Systems of Linear Inequalities in Two Variables Practice Questions — SAT

Question 1Question →

A digital marketing firm allocates its monthly advertising budget between search engine campaigns and social media campaigns. Let $x$ represent the amount, in thousands of dollars, spent on search engine campaigns, and let $y$ represent the amount, in thousands of dollars, spent on social media campaigns. The system of inequalities below represents the firm's monthly constraints:

\begin{aligned} x &\ge 2y \\ x + y &\le 25 \\ 3x + 4y &\le 80 \\ y &\ge 3 \end{aligned}

Based on these constraints, what is the maximum possible amount, in thousands of dollars, the firm can spend on social media campaigns?

Show answer & explanation

Answer: 8

Answer

8

To find the maximum possible value of

y

, we analyze the boundaries of the feasible region. The boundary lines of the system are

x = 2y

,

x + y = 25

,

3x + 4y = 80

, and

y = 3

. We can find the upper limit of

y

by combining the inequalities

x \ge 2y

and

3x + 4y \le 80

. Multiplying the first inequality by

3

gives

3x \ge 6y

. Adding

4y

to both sides yields

3x + 4y \ge 10y

. Because

3x + 4y \le 80

, it follows that

10y \le 80

, which simplifies to

y \le 8

. We verify that the point

(16, 8)

satisfies the remaining inequalities:

16 + 8 = 24 \le 25

and

8 \ge 3

, which are both true. Thus, the maximum possible value of

y

is

8

.

Step-by-Step Solution

1

Relate the variables using the constraints to establish an upper bound for

y

.

Since

x \ge 2y

, multiplying both sides by

3

gives

3x \ge 6y

.

This allows us to express the

3x

term in the cost inequality in terms of

y

to determine the maximum boundary.

2

Substitute

3x \ge 6y

into the inequality

3x + 4y \le 80

.

We get

6y + 4y \le 3x + 4y \le 80

, which simplifies to

10y \le 80

.

This establishes a direct upper limit for

y

based on the intersection of the two active boundary lines.

3

Solve the inequality

10y \le 80

for

y

.

y \le 8

.

This determines that the maximum possible value for

y

under these constraints is

8

.

4

Verify that the point corresponding to

y = 8

satisfies all other constraints in the system.

When

y = 8

, the boundary

x = 2y

gives

x = 16

. Checking

(16, 8)

against all inequalities:
-

16 \ge 2(8) \implies 16 \ge 16

(True)
-

16 + 8 \le 25 \implies 24 \le 25

(True)
-

3(16) + 4(8) \le 80 \implies 80 \le 80

(True)
-

8 \ge 3

(True)

We must verify that the optimal vertex lies within the feasible region defined by all four inequalities.

Key Concept

Maximizing a coordinate value within a bounded feasible region defined by a system of linear inequalities.

Question 2Question →

\begin{cases} y \geq -2x + 4 \\ y < \frac{1}{2}x + 2 \end{cases}

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

(1, 2)

(0, -1)

(2, 3)

(0, 3)

Show answer & explanation

Answer: $(1, 2)$

Answer

The ordered pair

(1, 2)

is a solution to the system of inequalities.

Substituting the coordinates of the point stating

(1, 2)

into the system of inequalities produces two true statements:

2 \geq -2(1) + 4

simplifies to

2 \geq 2

, which is true; and

2 < \frac{1}{2}(1) + 2

simplifies to

2 < 2.5

, which is also true. Therefore, the ordered pair

(1, 2)

is a solution to the system.

Step-by-Step Solution

1

Substitute the coordinates of each option into the first inequality,

y \geq -2x + 4

.

For

(1, 2)

,

2 \geq 2

(True). For

(0, -1)

,

-1 \geq 4

(False). For

(2, 3)

,

3 \geq 0

(True). For

(0, 3)

,

3 \geq 4

(False).

To determine which ordered pairs satisfy the first inequality.

2

Substitute the coordinates of the remaining candidate pairs,

(1, 2)

and

(2, 3)

, into the second inequality,

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

.

