with coordinates $(x, y)$ lies in the solution set of the system of inequalities below.

y \leq -x + 7

y \geq 2x - 5

y \geq 0

What is the maximum possible value of $x$ ?

3

4

7

2.5

Show answer & explanation

Answer: 4

Answer

The maximum possible value of

x

is 4.

To find the maximum possible value of

x

in the solution set, we analyze the boundaries of the system. The inequality

y \leq -x + 7

can be rewritten as

x \leq 7 - y

. The inequality

y \geq 2x - 5

can be rewritten as

x \leq 0.5y + 2.5

. For any point in the solution set,

x

must be less than or equal to both

7 - y

and

0.5y + 2.5

. Since

y \geq 0

, the maximum value of

x

will occur where these two upper bounds are equal:

7 - y = 0.5y + 2.5

, which simplifies to

1.5y = 4.5

, or

y = 3

. Substituting

y = 3

back into either equation gives

x = 4

. For any

y > 3

,

x \leq 7 - y < 4

. For any

y < 3

,

x \leq 0.5y + 2.5 < 4

. Thus, the maximum possible value of

x

is 4.

Step-by-Step Solution

1

Express the boundaries of

x

in terms of

y

from the given inequalities.

x \leq 7 - y

and

x \leq 0.5y + 2.5

To find the upper limits on

x

, we solve each inequality for

x

.

2

Find the intersection point of the two boundary lines by setting the expressions equal to each other.

7 - y = 0.5y + 2.5 \implies 1.5y = 4.5 \implies y = 3

. Substituting

y = 3

gives

x = 4

.

The maximum value of

x

occurs at the intersection of the two boundary constraints since one boundary increases with

y

and the other decreases with

y

.

3

Verify that the intersection point

(4, 3)

satisfies all three inequalities.

3 \leq -4 + 7

(true),

3 \geq 2(4) - 5

(true), and

3 \geq 0

(true).

Ensuring the point lies within the solution set confirms it is a valid maximum.

Key Concept

Systems of Linear Inequalities in Two Variables

Question 23Question →

A shipping company uses a cargo plane to transport two types of cargo containers: Type A and Type B. Let $x$ represent the number of Type A containers, and let $y$ represent the number of Type B containers. The constraints on the shipment are represented by the system of inequalities below:

\begin{aligned} x + y &\leq 45 \\ 3x + 5y &\leq 165 \end{aligned}

If the plane must carry at least $10$ Type A containers, what is the maximum number of Type B containers the plane can transport?

25

27

35

39

Show answer & explanation

Answer: 27

Answer

27

The correct answer of 27 is found by substituting the minimum value of

x

, which is 10, into the second inequality constraint. This yields

3(10) + 5y \leq 165

, which simplifies to

5y \leq 135

, or

y \leq 27

. Since the coordinate pair (10, 27) also satisfies the first inequality constraint (

10 + 27 = 37 \leq 45

), 27 is the maximum possible value for

y

.

Step-by-Step Solution

1

Identify the given constraints and the variable to maximize.

We are given the system of inequalities

x + y \leq 45

and

3x + 5y \leq 165

, with

x \geq 10

, and we want to find the maximum possible value of

y

.

This establishes the mathematical model and boundaries for the problem.

2

Express

y

in terms of

x

for both inequalities to see which constraint is more restrictive.

From the first inequality:

y \leq 45 - x

. From the second inequality:

3x + 5y \leq 165 \implies 5y \leq 165 - 3x \implies y \leq 33 - 0.6x

.

This allows us to analyze

y

as a function of

x

under both constraints.

3

Substitute the minimum possible value of

x

, which is 10, into both boundary expressions for

y

.

Under the first constraint:

y \leq 45 - 10 = 35

. Under the second constraint:

y \leq 33 - 0.6(10) = 27

.

Since both boundary lines have negative slopes, the maximum value of

y

will occur when

x

is at its minimum value of 10.

4

Determine the most restrictive upper bound for

y

when

x = 10

.

The value of

y

must satisfy both

y \leq 35

and

y \leq 27

, so the maximum possible value is 27.

A solution to a system of inequalities must satisfy all inequalities in the system simultaneously.

Key Concept

Solving systems of linear inequalities in context by evaluating boundary conditions and identifying constraints.

Question 24Question →

In the $xy$ -plane, a point $(x, y)$ is in the solution set of the system of inequalities below:

x - 2y \leq -6

x + y \leq 11

y \leq 5

What is the maximum possible value of $x$ ?

Show answer & explanation

Answer: 4

Answer

The maximum possible value of

x

is

4

.

The maximum possible value of

x

is

4

. For all points in the solution set where

y \leq 5

, the active upper bound is

x \leq 2y - 6

. Since

2y - 6

increases as

y

increases, the maximum value of

x

occurs at the maximum boundary

y = 5

, giving

x = 2(5) - 6 = 4

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the first two inequalities.

x \leq 2y - 6

and

x \leq 11 - y

To isolate the variable

x

and analyze how its upper bound is constrained by

y

.

