Jillian is selling boxes of cookies to raise money for her basketball team. The 10 oz. box costs $3.50, while the 16 oz. box costs $5.00. At the end of one week, she collected $97.50, selling a total of 24 boxes. The system of equations that models her sales is below.

Question

x+ y= 24

3.50x + 5.00y = 97.50

Solve the system of equations. How many 10 oz. boxes were sold?

6
9
12
15

2. What is the solution to this system of linear equations?

2x + y = 1

3x – y = –6

(–1, 3)
(1, –1)
(2, 3)
(5, 0)

3. Mira picked two numbers from a bowl. The difference of the two numbers was 4, and the sum of one-half of each number was 18. The system that represents Mira’s numbers is below.

x – y = 4

One-halfx + One-halfy = 18

Which two numbers did Mira pick?

10 and 8
18 and 4
20 and 16
40 and 32

4. Two systems of equations are shown. The first equation in system B is the original equation in system A. The second equation in system B is the sum of that equation and a multiple of the second equation in system A.
A.

1
2
𝑥
+
3
𝑦
=
11
→
1
2
𝑥
+
3
𝑦
=
11

5
𝑥
−
𝑦
=
17
→
15
𝑥
−
3
𝑦
=
51

15
𝑥
−
3
𝑦
=
51
‾

15
1
2
𝑥
=
62

B.

1
2
𝑥
+
3
𝑦
=
11

What is the solution to both systems A and B?

(3, 4)
(3, 5)
(4, 3)
(5, 3)

5. What is the solution to this system of linear equations?

y − 4x = 7

2y + 4x = 2

(3, 1)
(1, 3)
(3, −1)
(−1, 3)

6. The solution to the system of equation below is (−2, −1).

2x − 3y = −1
11x − 9y = −13

When the first equation is multiplied by −3, the sum of the two equations is equivalent to 5x = −10.

Which system of equations will also have a solution of (−2, −1)?

5x = −10
11x − 9y = −13


2x + 9y = −1
11x − 9y = −13


−6x + 9y = −1
11x − 9y = −13


−6x + 3y = −1
5x = −10


7. A soccer team ordered 12 jerseys and 12 pairs of shorts, for a total of $156. Later, they had to order 4 more jerseys and 6 more pairs of shorts, for a total of $62.

The system of equations that can be used to find x, the cost of each jersey, and y, the cost of each pair of shorts is shown.

12x + 12y = 156
4x + 6y = 62
What is the cost of each jersey?

$5
$8
$12
$13

8. What is the solution to this system of linear equations?

y − x = 6

y + x = −10

(−2, −8)
(−8, −2)
(6, −10)
(−10, 6)

9. Jose asks his friends to guess the higher of two grades he received on his math tests. He gives them two hints. The difference of the two grades is 16. The sum of one-eighth of the higher grade and one-half of the lower grade is 52. The system that represents his scores is below.

x – y = 16

StartFraction 1 over 8 EndFraction x plus StartFraction one-half EndFraction y equals 52.x + StartFraction one-half EndFraction x plus StartFraction one-half EndFraction y equals 18.y = 52

What is the higher grade of Jose’s two tests?

48
52
80
96

10. Jarred sells DVDs. His inventory shows that he has a total of 3,500 DVDs. He has 2,342 more contemporary titles than classic titles. Let x represent the number of contemporary titles and y represent the number of classic titles. The system of equations models the given information for both types of DVDs.

x + y = 3,500

x – y = 2,342

Solve the system of equations. How many contemporary titles does Jarred have?

1,158
1,737
2,342
2,921

Answer 1

Let's solve each of these problems step by step.

1. Jillian's Cookie Sales: The system of equations is: \[ \begin{align*} x + y &= 24 \quad \text{(1)} \ 3.50x + 5.00y &= 97.50 \quad \text{(2)} \end{align*} \] From equation (1), we can express \(y\): \[ y = 24 - x \] Substituting \(y\) in equation (2): \[ 3.50x + 5.00(24 - x) = 97.50 \ 3.50x + 120 - 5.00x = 97.50 \ -1.50x + 120 = 97.50 \ -1.50x = -22.50 \ x = 15 \] Therefore, the number of 10 oz. boxes sold is 15.

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2. Second System of Equations: The equations are: \[ \begin{align*} 2x + y &= 1 \quad \text{(1)} \ 3x - y &= -6 \quad \text{(2)} \end{align*} \] Adding both equations to eliminate \(y\): \[ (2x + y) + (3x - y) = 1 + (-6) \ 5x = -5 \ x = -1 \] Substitute \(x\) into equation (1): \[ 2(-1) + y = 1 \ -2 + y = 1 \ y = 3 \] The solution is (–1, 3).