For

(1, 2)

,

2 < 2.5

(True). For

(2, 3)

,

3 < 3

(False).

To determine which of the remaining candidates also satisfies the second inequality.

3

Identify the ordered pair that satisfies both inequalities in the system.

Only

(1, 2)

satisfies both inequalities.

An ordered pair is a solution to a system of inequalities if and only if it satisfies all inequalities in the system.

Key Concept

Verifying solutions to a system of linear inequalities by substitution.

Question 3Question →

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

\begin{aligned} y &\geq x + 1 \\ y &> -2x + 4 \end{aligned}

(1, 3)

(3, 1)

(0, 2)

(-1, 5)

Show answer & explanation

Answer: $(1, 3)$

Answer

The ordered pair

(1, 3)

The ordered pair

(1, 3)

is the correct answer because substituting

x = 1

and

y = 3

into the system of inequalities results in two true statements:

3 \geq 1 + 1

simplifies to

3 \geq 2

(true), and

3 > -2(1) + 4

simplifies to

3 > 2

(true).

Step-by-Step Solution

1

Substitute the coordinates of the ordered pair

(1, 3)

into the first inequality,

y \geq x + 1

.

3 \geq 1 + 1 \Rightarrow 3 \geq 2

, which is true.

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

2

Substitute the coordinates of the ordered pair

(1, 3)

into the second inequality,

y > -2x + 4

.

3 > -2(1) + 4 \Rightarrow 3 > 2

, which is true.

The ordered pair must also satisfy the second inequality to be a solution to the system.

3

Verify that the other ordered pairs do not satisfy both inequalities.

None of the other options satisfy both inequalities:

(3, 1)

fails the first inequality,

(0, 2)

fails the second inequality, and

(-1, 5)

fails the second inequality.

Ensuring that there is exactly one correct solution.

Key Concept

A solution to a system of linear inequalities in two variables is an ordered pair

(x, y)

that makes all inequalities in the system true.

Alternative Method

Alternatively, you can graph the two inequalities on a coordinate plane. The region representing the solution to the system is the overlapping shaded area of both inequalities. By plotting each point, you can visually identify which point lies within this double-shaded region.

Estimated Time:1m 0s

Question 4Question →

A student wants to buy $x$ notebooks and $y$ pens. The student must buy at least $3$ notebooks. Each notebook costs $3$ dollars and each pen costs $2$ dollars. If the student can spend a maximum of $15$ dollars, what is the maximum number of pens the student can buy?

Show answer & explanation

Answer: 3

Answer

3

The correct answer is 3. The constraints are represented by the system of inequalities

x \geq 3

and

3x + 2y \leq 15

. To find the maximum number of pens

y

, we minimize the number of notebooks

x

by setting

x = 3

. Substituting

x = 3

into the budget inequality gives

3(3) + 2y \leq 15

, which simplifies to

9 + 2y \leq 15

. Subtracting

9

from both sides results in

2y \leq 6

. Dividing by

2

yields

y \leq 3

. Therefore, the maximum number of pens the student can buy is 3.

Step-by-Step Solution

1

Set up the system of inequalities from the given constraints.

The system of inequalities is

x \geq 3

and

3x + 2y \leq 15

, where

x

represents the number of notebooks and

y

represents the number of pens.

To represent the minimum number of notebooks required and the total budget limit.

2

Substitute the minimum possible value of

x

into the budget inequality to maximize

y

.

Setting

x = 3

gives

3(3) + 2y \leq 15

, which simplifies to

9 + 2y \leq 15

.

Since the cost per notebook is positive, minimizing the number of notebooks leaves the maximum amount of budget for purchasing pens.

3

Solve the inequality for

y

.

2y \leq 6 \implies y \leq 3

.

To find the upper bound for the number of pens the student can buy.

Key Concept

Solving a system of linear inequalities in a real-world context to find a maximum value.

Question 5Question →

In the $xy$ -plane, a point $(x, y)$ is a solution to the system of inequalities below.

y \leq -3x + 14

y \leq 4x - 7

What is the maximum possible value of $y$ ?