2

Apply the third inequality constraint,

y \leq 5

, to find the limits on these upper bounds.

2y - 6 \leq 4

and

11 - y \geq 6

Since

y

cannot exceed

5

, the value of

2y - 6

is maximized when

y = 5

, and

11 - y

is minimized when

y = 5

.

3

Compare the two upper bounds to determine the active constraint for the domain

y \leq 5

.

Since

2y - 6 \leq 4

and

11 - y \geq 6

, the active bound is

x \leq 2y - 6

.

For any point to satisfy the system,

x

must be less than or equal to both bounds, meaning it is restricted by the smaller of the two bounds.

4

Calculate the maximum value of

x

at the boundary point.

x = 4

at the point

(4, 5)

The function

2y - 6

is increasing with respect to

y

, so its maximum value occurs at the largest possible value of

y

, which is

5

.

Key Concept

Finding the maximum value of a coordinate within a system of linear inequalities.

Question 25Question →

In the $xy$ -plane, a point $(x, y)$ lies in the solution set of the system of inequalities below.

$y \leq -x + 8$
$y \leq 2x + 2$

What is the maximum possible value of $y$ ?

Show answer & explanation

Answer: 6

Answer

The maximum possible value of

y

is 6.

The solution set to the system of inequalities is the region in the coordinate plane that lies on or below both boundary lines,

y = -x + 8

and

y = 2x + 2

. The maximum

y

-value in this region occurs at the intersection of the two lines. Solving the system of equations by setting

-x + 8 = 2x + 2

gives

3x = 6

, which simplifies to

x = 2

. Substituting

x = 2

back into either equation yields

y = 6

. Therefore, the maximum possible value of

y

is 6.

Step-by-Step Solution

1

Set the two boundary equations equal to each other to find the

x

-coordinate of the intersection point.

-x + 8 = 2x + 2 \implies 3x = 6 \implies x = 2

Since the solution region is bounded from above by both inequalities, the maximum value of

y

must occur at the intersection of the two boundary lines.

2

Substitute the

x

-value back into one of the equations to find the corresponding

y

-value.

y = 2(2) + 2 = 6

This determines the

y

-coordinate of the intersection point, which represents the maximum height of the shaded region.

Key Concept

Systems of Linear Inequalities in Two Variables

Estimated Time:45s

Question 26Question →

Consider the system of inequalities shown below:

\begin{aligned} y &\geq 2x + 1 \\ y &< -x + 6 \end{aligned}

Which coordinate pair $(x, y)$ is a solution to the system?

(2, 3)

(1, 4)

(1, 5)

(2, 5)

Show answer & explanation

Answer: $(1, 4)$

Answer

The coordinate pair

(1, 4)

is a solution to the system of inequalities.

The coordinate pair

(1, 4)

is the correct answer because substituting

x = 1

and

y = 4

into both inequalities yields true statements:

4 \geq 2(1) + 1 \Rightarrow 4 \geq 3

is true, and

4 < -1 + 6 \Rightarrow 4 < 5

is true.

Step-by-Step Solution

1

Substitute the coordinates of the correct pair,

(1, 4)

, into the first inequality,

y \geq 2x + 1

.

4 \geq 2(1) + 1 \Rightarrow 4 \geq 3

, which is true.

To be a solution to the system, the point must satisfy the first inequality.

2

Substitute the same coordinates,

(1, 4)

, into the second inequality,

y < -x + 6

.

4 < -1 + 6 \Rightarrow 4 < 5

, which is true.

The point must also satisfy the second inequality to be a solution to the entire system.

3

Conclude that since

(1, 4)

satisfies both inequalities, it is a solution to the system.

The point

(1, 4)

is in the solution set.

A coordinate pair is a solution to a system of inequalities if and only if it satisfies every inequality in the system.

Key Concept

Testing coordinate pairs in a system of linear inequalities by substitution.

Estimated Time:45s

Question 27Question →

A public library is purchasing new books for its collection. The library plans to buy paperback books for $6 each and hardcover books for$ 20 each. The library has a budget of at most $1,200 for this purchase, wants to buy at most 120 total books, and must purchase at least 45 hardcover books. What is the maximum number of paperback books the library can purchase?

Show answer & explanation

Answer: 50

Answer

The maximum number of paperback books the library can purchase is 50.

The correct maximum number of paperback books is 50. By expressing the budget constraint as

p \le 200 - \frac{10}{3}h

, we see that the number of paperback books is maximized when the number of hardcover books,

h

, is minimized. Since the library must buy at least 45 hardcover books, we substitute

h = 45

into the inequality to get

p \le 50

. This combination also satisfies the total book constraint because

50 + 45 = 95

, which is less than or equal to 120.

Step-by-Step Solution

1

Set up the system of inequalities representing the library's constraints.

Let

p

be the number of paperbacks and

h

be the number of hardcovers. The constraints are

6p + 20h \le 1200

,

p + h \le 120

, and

h \ge 45

.

This translates the word problem into mathematical inequalities.

2

Isolate the variable

p

in the budget inequality to express it in terms of

h

.

p \le 200 - \frac{10}{3}h

This shows how the number of paperbacks depends on the number of hardcovers purchased.