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3. Mira's Numbers: The system of equations is: \[ \begin{align*} x - y &= 4 \quad \text{(1)} \ \frac{1}{2}x + \frac{1}{2}y &= 18 \quad \text{(2)} \end{align*} \] Equation (2) can be rewritten as: \[ x + y = 36 \] Now, we solve: \[ \begin{align*} x - y &= 4 \quad \text{(3)} \ x + y &= 36 \quad \text{(4)} \end{align*} \] Adding (3) and (4): \[ 2x = 40 \implies x = 20 \] Substitute \(x\) into (4): \[ 20 + y = 36 \implies y = 16 \] The two numbers Mira picked are 20 and 16.

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4. System A and B: The equations: \[ \begin{align*} \frac{1}{2}x + 3y &= 11 \quad \text{(1)} \ 5x - y &= 17 \quad \text{(2)} \end{align*} \] From (2), rearranging gives: \[ y = 5x - 17 \] Substitute \(y\) into (1): \[ \frac{1}{2}x + 3(5x - 17) = 11 \ \frac{1}{2}x + 15x - 51 = 11 \ 15.5x = 62 \ x = 4 \] Substitute back for \(y\): \[ y = 5(4) - 17 = 20 - 17 = 3 \] The solution is (4, 3).

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5. Linear Equations: The equations are: \[ \begin{align*} y - 4x &= 7 \quad \text{(1)} \ 2y + 4x &= 2 \quad \text{(2)} \end{align*} \] From (1): \[ y = 4x + 7 \] Substitute into (2): \[ 2(4x + 7) + 4x = 2 \ 8x + 14 + 4x = 2 \ 12x = -12 \ x = -1 \] Substitute back for \(y\): \[ y = 4(-1) + 7 = -4 + 7 = 3 \] The solution is (–1, 3).

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6. Equations with Solution \( (−2, −1) \): You are given several systems to evaluate. The first is: \[ 5x = -10 \quad (1)\ 11x - 9y = -13 \quad (2) \] Substitute the values \(x = -2\) and \(y = -1\) to check if they work: \[ 11(-2) - 9(-1) = -13 \ -22 + 9 = -13 \ Yes. \] Check others similarly: The compatible system is \( -6x + 3y = -1, ; 5x = -10\).

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7. Costs of Jerseys and Shorts: Using the equations: \[ \begin{align*} 12x + 12y &= 156 \quad \text{(1)} \ 4x + 6y &= 62 \quad \text{(2)} \end{align*} \] From (1): \[ x + y = 13 \implies y = 13 - x \] Substitute into (2): \[ 4x + 6(13 - x) = 62 \ 4x + 78 - 6x = 62 \ -2x + 78 = 62 \ -2x = -16 \ x = 8 \] Thus, the cost of each jersey is $8.

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8. System of Linear Equations: The equations are: \[ \begin{align*} y - x &= 6 \quad \text{(1)} \ y + x &= -10 \quad \text{(2)} \end{align*} \] Adding both: \[ 2y = -4 \implies y = -2 \] Substitute \(y\) into (1): \[ -2 - x = 6 \implies -x = 8 \implies x = -8. \] The solution is (−8, −2).

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9. Jose's Test Grades: The system is: \[ \begin{align*} x - y &= 16 \quad \text{(1)} \ \frac{1}{8}x + \frac{1}{2}y &= 52 \quad \text{(2)} \end{align*} \] Expressing \(y\): \[ y = x - 16 \] Substitute into (2): \[ \frac{1}{8}x + \frac{1}{2}(x - 16) = 52 \ \frac{1}{8}x + \frac{1}{2}x - 8 = 52 \ \frac{1}{8}x + \frac{4}{8}x = 60 \ \frac{5}{8}x = 60 \implies x = 96 \] The higher grade is 96.

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10. Jarred's DVDs: The equations: \[ \begin{align*} x + y &= 3500 \quad \text{(1)} \ x - y &= 2342 \quad \text{(2)} \end{align*} \] Adding both: \[ 2x = 5842 \implies x = 2921 \] Therefore, Jarred has 2,921 contemporary titles.

In summary, here are the answers to the respective questions:

1. 15
2. (–1, 3)
3. 20 and 16
4. (4, 3)
5. (–1, 3)
6. −6x + 3y = −1
7. $8
8. (−8, −2)
9. 96
10. 2,921

Answer 2

6.
5x = −10
11x − 9y = −13

Answer 3

For the system of equations given below:

1. \(5x = -10\)
2. \(11x - 9y = -13\)

Let's solve it step by step:

Step 1: Solve for \(x\) in the first equation

From the first equation \(5x = -10\): \[ x = \frac{-10}{5} = -2 \]

Step 2: Substitute \(x\) into the second equation

Now substitute \(x = -2\) into the second equation: \[ 11(-2) - 9y = -13 \] This simplifies to: \[ -22 - 9y = -13 \]

Step 3: Solve for \(y\)

Now, isolate \(y\): \[ -9y = -13 + 22 \ -9y = 9 \ y = \frac{9}{-9} = -1 \]

Solution

The solution to the system of equations is: \[ (x, y) = (-2, -1) \]

This means the solution to the system:

1. \(5x = -10\)
2. \(11x - 9y = -13\) is \(\boxed{(-2, -1)}\).

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