Show answer & explanation

Answer: 5

Answer

The maximum possible value of

y

is 5.

The maximum value of

y

is found at the intersection of the two boundary lines,

y = -3x + 14

and

y = 4x - 7

. Solving the system of equations gives

x = 3

and

y = 5

. Because the inequalities are both of the 'less than or equal to' type, any point in the solution set must lie on or below both lines. For

x < 3

, the value of

y

is limited by

y \leq 4x - 7

, which is strictly less than 5. For

x > 3

, the value of

y

is limited by

y \leq -3x + 14

, which is also strictly less than 5. Thus, the maximum value of

y

is 5.

Step-by-Step Solution

1

Set the two boundary equations equal to find their intersection point.

-3x + 14 = 4x - 7

The maximum value of

y

under these constraints occurs at the intersection of the two boundary lines.

2

Solve the equation for

x

.

x = 3

Isolating

x

determines the x-coordinate of the intersection point.

3

Substitute

x = 3

back into one of the boundary equations to solve for

y

.

y = 5

This yields the y-coordinate of the intersection point, which is the upper bound for

y

in the solution set.

Key Concept

Determining the maximum value of a variable in the solution set of a system of linear inequalities by finding the intersection of their boundary lines.

Estimated Time:1m 30s

Question 6Question →

A graphic designer is exporting images for a website. There are two types of images: standard images and high-resolution images. Each standard image, $x$ , requires 3 megabytes (MB) of storage and takes 2 seconds to upload. Each high-resolution image, $y$ , requires 8 MB of storage and takes 5 seconds to upload. The designer must limit the total storage of the exported images to at most 120 MB and the total upload time to at least 60 seconds. Which of the following systems of inequalities models this situation?

3x + 2y \leq 120

8x + 5y \geq 60

3x + 8y \geq 120

2x + 5y \leq 60

8x + 3y \leq 120

5x + 2y \geq 60

3x + 8y \leq 120

2x + 5y \geq 60

Show answer & explanation

Answer: $3x + 8y \leq 120$
$2x + 5y \geq 60$

Answer

The system of inequalities is

3x + 8y \leq 120

and

2x + 5y \geq 60

.

The correct system models the limits correctly. The storage constraint is represented by

3x + 8y \leq 120

since 'at most' corresponds to a less-than-or-equal-to sign. The upload time constraint is represented by

2x + 5y \geq 60

since 'at least' corresponds to a greater-than-or-equal-to sign.

Step-by-Step Solution

1

Set up the inequality for the storage constraint.

3x + 8y \leq 120

Each standard image uses 3 megabytes and each high-resolution image uses 8 megabytes. The total storage must be at most (less than or equal to) 120 megabytes.

2

Set up the inequality for the upload time constraint.

2x + 5y \geq 60

Each standard image takes 2 seconds and each high-resolution image takes 5 seconds to upload. The total upload time must be at least (greater than or equal to) 60 seconds.

3

Combine both inequalities into a system.

The system containing

3x + 8y \leq 120

and

2x + 5y \geq 60

.

Both constraints must be satisfied simultaneously.

Key Concept

Translating verbal constraints into systems of linear inequalities in two variables.

Question 7Question →

A system of inequalities is shown below.

\begin{aligned} y &\leq -3x + 15 \\ y &\leq 2x + 2 \end{aligned}

If

(x, y)

is a solution to the system, what is the maximum possible integer value of

y

?

6

7

8

15

Show answer & explanation

Answer: 7

Answer

7

The maximum value of

y

in the system occurs at the intersection of the boundary lines

y = -3x + 15

and

y = 2x + 2

. Solving for the intersection gives

x = 2.6

and

y = 7.2

. Since any point in the solution set must satisfy

y \leq 7.2

, the largest possible integer value for

y

is 7. We can confirm this is achievable because when

y = 7

,

x

can be any value in the range

[2.5, 2.67]

(such as

x=2.6

), which satisfies both inequalities.