3

Maximize

p

by using the minimum possible value of

h

.

p \le 200 - \frac{10}{3}(45) = 50

To maximize

p

, we must minimize

h

because buying more hardcovers decreases the remaining budget for paperbacks. The minimum value for

h

is 45.

4

Verify if the solution satisfies the remaining total books constraint.

50 + 45 = 95 \le 120

, which is true.

This ensures the solution is feasible and does not violate any other constraints.

Key Concept

Systems of Linear Inequalities in Two Variables

Question 28Question →

A beverage company produces a fruit juice blend using apple juice and cranberry juice. Let $x$ represent the volume, in liters, of apple juice, and let $y$ represent the volume, in liters, of cranberry juice used in one batch of the blend. The production constraints for each batch are modeled by the system of inequalities below:

x + y \geq 100

y \geq \frac{1}{3}x

y \leq -\frac{1}{2}x + 75

What is the maximum possible volume, in liters, of cranberry juice that can be used in a single batch?

Show answer & explanation

Answer: 50

Answer

50

To find the maximum possible value of

y

, we examine the bounds of the feasible region. Rewriting the inequalities in terms of

x

gives

x \geq 100 - y

and

x \leq 150 - 2y

. Since

x

must satisfy both inequalities simultaneously, the lower bound must be less than or equal to the upper bound, meaning

100 - y \leq 150 - 2y

. Solving this inequality for

y

gives

y \leq 50

. The point

(50, 50)

also satisfies the remaining constraint

y \geq \frac{1}{3}x

, confirming that 50 is the maximum possible value.

Step-by-Step Solution

1

Isolate

x

in the first inequality:

x + y \geq 100

becomes

x \geq 100 - y

.

x \geq 100 - y

This establishes a lower bound for the volume of apple juice in terms of the volume of cranberry juice.

2

Isolate

x

in the third inequality:

y \leq -\frac{1}{2}x + 75

can be rewritten by subtracting 75 from both sides to get

y - 75 \leq -\frac{1}{2}x

, then multiplying by

-2

and reversing the inequality sign to get

x \leq 150 - 2y

.

x \leq 150 - 2y

This establishes an upper bound for the volume of apple juice in terms of the volume of cranberry juice.

3

Combine the lower and upper bounds of

x

:

100 - y \leq x \leq 150 - 2y

. For a valid solution to exist, the lower bound must be less than or equal to the upper bound.

100 - y \leq 150 - 2y

This allows us to solve directly for the constraint on the cranberry juice volume

y

.

4

Solve the inequality

100 - y \leq 150 - 2y

by adding

2y

to both sides and subtracting 100 from both sides.

y \leq 50

This determines that the maximum possible value for

y

is 50.

5

Substitute

y = 50

back into the compound inequality to find

x

:

100 - 50 \leq x \leq 150 - 2(50) \implies 50 \leq x \leq 50

, which means

x = 50

. Check if the point

(50, 50)

satisfies the remaining inequality

y \geq \frac{1}{3}x

:

50 \geq \frac{1}{3}(50) \approx 16.67

, which is true.

The point

(50, 50)

is a valid solution to the system.

This confirms that 50 is a achievable value within the system of inequalities.

Key Concept

Finding the boundary limits and optimization points within a system of linear inequalities in two variables.

Question 29Question →

In the $xy$ -plane, the region $R$ consists of all points $(x, y)$ that satisfy the system of inequalities below:

\begin{aligned} 3x - 4y &\geq -12 \\ y &\geq 2x - 7 \\ x + 2y &\geq 2 \end{aligned}

If $(x, y)$ is a point in region $R$ , what is the maximum possible value of the expression $x - 2y$ ?

2.0

4.4

5.0

-5.2

Show answer & explanation

Answer: 4.4

Answer

4.4

The maximum value of a linear expression over a closed, bounded polygonal region must occur at one of the vertices of that region. By solving the systems of equations for each pair of boundary lines, the vertices of the triangular region are found to be

(8, 9)

,

(3.2, -0.6)

, and

(-1.6, 1.8)

. Evaluating the expression

x - 2y

at each vertex gives

-10

at

(8, 9)

,

4.4

at

(3.2, -0.6)

, and

-5.2

at

(-1.6, 1.8)

. Comparing these values,

4.4

is the maximum possible value.

Step-by-Step Solution

1

Find the intersection points of the boundary equations by solving the three systems of two linear equations.

System 1 (

3x - 4y = -12

and

y = 2x - 7

) yields

(8, 9)

.
System 2 (

y = 2x - 7

and

x + 2y = 2

) yields

(3.2, -0.6)

.
System 3 (

3x - 4y = -12

and

x + 2y = 2

) yields

(-1.6, 1.8)

.

The boundaries of the region are defined by these lines, so the vertices of the region occur at their intersections.

2

Verify that the region defined by the inequalities is the interior of the triangle formed by these vertices by testing a point inside, such as

(2, 2)

.

