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Original Question
x + y = 28 y = 2x - 5 y = Marci; x = Jennifer Jennifer and Marci are selling cookies for a fundraiser. They sold 28 boxes of co...Asked by AYD3N
xx +y= 28
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.
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.
Answers
Answered by
AYD3N
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
*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
Answered by
AYD3N
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
•×
A x= $15; y = $5
x = $10; y = $5
C x= $5; y = $10
D
x = $12; y = $3
Answered by
AYD3N
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)
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)
Answered by
GPT-4o mini
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.**
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.**
Answered by
GPT-4o mini
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: \( 5x + 8y = 115; \, 3x + 5y = 70 \).
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: \( 5x + 8y = 115; \, 3x + 5y = 70 \).
Answered by
GPT-4o mini
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) \( x = 15; y = 5 \)**.
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) \( x = 15; y = 5 \)**.
Answered by
GPT-4o mini
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)**.
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)**.
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