Step-by-Step Solution

1

Find the intersection point of the boundary lines.

Set the two equations equal to find the

x

-coordinate:

-3x + 15 = 2x + 2 \implies 5x = 13 \implies x = 2.6

.

The maximum value of

y

in a system bounded from above by two lines with opposing slopes occurs at their point of intersection.

2

Calculate the corresponding maximum

y

-value.

Substitute

x = 2.6

into either boundary equation:

y = 2(2.6) + 2 = 7.2

.

This determines the absolute upper bound of

y

for any coordinate pair in the solution set.

3

Determine the maximum integer value of

y

.

Since

y \leq 7.2

, the largest integer value that

y

can take is 7.

The question asks specifically for the maximum integer value of

y

, and since

7 \leq 7.2

, a solution exists at this

y

-value.

4

Verify that a solution exists for

y = 7

.

For

y = 7

, the system requires

7 \leq -3x + 15 \implies x \leq 2.67

and

7 \leq 2x + 2 \implies x \geq 2.5

. The interval

[2.5, 2.67]

is non-empty, so valid solutions (e.g.,

(2.6, 7)

) exist.

This confirms that

y = 7

is attainable within the feasible region.

Key Concept

Optimization of variables within systems of linear inequalities

Question 8Question →

A graphic designer allocates their weekly work time between designing layouts, which takes $d$ hours, and client consultations, which takes $c$ hours. The designer works a maximum of 35 hours per week. Additionally, the designer spends at least twice as much time designing layouts as they do in client consultations. Which of the following systems of inequalities represents this situation?

\begin{aligned} d + c &\leq 35 \\ d &\geq 2c \end{aligned}

\begin{aligned} d + c &\geq 35 \\ d &\leq 2c \end{aligned}

\begin{aligned} d + c &\leq 35 \\ c &\geq 2d \end{aligned}

\begin{aligned} d + c &\geq 35 \\ c &\leq 2d \end{aligned}

Show answer & explanation

Answer: $\begin{aligned} d + c &\leq 35 \\ d &\geq 2c \end{aligned}$

Answer

The system with the inequalities

d + c \leq 35

and

d \geq 2c

The system containing the inequalities

d + c \leq 35

and

d \geq 2c

is correct. The constraint 'works a maximum of 35 hours per week' means the sum of the hours,

d + c

, must be less than or equal to 35 (

d + c \leq 35

). The constraint 'spends at least twice as much time designing layouts as they do in client consultations' means that design hours,

d

, must be greater than or equal to twice the consultation hours,

c

(

d \geq 2c

).

Step-by-Step Solution

1

Identify the inequality representing the limit on total weekly hours.

d + c \leq 35

The phrase 'works a maximum of 35 hours per week' means the total hours spent on designing (

d

) and consulting (

c

) cannot exceed 35.

2

Identify the inequality representing the relative relationship between designing and consulting hours.

d \geq 2c

The phrase 'at least twice as much time designing layouts as they do in client consultations' means the design hours (

d

) must be greater than or equal to two times the consultation hours (

2c

).

3

Combine the individual inequalities into a system.

The system consists of

d + c \leq 35

and

d \geq 2c

.

Both conditions must be satisfied simultaneously to represent the designer's weekly schedule constraint.

Key Concept

Systems of Linear Inequalities in Two Variables

Estimated Time:1m 0s

Question 9Question →

A closed triangular region in the coordinate plane is defined by the following system of linear inequalities:

\begin{aligned} y &\geq 2x - 4 \\ y &\leq -x + 8 \\ x &\geq 1 \end{aligned}

What is the maximum possible value of the expression $2x + y$ for any point $(x, y)$ that lies within or on the boundary of this region?

Show answer & explanation

Answer: 12

Answer

The maximum possible value of the expression

2x + y

is 12.