The point

(2, 2)

satisfies all three inequalities:

3(2) - 4(2) = -2 \geq -12

(True)

2 \geq 2(2) - 7 = -3

(True)

2 + 2(2) = 6 \geq 2

(True)

This confirms that the region is bounded and the vertices are the extreme points of the closed triangular region.

3

Evaluate the expression

x - 2y

at each of the three vertices.

At

(8, 9)

:

8 - 2(9) = -10

At

(3.2, -0.6)

:

3.2 - 2(-0.6) = 3.2 + 1.2 = 4.4

At

(-1.6, 1.8)

:

-1.6 - 2(1.8) = -1.6 - 3.6 = -5.2

A linear expression defined over a closed, bounded polygonal region reaches its maximum and minimum values at the vertices of that region.

4

Compare the evaluated values to identify the maximum value.

The maximum value is

4.4

.

Among the three candidate values (

-10

,

4.4

, and

-5.2

),

4.4

is the greatest.

Key Concept

Linear programming vertex method for systems of linear inequalities.

Estimated Time:3m 0s

Question 30Question →

A community garden has a total area of $1,200$ square feet available for planting tomatoes and squash. Each tomato plant requires $8$ square feet of space, and each squash plant requires $15$ square feet of space. The garden coordinators want to plant at least $100$ plants in total, but they want the number of squash plants to be no more than twice the number of tomato plants. If $t$ represents the number of tomato plants and $s$ represents the number of squash plants, which of the following systems of inequalities represents this situation?

\begin{cases} 8t + 15s \geq 1,200 \\ t + s \leq 100 \\ s \leq 2t \end{cases}

\begin{cases} 8t + 15s \leq 1,200 \\ t + s \geq 100 \\ t \leq 2s \end{cases}

\begin{cases} 8t + 15s \leq 1,200 \\ t + s \geq 100 \\ s \leq 2t \end{cases}

\begin{cases} 8t + 15s \leq 1,200 \\ t + s \geq 100 \\ s \geq 2t \end{cases}

Show answer & explanation

Answer: $\begin{cases} 8t + 15s \leq 1,200 \\ t + s \geq 100 \\ s \leq 2t \end{cases}$

Answer

The system of inequalities containing the constraints

8t + 15s \leq 1,200

,

t + s \geq 100

, and

s \leq 2t

.

The correct system of inequalities maps the garden's constraints accurately:

8t + 15s \leq 1,200

limits the space usage to a maximum of

1,200

square feet;

t + s \geq 100

ensures there are at least

100

plants in total; and

s \leq 2t

restricts the number of squash plants to at most twice the number of tomato plants.

Step-by-Step Solution

1

Translate the area constraint.

8t + 15s \leq 1,200

Since the total area available is

1,200

square feet and each tomato plant requires

8

square feet while each squash plant requires

15

square feet, the combined space used by

t

tomato plants and

s

squash plants cannot exceed

1,200

.

2

Translate the total plant count constraint.

t + s \geq 100

The coordinators want at least

100

plants, meaning the sum of the tomato and squash plants must be greater than or equal to

100

.

3

Translate the plant ratio constraint.

s \leq 2t

The number of squash plants (

s

) must be no more than (less than or equal to) twice the number of tomato plants (

2t

).

4

Combine the inequalities into a single system.

\begin{cases} 8t + 15s \leq 1,200 \\ t + s \geq 100 \\ s \leq 2t \end{cases}

Grouping these three constraints forms the mathematical model that represents the given scenario.

Key Concept

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

Question 31Question →

A solution to a system of inequalities in the $xy$ -plane is represented by the point $(2, y)$ . If the system is defined by:

y \geq 4x - 2

y \leq -2x + 10

what is the value of $y$ ?

Show answer & explanation

Answer: 6

Answer

The value of

y

must be exactly 6 because it is bounded both below and above by 6.

Substituting

x = 2

into the system of inequalities yields two constraints:

y \geq 4(2) - 2

, which simplifies to

y \geq 6

, and

y \leq -2(2) + 10

, which simplifies to

y \leq 6

. The only value of

y

that satisfies both statements is 6.

Step-by-Step Solution

1

Substitute

x = 2

into the first inequality

y \geq 4x - 2

y \geq 6

To determine the constraint on the

y

-value from the first inequality at the coordinate

x = 2

.

2

Substitute

x = 2

into the second inequality

y \leq -2x + 10

y \leq 6

To determine the constraint on the

y

-value from the second inequality at the coordinate

x = 2

.

3

Find the value of

y

that satisfies both

y \geq 6

and

y \leq 6

y = 6

For a value to be simultaneously greater than or equal to 6 and less than or equal to 6, it must be equal to 6.

Key Concept

Evaluating a system of linear inequalities at a specific coordinate to determine a unique boundary value

Estimated Time:45s

Question 32Question →

In the

xy

-plane, a point with coordinates

(x, y)

lies in the solution set of the system of inequalities below:

\begin{aligned} 2(x - 5) + y &\leq 16 \\ x + y &\leq 25 \end{aligned}

If

y \geq 8

, what is the maximum possible integer value of

x

?

