xx +y= 28

Question

y =2x-5
y = Marci; x = Jennifer
Jennifer and Marci are selling cookies for a fundraiser. They sold 28 boxes of cookies together. Marci sold 5 less than twice the number that Jennifer sold. How many boxes of cookies did each girl sell? Use system of equations to solve the problem.
Ix
A
Marci sold 11 boxes and Jennifer sold 17 boxes.
Marci sold 13 boxes and Jennifer sold 15 boxes.
Ix
C
Marci sold 15 boxes and Jennifer sold 13 boxes.
Marci sold 17 boxes and Jennifer sold 11 boxes.

Answer 1

To solve the problem, we can use the given equations to form a system of equations. The equations are:

1. \( x + y = 28 \) (where \( x \) is the number of boxes sold by Jennifer and \( y \) is the number of boxes sold by Marci)
2. \( y = 2x - 5 \) (Marci sold 5 less than twice the number that Jennifer sold)

Now we can substitute the second equation into the first equation.

Substituting \( y \) from the second equation into the first equation:

\[ x + (2x - 5) = 28 \]

Combine like terms:

\[ 3x - 5 = 28 \]

Now, add 5 to both sides:

\[ 3x = 33 \]

Next, divide both sides by 3:

\[ x = 11 \]

Now that we have \( x \) (the number of boxes sold by Jennifer), we can find \( y \) by substituting \( x \) back into the equation for \( y \):

\[ y = 2(11) - 5 = 22 - 5 = 17 \]

So, Jennifer sold \( 11 \) boxes, and Marci sold \( 17 \) boxes.

To confirm:

• The total sold should equal 28: \( 11 + 17 = 28 \) (correct)
• Marci should have sold 5 less than twice Jennifer’s sales: \( 2(11) - 5 = 17 \) (correct)

Thus, the solution is:

• Marci sold 17 boxes and Jennifer sold 11 boxes.

Answer 2

A discount store is selling 5 small tables with 8 chairs for $115. Three tables with 5 chairs cost $70.
*x Which system of linear equations could be used to find the cost of each table (x) and the cost of each chair (y)?

x A 3X + 8y = $70; 8x + 3y = $115

×B
5x + 8y = $115; 3x + 5y = $70

×C
5x + 8y = $115; 2x - 5y = $70

× D
8x + 5y = $115; 5x + 3y = $70

Answer 3

To find the cost of each table (denoted as \( x \)) and the cost of each chair (denoted as \( y \)), we can set up the following equations based on the information provided:

1. From the first scenario: 5 small tables with 8 chairs cost $115. This can be expressed with the equation:

   \[ 5x + 8y = 115 \]

2. From the second scenario: 3 small tables with 5 chairs cost $70. This can be expressed with the equation:

   \[ 3x + 5y = 70 \]

Thus, the correct system of linear equations is:

\[ \begin{aligned}

1. & \quad 5x + 8y = 115 \
2. & \quad 3x + 5y = 70 \ \end{aligned} \]

Looking at the options:

• Option B is the correct choice: \( 5x + 8y = 115; , 3x + 5y = 70 \).

Answer 4

1x Determine the cost of each table (x) and the cost of each chair (y).
•×

A x= $15; y = $5

x = $10; y = $5

C x= $5; y = $10

D
x = $12; y = $3

Answer 5

To determine the cost of each table (\( x \)) and the cost of each chair (\( y \)), we'll solve the system of equations we derived:

1. \( 5x + 8y = 115 \) (Equation 1)
2. \( 3x + 5y = 70 \) (Equation 2)

Let’s solve these equations step by step.

Step 1: Solve Equation 1 for \( y \)

From the first equation: \[ 5x + 8y = 115 \] To isolate \( y \), we can rearrange it: \[ 8y = 115 - 5x \] \[ y = \frac{115 - 5x}{8} \]

Step 2: Substitute \( y \) in Equation 2

Now substitute \( y \) into the second equation: \[ 3x + 5\left(\frac{115 - 5x}{8}\right) = 70 \] Multiply through by 8 to eliminate the fraction: \[ 24x + 5(115 - 5x) = 560 \] Distributing: \[ 24x + 575 - 25x = 560 \] Combine like terms: \[ -1x + 575 = 560 \] To isolate \( x \), subtract 575 from both sides: \[ -1x = 560 - 575 \] \[ -1x = -15 \] Thus: \[ x = 15 \]

Step 3: Substitute \( x \) back to find \( y \)

Now substitute \( x = 15 \) back into the equation for \( y \): \[ y = \frac{115 - 5(15)}{8} \] Calculating: \[ y = \frac{115 - 75}{8} = \frac{40}{8} = 5 \]

Conclusion

Thus, the solutions are:

• \( x = 15 \) (cost of each table)
• \( y = 5 \) (cost of each chair)

The correct answer is: A) \( x = 15; y = 5 \).

Answer 6

x = y - 3
x + 3y = 13
What is the solution to the system of equations?

(1, 4)

4× B（4,1）

4×C （7,4）

(2.5, 5.5)

1x E
(5.5, 2.5)

Answer 7

To find the solution to the system of equations given by:

1. \( x = y - 3 \) (Equation 1)
2. \( x + 3y = 13 \) (Equation 2)

We can use substitution or elimination. Since Equation 1 already expresses \( x \) in terms of \( y \), we can substitute that into Equation 2.

Step 1: Substitute Equation 1 into Equation 2

Substituting \( x \) from Equation 1 into Equation 2:

\[ (y - 3) + 3y = 13 \]

Step 2: Combine like terms

Combine the \( y \) terms:

\[ y - 3 + 3y = 13 \] \[ 4y - 3 = 13 \]

Step 3: Solve for \( y \)

Add 3 to both sides:

\[ 4y = 16 \]

Now, divide by 4:

\[ y = 4 \]

Step 4: Substitute back to find \( x \)

Now substitute \( y = 4 \) back into Equation 1 to find \( x \):

\[ x = y - 3 \] \[ x = 4 - 3 = 1 \]

Conclusion

So the solution to the system of equations is:

\[ (x, y) = (1, 4) \]

Thus, the correct answer is (1, 4).