To find the maximum possible value of the expression

2x + y

subject to the given system of inequalities, we first identify the boundary lines and find the vertices of the bounded triangular region in the coordinate plane. The boundary lines are

y = 2x - 4

,

y = -x + 8

, and

x = 1

. The intersection of

y = 2x - 4

and

y = -x + 8

occurs at

x = 4, y = 4

, which gives vertex

(4, 4)

. The intersection of

y = 2x - 4

and

x = 1

occurs at

(1, -2)

. The intersection of

y = -x + 8

and

x = 1

occurs at

(1, 7)

. Evaluating the linear expression

2x + y

at these three vertices gives

2(4) + 4 = 12

,

2(1) - 2 = 0

, and

2(1) + 7 = 9

. By the corner point theorem, the maximum value of a linear function on a closed polygonal region occurs at one of the vertices. Comparing the values, the maximum possible value is 12.

Step-by-Step Solution

1

Find the vertex formed by the intersection of the boundary lines

y = 2x - 4

and

y = -x + 8

.

Vertex

A(4, 4)

Setting the two equations equal:

2x - 4 = -x + 8 \implies 3x = 12 \implies x = 4

. Substituting

x = 4

back into either equation yields

y = 4

.

2

Find the vertex formed by the intersection of the boundary line

y = 2x - 4

and the vertical line

x = 1

.

Vertex

B(1, -2)

Substituting

x = 1

into

y = 2x - 4

gives

y = 2(1) - 4 = -2

.

3

Find the vertex formed by the intersection of the boundary line

y = -x + 8

and the vertical line

x = 1

.

Vertex

C(1, 7)

Substituting

x = 1

into

y = -x + 8

gives

y = -1 + 8 = 7

.

4

Evaluate the expression

2x + y

at each of the three vertices.

At

A(4, 4)

, the value is

12

; at

B(1, -2)

, the value is

0

; at

C(1, 7)

, the value is

9

.

According to the Corner Point Theorem of linear programming, the maximum or minimum of a linear objective function on a closed bounded region must occur at one of the vertices.

5

Identify the maximum value from the evaluated points.

The maximum value is 12.

Comparing the values 12, 0, and 9 shows that 12 is the largest value.

Key Concept

Linear Programming and Systems of Inequalities

Question 10Question →

In the $xy$ -plane, the solution set to the system of inequalities below is a bounded region.

\begin{aligned} y &\leq -2x + 21 \\ y &\leq -\frac{1}{2}x + 9 \\ y &\geq -x + 5 \\ x &\geq 2 \\ y &\geq 1 \end{aligned}

What is the maximum possible value of $3x + 2y$ for a point $(x, y)$ in this region?

22

31

32

34

Show answer & explanation

Answer: 34

Answer

34

The correct value is 34. The solution set to the system of inequalities is a bounded pentagonal region. Evaluating the expression

3x + 2y

at each vertex of this region yields:

12

at

(2, 3)

,

22

at

(2, 8)

,

34

at

(8, 5)

,

32

at

(10, 1)

, and

14

at

(4, 1)

. Comparing these values shows that the maximum value is 34.

Step-by-Step Solution

1

Understand the behavior of the objective function over the region

The maximum value of a linear expression

3x + 2y

over a closed, bounded polygonal region must occur at one of the vertices (corner points) of the region.

This is a fundamental theorem of linear programming, which simplifies the search for the maximum value to only the boundary intersections of the system.

2

Determine the vertices of the feasible region

The boundary lines are

y = -2x + 21

,

y = -\frac{1}{2}x + 9

,

y = -x + 5

,

x = 2

, and

y = 1

. Finding their intersections that satisfy all inequalities yields five vertices:

(2, 3)

,

(2, 8)

,

(8, 5)

,

(10, 1)

, and

(4, 1)

.

Solving the pairwise equations of lines that bound the shaded region determines the exact coordinates of all corner points.

3

Evaluate the expression

3x + 2y

at each of the five vertices

At

(2, 3)

:

3(2) + 2(3) = 12

At

(2, 8)

:

3(2) + 2(8) = 22

At

(8, 5)

:

3(8) + 2(5) = 34

At

(10, 1)

:

3(10) + 2(1) = 32

At

(4, 1)

:

3(4) + 2(1) = 14

Calculating the value at each candidate vertex allows us to compare and find the absolute maximum.