6

9

13

17

Show answer & explanation

Answer: 9

Answer

9

To find the maximum possible integer value of

x

in the system's solution set under the constraint

y \geq 8

, we find the boundaries for

x

in both inequalities. Simplifying the first inequality,

2(x - 5) + y \leq 16

, gives

2x - 10 + y \leq 16

, which simplifies to

x \leq 13 - 0.5y

. The second inequality,

x + y \leq 25

, simplifies to

x \leq 25 - y

. Since both boundary lines have negative slopes, the maximum value of

x

for any point in the solution set occurs at the minimum value of

y

, which is

y = 8

. Substituting

y = 8

into both simplified inequalities gives

x \leq 9

and

x \leq 17

. To satisfy both inequalities simultaneously,

x

must be less than or equal to the more restrictive upper bound, which is 9. Thus, the maximum possible integer value of

x

is 9.

Step-by-Step Solution

1

Simplify the first inequality to isolate x.

2(x - 5) + y \leq 16 \implies 2x - 10 + y \leq 16 \implies 2x \leq 26 - y \implies x \leq 13 - 0.5y

To determine the boundary condition imposed on x by the first inequality.

2

Simplify the second inequality to isolate x.

x + y \leq 25 \implies x \leq 25 - y

To determine the boundary condition imposed on x by the second inequality.

3

Determine the maximum value of x under the condition

y \geq 8

.

For the first inequality, the maximum value of x occurs at the minimum value of y (

y = 8

), which gives

x \leq 13 - 0.5(8) = 9

. For the second inequality, the maximum value of x occurs at

y = 8

, which gives

x \leq 25 - 8 = 17

.

Because both boundary lines have negative slopes, the maximum value of x is achieved when y is at its minimum.

4

Find the intersection of the constraints to find the maximum possible value of x that satisfies both inequalities.

Since x must be less than or equal to 9 and less than or equal to 17, the maximum integer value that satisfies both conditions is 9.

The solution set must satisfy all inequalities in the system simultaneously.

Key Concept

Systems of Linear Inequalities in Two Variables

Question 33Question →

A student works two part-time jobs: tutoring, which pays $20$ dollars per hour, and working at a bookstore, which pays $12$ dollars per hour. The student can work at most $15$ hours per week and wants to earn at least $220$ dollars per week. If the student works a whole number of hours at each job, what is the minimum number of hours the student must tutor to meet these requirements?

Show answer & explanation

Answer: 5

Answer

5

The system of inequalities modeling the scenario consists of

x + y \leq 15

(representing the limit on total hours) and

20x + 12y \geq 220

(representing the earning target), where

x

is the hours of tutoring and

y

is the hours at the bookstore. To find the minimum hours of tutoring, we assume the maximum bookstore hours

y = 15 - x

and substitute it into the earnings equation:

20x + 12(15 - x) \geq 220

. Simplifying yields

8x + 180 \geq 220

, which reduces to

8x \geq 40

, or

x \geq 5

. The minimum integer value that satisfies this condition is 5.

Step-by-Step Solution

1

Define variables and set up the constraint for total hours worked.

x + y \leq 15

Let

x

be the number of hours tutoring and

y

be the number of hours working at the bookstore. The total hours cannot exceed 15.

2

Set up the constraint for the minimum weekly earnings.

20x + 12y \geq 220

Tutoring pays

20

dollars per hour and the bookstore pays

12

dollars per hour, and the total earnings must be at least

220

dollars.

3

Substitute the maximum value of

y

in terms of

x

into the earnings inequality.

20x + 12(15 - x) \geq 220

To minimize

x

, we must maximize

y

. From

x + y \leq 15

, the maximum value of

y

is

15 - x

.

4

Solve the inequality for

x

.

x \geq 5

Distribute and simplify:

20x + 180 - 12x \geq 220 \Rightarrow 8x + 180 \geq 220 \Rightarrow 8x \geq 40 \Rightarrow x \geq 5

.

Key Concept

Solving systems of linear inequalities to find optimal boundary values in context.

Estimated Time:1m 30s

Question 34Question →

A manufacturer produces standard chairs and deluxe chairs. Each standard chair requires $2$ hours of assembly and $1$ hour of finishing. Each deluxe chair requires $3$ hours of assembly and $2$ hours of finishing. The manufacturer has a maximum of $240$ hours available for assembly and a maximum of $150$ hours available for finishing each day. Additionally, the manufacturer must produce at least $10$ deluxe chairs daily. If $x$ represents the number of standard chairs produced daily and $y$ represents the number of deluxe chairs produced daily, which of the following systems of inequalities best represents this situation?

\begin{aligned} 2x + 3y &\geq 240 \\ x + 2y &\geq 150 \\ y &\geq 10 \end{aligned}

\begin{aligned} 3x + 2y &\leq 240 \\ 2x + y &\leq 150 \\ y &\geq 10 \end{aligned}

\begin{aligned} 2x + 3y &\leq 240 \\ x + 2y &\leq 150 \\ y &\geq 10 \end{aligned}

\begin{aligned} 2x + 3y &\leq 240 \\ x + 2y &\leq 150 \\ y &\leq 10 \end{aligned}

Show answer & explanation

Answer: $\begin{aligned} 2x + 3y &\leq 240 \\ x + 2y &\leq 150 \\ y &\geq 10 \end{aligned}$

Answer

The system containing

2x + 3y \leq 240

,

x + 2y \leq 150

, and

y \geq 10

.

The correct system represents each constraint accurately. The assembly time requirement is

2x + 3y \leq 240

because standard chairs consume

2

hours each, deluxe chairs consume

3

hours each, and the total cannot exceed

240

. The finishing time requirement is

x + 2y \leq 150

since standard chairs consume

1

hour each, deluxe chairs consume

2

hours each, and the total cannot exceed

150

. Finally, the requirement to produce at least

10

deluxe chairs means

y \geq 10

.

Step-by-Step Solution

1

Set up the inequality for assembly time.

2x + 3y \leq 240

Each standard chair (

x

) needs

2

hours and each deluxe chair (

y

) needs

3

hours, with a maximum limit of

240

hours.

2

Set up the inequality for finishing time.

x + 2y \leq 150

Each standard chair (

x

) needs

1

hour and each deluxe chair (

y

) needs

2

hours, with a maximum limit of

150

hours.

3

Set up the inequality for the minimum production limit of deluxe chairs.

y \geq 10

The manufacturer must produce at least

10

deluxe chairs, meaning the quantity must be greater than or equal to

10

.

4

Combine the individual inequalities into a single system.

The final system is

2x + 3y \leq 240

,

x + 2y \leq 150

, and

y \geq 10

.

All three conditions must be satisfied simultaneously.

Key Concept

Systems of Linear Inequalities in Two Variables

Question 35Question →

A delivery driver is loading a van with small boxes and large boxes. Each small box weighs $10$ pounds, and each large box weighs $35$ pounds. The total weight of the boxes in the van cannot exceed $1,000$ pounds. The driver must load at least $4$ times as many small boxes as large boxes. If the driver loads at least $10$ large boxes, what is the maximum number of large boxes that the driver can load?

Show answer & explanation

Answer: 13

Answer

13

To find the maximum number of large boxes,

y

, we set up the system of inequalities:

10x + 35y \le 1000

and

x \ge 4y

. To maximize

y

, we minimize

x

by setting

x = 4y

. Substituting this into the weight constraint gives

10(4y) + 35y \le 1000

, which simplifies to

75y \le 1000

, or

y \le 13.33

. The largest integer satisfying this inequality is 13, which also satisfies

y \ge 10

.

Step-by-Step Solution

1

Define variables for the quantities of each type of box.

Let

x

be the number of small boxes and

y

be the number of large boxes, where

x

and

y

are non-negative integers.

Establishing variables is necessary to translate the verbal constraints into algebraic inequalities.

2

Translate the given constraints into a system of linear inequalities.

The weight limit gives

10x + 35y \le 1000

. The requirement for at least 4 times as many small boxes as large boxes gives

x \ge 4y

. The minimum of 10 large boxes gives

y \ge 10

.

Representing the scenario mathematically allows us to find the boundaries of the solution set.

3

Express the weight inequality in terms of a single variable by using the boundary condition of the second inequality.

To maximize

y

, we want to minimize the weight contributed by the small boxes,

x

. The minimum value of

x

for any

y

is

x = 4y

. Substituting

x = 4y

into the weight inequality gives

10(4y) + 35y \le 1000

.

Finding the extreme case (minimum number of small boxes) provides the upper limit for the number of large boxes.

4

Simplify the inequality and solve for

y

.

40y + 35y \le 1000 \implies 75y \le 1000 \implies y \le \frac{1000}{75} \approx 13.33

.

This calculation determines the upper algebraic bound for the number of large boxes.

5

Determine the maximum integer value for

y

that satisfies the system.

Since the number of boxes must be an integer, the maximum integer value less than or equal to

13.33

is

13

. Since

13 \ge 10

, this satisfies all constraints.

Real-world quantities like boxes must be whole numbers, so we take the largest integer within the solution set.

Key Concept

Solving systems of linear inequalities in two variables to optimize a value under constraints.

Question 36Question →

In the $xy$ -plane, a point with coordinates $(a, b)$ lies in the solution set of the system of inequalities below:

\begin{aligned} y &\ge -2x + 8 \\ y &< \frac{1}{2}x + 3 \end{aligned}

If $a$ and $b$ are integers, what is the minimum possible value of $a$ ?

1

2

3

4

Show answer & explanation

Answer: 3

Answer

3

The correct value is 3. Substituting

x = 3

into the system gives

y \ge 2

and

y < 4.5

. Since the coordinate

b

must be an integer,

b

can be 2, 3, or 4, showing that a valid integer coordinate pair exists. For any integer

x < 3

, there are no integer solutions for

y

.

Step-by-Step Solution

1

Find the intersection point of the boundary lines of the system of inequalities by setting their corresponding equations equal.