4

Compare the evaluated values

The maximum value is

34

, occurring at the vertex

(8, 5)

.

Comparing all calculated values shows that 34 is the largest possible value.

Key Concept

Linear Programming and Bounded Systems of Inequalities

Estimated Time:3m 0s

Question 11Question →

In the

xy

-plane, a system of inequalities is defined as follows:

\begin{aligned} y &\geq 2x - 4 \\ y &\leq -x + 8 \\ y &\geq \frac{1}{2}x \end{aligned}

How many points

(x, y)

with integer coordinates satisfy this system of inequalities?

Show answer & explanation

Answer: 5

Answer

The total number of points with integer coordinates that satisfy the system is 5.

The system of inequalities defines a closed triangular region in the coordinate plane. Finding the vertices of this triangle gives the horizontal boundaries for

x

, which are

\frac{8}{3} \approx 2.67

and

\frac{16}{3} \approx 5.33

. The only integers in this range are

x = 3

,

x = 4

, and

x = 5

. Testing each of these integers in the inequalities shows that when

x = 3

,

y

can be

2, 3, 4,

or

5

(4 points); when

x = 4

,

y

can only be

4

(1 point); and when

x = 5

, there are no valid solutions. Adding these together gives a total of 5 points.

Step-by-Step Solution

1

Find the intersection points of the three boundary lines to define the vertices of the solution region.

The intersection of

y = 2x - 4

and

y = \frac{1}{2}x

is at

x = \frac{8}{3} \approx 2.67, y = \frac{4}{3} \approx 1.33

.
The intersection of

y = -x + 8

and

y = \frac{1}{2}x

is at

x = \frac{16}{3} \approx 5.33, y = \frac{8}{3} \approx 2.67

.
The intersection of

y = 2x - 4

and

y = -x + 8

is at

x = 4, y = 4

.

Finding the vertices determines the exact boundaries of the solution region in the coordinate plane.

2

Identify the integer

x

-coordinates that lie within the horizontal span of the region.

The

x

-coordinates of the region range from

\frac{8}{3} \approx 2.67

to

\frac{16}{3} \approx 5.33

. The integers in this interval are

x = 3

,

x = 4

, and

x = 5

.

Since both coordinates must be integers, we only need to test integer values of

x

that fall within the boundaries of the region.

3

Find the integer

y

-values for each candidate

x

-value.

For

x = 3

: The inequalities require

y \geq 2(3) - 4 = 2

,

y \leq -3 + 8 = 5

, and

y \geq \frac{1}{2}(3) = 1.5

. Thus,

2 \leq y \leq 5

. The integer solutions are

y = 2, 3, 4, 5

(4 points).
For

x = 4

: The inequalities require

y \geq 2(4) - 4 = 4

,

y \leq -4 + 8 = 4

, and

y \geq \frac{1}{2}(4) = 2

. Thus,

4 \leq y \leq 4

, which means

y = 4

(1 point).
For

x = 5

: The inequalities require

y \geq 2(5) - 4 = 6

and

y \leq -5 + 8 = 3

. No real number

y

can satisfy both

y \geq 6

and

y \leq 3

(0 points).

Evaluating the system at each candidate

x

-value reveals the set of matching integer

y

-values.

4

Sum the number of points found for each integer

x

-value.

4 points (when

x = 3

) + 1 point (when

x = 4

) + 0 points (when

x = 5

) = 5 points.

This yields the total number of integer coordinate pairs

(x, y)

that satisfy the system of inequalities.

Key Concept

Analyzing a bounded region defined by a system of linear inequalities to find discrete integer solutions (lattice points).

Alternative Method

Graph the three boundary lines on a grid:

y = 2x - 4

(solid line, shaded above),

y = -x + 8

(solid line, shaded below), and

y = \frac{1}{2}x

(solid line, shaded above). Identify the triangular intersection region on the grid and count the grid intersections (lattice points) that lie within or on the boundaries of this shaded triangle.