-2x + 8 = \frac{1}{2}x + 3 \implies 5 = 2.5x \implies x = 2

This determines the x-coordinate where the boundaries of the two inequality regions intersect.

2

Evaluate the system of inequalities at the intersection point

x = 2

.

y \ge -2(2) + 8 \implies y \ge 4

and

y < \frac{1}{2}(2) + 3 \implies y < 4

. This yields

4 \le y < 4

, which has no solution.

This checks if any valid solution exists at the boundary intersection point.

3

Test consecutive integer values of

x

greater than 2 to find the first value that allows at least one integer

y

.

Substituting

x = 3

gives

y \ge -2(3) + 8 \implies y \ge 2

and

y < \frac{1}{2}(3) + 3 \implies y < 4.5

. The integers that satisfy

2 \le y < 4.5

are

y = 2, 3,

and

4

.

Since

a = 3

is the smallest integer greater than 2 that yields integer solutions for

y

, it is the minimum possible value.

Key Concept

Identifying integer coordinate solutions in systems of linear inequalities

Estimated Time:2m 0s

Question 37Question →

An artist creates custom oil paintings and charcoal sketches.

* Each oil painting, $x$ , requires $4$ hours of setup time and $2$ hours of finishing time.
* Each charcoal sketch, $y$ , requires $1$ hour of setup time and $3$ hours of finishing time.

Each week, the artist can spend at most $20$ hours on setup time and at most $30$ hours on finishing time. Which of the following systems of inequalities represents this situation?

\begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \end{aligned}

\begin{aligned} 4x + 2y &\le 20 \\ x + 3y &\le 30 \end{aligned}

\begin{aligned} 4x + y &\le 20 \\ 2x + 3y &\le 30 \end{aligned}

\begin{aligned} 4x + y &\le 30 \\ 2x + 3y &\le 20 \end{aligned}

Show answer & explanation

Answer: $\begin{aligned} 4x + y &\le 20 \\ 2x + 3y &\le 30 \end{aligned}$

Answer

The system of inequalities consisting of the inequalities

4x + y \le 20

and

2x + 3y \le 30

The variable

x

represents the number of oil paintings and

y

represents the number of charcoal sketches. Setup time constraint: each oil painting requires

4

hours and each charcoal sketch requires

1

hour. Thus, the total setup time is

4x + y

hours. Since the artist can spend at most

20

hours on setup time, this is represented by the inequality

4x + y \le 20

. Finishing time constraint: each oil painting requires

2

hours and each charcoal sketch requires

3

hours. Thus, the total finishing time is

2x + 3y

hours. Since the artist can spend at most

30

hours on finishing time, this is represented by the inequality

2x + 3y \le 30

. Combining these gives the correct system.

Step-by-Step Solution

1

Identify the variables and setup times for each type of artwork to write the setup time inequality.

The total setup time is

4x + y

hours. Since the limit is at most

20

hours, the inequality is

4x + y \le 20

.

To establish the mathematical representation of the setup time constraint.

2

Identify the variables and finishing times for each type of artwork to write the finishing time inequality.

The total finishing time is

2x + 3y

hours. Since the limit is at most

30

hours, the inequality is

2x + 3y \le 30

.

To establish the mathematical representation of the finishing time constraint.

3

Combine the two inequalities into a system.

The system of inequalities is

4x + y \le 20

and

2x + 3y \le 30

.

To find the system that represents both constraints simultaneously.

Key Concept

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

Question 38Question →

A bounded region in the $xy$ -plane is defined by the following system of inequalities:

\begin{aligned} y &\ge \frac{1}{2}x + 2 \\ y &\le 2x - 1 \\ y &\le -x + 11 \end{aligned}

What is the maximum $y$ -coordinate of any point within this region?

Show answer & explanation

Answer: 7

Answer

The maximum y-coordinate of any point in the bounded region is 7.

The solution region is a triangle with vertices at

(2, 3)

,

(4, 7)

, and

(6, 5)

. Since all inequalities are less-than-or-equal-to or greater-than-or-equal-to, the boundary points are included in the solution set. The maximum y-value occurs at the vertex

(4, 7)

, which gives a maximum y-coordinate of 7.

Step-by-Step Solution

1

Determine the equations of the boundary lines.

The boundary equations are

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

,

y = 2x - 1

, and

y = -x + 11

.

These equations represent the boundaries of the system of inequalities.

2

Find the intersection point of

y = 2x - 1

and

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

.

Solving

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

yields

x = 2

, which gives

y = 3

. The intersection is

(2, 3)

.

This is one vertex of the bounded region.

3

Find the intersection point of

y = 2x - 1

and

y = -x + 11

.

Solving

2x - 1 = -x + 11

yields

x = 4

, which gives

y = 7

. The intersection is

(4, 7)

.

This is the second vertex of the bounded region.

4

Find the intersection point of

y = -x + 11

and

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

.

Solving

-x + 11 = \frac{1}{2}x + 2

yields

x = 6

, which gives

y = 5

. The intersection is

(6, 5)

.

This is the third vertex of the bounded region.

5

Compare the y-coordinates of the three vertices of the bounded region.

The y-coordinates of the vertices

(2, 3)

,

(4, 7)

, and

(6, 5)

are 3, 7, and 5, respectively. The maximum value is 7.

The maximum value of a linear coordinate over a bounded convex region occurs at one of its vertices.

Key Concept

Finding the vertices of a bounded region defined by a system of linear inequalities and optimizing a coordinate value.

Question 39Question →

A landscaping company is planting $x$ maple trees and $y$ pine trees in a park. The number of trees of each type must satisfy the system of inequalities below:

\begin{aligned} 30x + 40y &\le 360 \\ x + y &\ge 10 \\ x &\ge 4 \end{aligned}

What is the maximum number of pine trees the company can plant?

Show answer & explanation

Answer: 6

Answer

The maximum number of pine trees the company can plant is 6.

To find the maximum number of pine trees,

y

, we look at the boundary constraints. The constraint

x \ge 4

states that at least

4

maple trees must be planted. Since planting fewer maple trees leaves more of the budget for pine trees, we minimize

x

by setting

x = 4

. Substituting this value into the budget inequality

30x + 40y \le 360

gives

120 + 40y \le 360

. Solving for

y

yields

40y \le 240

, which simplifies to

y \le 6

. We then verify that the solution

(4, 6)

satisfies the total tree constraint

x + y \ge 10

, which it does since

4 + 6 = 10

. Thus, the maximum number of pine trees is 6.

Step-by-Step Solution

1

Substitute the minimum possible value of

x

into the first inequality.

Since

x \ge 4

, the smallest possible value for

x

is

4

. Substituting

x = 4

into

30x + 40y \le 360

yields:

30(4) + 40y \le 360

120 + 40y \le 360

To maximize the value of

y

under the resource constraint, we must minimize the value of

x

.

2

Solve the inequality for

y

.

40y \le 240

y \le 6

This establishes that the maximum possible value for

y

based on the budget constraint is

6

.

3

Verify that

(4, 6)

satisfies all inequalities in the system.

Checking the second inequality:

x + y \ge 10 \implies 4 + 6 = 10 \ge 10

, which is true. The third inequality

x \ge 4 \implies 4 \ge 4

is also true.

A coordinate pair must satisfy all inequalities in the system to be a valid solution.

Key Concept

To find the maximum value of a variable in a system of inequalities with constraints, analyze the boundary lines and the intersection points of the feasible region.

Question 40Question →

A community food bank is preparing two types of relief packages: standard boxes and family boxes. Let $x$ represent the number of standard boxes and $y$ represent the number of family boxes. The food bank needs to prepare at least 120 boxes in total. Each standard box contains 3 cans of soup and 2 packages of grain. Each family box contains 6 cans of soup and 5 packages of grain. The food bank has a maximum of 600 cans of soup and a maximum of 450 packages of grain available. Which of the following systems of inequalities represents this situation?

\begin{aligned} x + y &\le 120 \\ 3x + 6y &\ge 600 \\ 2x + 5y &\ge 450 \end{aligned}

\begin{aligned} x + y &\ge 120 \\ 6x + 3y &\le 600 \\ 5x + 2y &\le 450 \end{aligned}

\begin{aligned} x + y &\ge 120 \\ 3x + 6y &\le 600 \\ 2x + 5y &\le 450 \end{aligned}

\begin{aligned} x + y &\ge 120 \\ 3x + 2y &\le 600 \\ 6x + 5y &\le 450 \end{aligned}

Show answer & explanation

Answer: $\begin{aligned} x + y &\ge 120 \\ 3x + 6y &\le 600 \\ 2x + 5y &\le 450 \end{aligned}$

Answer

The system of inequalities with

x + y \ge 120

,

3x + 6y \le 600

, and

2x + 5y \le 450

.

The correct system of inequalities translates the constraint of preparing at least 120 boxes as

x + y \ge 120

. The maximum limit of 600 cans of soup translates as

3x + 6y \le 600

, since each standard box contains 3 cans and each family box contains 6. The maximum limit of 450 packages of grain translates as

2x + 5y \le 450

, since each standard box contains 2 packages and each family box contains 5.

Step-by-Step Solution

1

Translate the total package constraint.

x + y \ge 120

The phrase 'at least 120 boxes in total' indicates that the sum of

x

and

y

must be greater than or equal to 120.

2

Translate the soup constraint.

3x + 6y \le 600

Since each standard box (

x

) has 3 cans of soup and each family box (

y

) has 6 cans of soup, the total soup used is

3x + 6y

. The food bank has 'a maximum of 600 cans,' so this total must be less than or equal to 600.

3

Translate the grain constraint.

2x + 5y \le 450

Since each standard box (

x

) has 2 packages of grain and each family box (

y

) has 5 packages of grain, the total grain used is

2x + 5y

. The food bank has 'a maximum of 450 packages,' so this total must be less than or equal to 450.

Key Concept

Systems of Linear Inequalities in Two Variables

Estimated Time:1m 30s

Systems of Linear Inequalities in Two Variables

Answer

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