Estimated Time:2m 30s

Question 12Question →

An online store sells small gift boxes for $10$ dollars each and large gift boxes for $15$ dollars each. A customer wants to buy a total of at most $10$ boxes. If the customer must spend at least $120$ dollars on the boxes, what is the minimum number of large gift boxes the customer must buy?

Show answer & explanation

Answer: 4

Answer

The minimum number of large gift boxes the customer must buy is 4.

Using the constraints from the problem, we set up the system of inequalities:

x + y \leq 10

and

10x + 15y \geq 120

. Expressing the first inequality as

x \leq 10 - y

and substituting it into the second gives

10(10 - y) + 15y \geq 120

. Simplifying this results in

100 + 5y \geq 120

, which simplifies to

5y \geq 20

, or

y \geq 4

. Therefore, the minimum number of large gift boxes the customer must buy is

4

.

Step-by-Step Solution

1

Define variables for the quantities of each box.

Let

x

be the number of small boxes and

y

be the number of large boxes.

Establishing variables is necessary to translate the word problem into algebraic inequalities.

2

Write the system of inequalities representing the constraints.

x + y \leq 10

and

10x + 15y \geq 120

The total number of boxes is at most

10

, and the total cost must be at least

120

dollars.

3

Express

x

in terms of

y

using the first inequality.

x \leq 10 - y

This allows substitution into the second inequality to solve for the target variable

y

.

4

Substitute

x \leq 10 - y

into the second inequality and simplify.

100 + 5y \geq 120

Solving the resulting single-variable inequality will determine the possible values of

y

.

5

Solve the inequality for

y

.

y \geq 4

Subtracting

100

and dividing by

5

isolates

y

to show its minimum possible value is

4

.

Key Concept

Systems of Linear Inequalities in Two Variables

Question 13Question →

A sports store orders standard helmets for $20$ dollars each and premium helmets for $50$ dollars each. The store can spend at most $2{}200$ dollars on this order. The distributor requires the store to order at least $60$ helmets in total. If the store decides to order at least $15$ premium helmets, what is the maximum number of standard helmets that the store can order?

Show answer & explanation

Answer: 72

Answer

The maximum number of standard helmets the store can order is 72.

By setting the number of premium helmets to its minimum allowed value of

15

to maximize the budget remaining for standard helmets, we find

20x \leq 1450

, which yields

x \leq 72.5

. Since standard helmets must be ordered in whole numbers, the maximum number is

72

, which also satisfies the minimum total order of

60

helmets (

72 + 15 = 87 \geq 60

).

Step-by-Step Solution

1

Define variables and translate constraints into inequalities

Let

x

be the number of standard helmets and

y

be the number of premium helmets. The constraints are

20x + 50y \leq 2200

,

x + y \geq 60

, and

y \geq 15

.

To represent the problem's mathematical relationships using a system of linear inequalities.

2

Solve for the upper limit of

x

using the budget constraint and the minimum value of

y

Since

20x \leq 2200 - 50y

,

x

is maximized when

y

is at its minimum value,

y = 15

. Substituting

y = 15

yields

20x + 750 \leq 2200 \implies 20x \leq 1450 \implies x \leq 72.5

.

To find the maximum possible value of standard helmets under the budget constraint.

3

Verify with the minimum total order constraint and determine the maximum integer value

Checking

x + y \geq 60

with

y = 15

gives

x + 15 \geq 60 \implies x \geq 45

. Since

x

must satisfy

45 \leq x \leq 72.5

and must be an integer, the maximum integer value is

72

.

To ensure the solution is physically possible as a whole number of items and satisfies all constraints.

Key Concept

Systems of Linear Inequalities in Two Variables

Question 14Question →

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 15Question →

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 16Question →

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 17Question →

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 18Question →

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 19Question →

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 20Question →

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.

Systems of Linear Inequalities in Two Variables

Answer

Step-by-Step Solution

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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