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x + 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.
Responses
A Marci sold 11 boxes and Jennifer sold 17 boxes.Marci sold 11 boxes and Jennifer sold 17 boxes.
B Marci sold 13 boxes and Jennifer sold 15 boxes.Marci sold 13 boxes and Jennifer sold 15 boxes.
C Marci sold 15 boxes and Jennifer sold 13 boxes.Marci sold 15 boxes and Jennifer sold 13 boxes.
D Marci sold 17 boxes and Jennifer sold 11 boxes.Marci sold 17 boxes and Jennifer sold 11 boxes.
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11 answers

To solve the problem using the system of equations given:

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

Now we can substitute the second equation into the first.

Substituting \( y \) in the first equation:

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

Combine like terms:

\[ 3x - 5 = 28 \]

Now, add 5 to both sides:

\[ 3x = 33 \]

Now divide by 3:

\[ x = 11 \]

Now we can substitute \( x = 11 \) back into the equation for \( y \):

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

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

Thus, the answer is: D. Marci sold 17 boxes and Jennifer sold 11 boxes.

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Solving a Linear System
A discount store is selling 5 small tables with 8 chairs for $115. Three tables with 5 chairs cost $70.
Question 1
Which system of linear equations could be used to find the cost of each table (x) and the cost of each chair (y)?
Responses
A 3x + 8y = $70; 8x + 3y = $1153x + 8y = $70; 8x + 3y = $115
B 5x + 8y = $115; 3x + 5y = $705x + 8y = $115; 3x + 5y = $70
C 5x + 8y = $115; 2x − 5y = $705x + 8y = $115; 2x − 5y = $70
D 8x + 5y = $115; 5x + 3y = $708x + 5y = $115; 5x + 3y = $70
Question 2
Determine the cost of each table (x) and the cost of each chair (y).
Responses
A x = $15; y = $5x = $15; y = $5
B x = $10; y = $5x = $10; y = $5
C x = $5; y = $10x = $5; y = $10
D x = $12; y = $3

To set up the system of linear equations based on the problem statements:

  1. The first statement is about 5 small tables and 8 chairs for $115. This can be written as: \[ 5x + 8y = 115 \]

  2. The second statement is about 3 tables and 5 chairs costing $70. This can be expressed as: \[ 3x + 5y = 70 \]

Thus, the system of equations to find the cost of each table \(x\) and the cost of each chair \(y\) is: \[ \begin{align*}

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

The options provided for the first question can now be evaluated. The correct system of equations is: B. \(5x + 8y = 115; \ 3x + 5y = 70\)

Now, we can solve the system of equations to determine the cost of each table and chair.

We can use substitution or elimination. Here, we will use the elimination method:

  1. Multiply the second equation by a number that allows the coefficients of \(y\) to be the same in both equations. We can multiply the second equation by 8: \[ 8(3x + 5y) = 8(70) \implies 24x + 40y = 560 \]

  2. Next, multiply the first equation by 5: \[ 5(5x + 8y) = 5(115) \implies 25x + 40y = 575 \]

Now the system of equations looks like this: \[ \begin{align*}

  1. & \quad 25x + 40y = 575 \
  2. & \quad 24x + 40y = 560 \end{align*} \]

  1. Subtract the second equation from the first: \[ (25x + 40y) - (24x + 40y) = 575 - 560 \] \[ x = 15 \]

  2. Now, plug \(x = 15\) back into one of the original equations (let's use the first equation): \[ 5(15) + 8y = 115 \implies 75 + 8y = 115 \] \[ 8y = 40 \implies y = 5 \]

Now we have found the values:

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

Thus, the response to question 2 is: A. \(x = 15; y = 5\)

A concession stand sells hamburgers (h) for $2 and hotdogs (d) for $1. On Friday night they sold a total of 300 hamburgers and hotdogs and made $420. How many of each did they sell?

Which system of equations matches the situation?
Responses
A h + d = 300
2h + d = 420
B h + d = 420
2h + d = 300
C h + d = 300
h + 2d = 420
D h + d = 720
2h + d = 300

To set up the system of equations based on the given situation:

  1. Total number of items sold: The concession stand sold a total of 300 hamburgers and hotdogs. This can be expressed as: \[ h + d = 300 \] where \(h\) is the number of hamburgers and \(d\) is the number of hotdogs.

  2. Total revenue: They made $420 from selling hamburgers at $2 each and hotdogs at $1 each. This can be expressed as: \[ 2h + d = 420 \]

Combining these two pieces of information, we get the following system of equations: \[ \begin{align*}

  1. & \quad h + d = 300 \
  2. & \quad 2h + d = 420 \end{align*} \]

Looking at the provided answer options, the correct choice is: A. \[ \begin{align*} h + d = 300 \ 2h + d = 420 \end{align*} \]

Now, let's solve this system to find the number of hamburgers and hotdogs sold.

  1. From the first equation, we can express \(d\) in terms of \(h\): \[ d = 300 - h \]

  2. Substitute \(d\) into the second equation: \[ 2h + (300 - h) = 420 \]

  3. Simplify and solve for \(h\): \[ 2h + 300 - h = 420 \] \[ h + 300 = 420 \] \[ h = 120 \]

  4. Now substitute \(h = 120\) back into the equation for \(d\): \[ d = 300 - 120 = 180 \]

Therefore, they sold 120 hamburgers and 180 hotdogs.

A group of 16 people bought tickets to a museum. They paid a total of $56. Each adult ticket costs $4 and each child ticket costs $2. Formulate and solve a system of equations to determine how many adult tickets and child tickets did they buy.
Responses
A 4 adult tickets, 12 child tickets4 adult tickets, 12 child tickets
B 8 adult tickets, 8 child tickets8 adult tickets, 8 child tickets
C 10 adult tickets, 6 child tickets10 adult tickets, 6 child tickets
D 12 adult tickets, 4 child tickets

To solve this problem, we can set up a system of equations based on the information given about the tickets.

Let:

  • \( a \) be the number of adult tickets
  • \( c \) be the number of child tickets

From the problem, we have the following:

  1. Total number of tickets: The total number of people (adult and child tickets) who bought tickets is 16: \[ a + c = 16 \]

  2. Total cost of tickets: The total amount paid is $56. Each adult ticket costs $4, and each child ticket costs $2: \[ 4a + 2c = 56 \]

Now we have the following system of equations to work with: \[ \begin{align*}

  1. & \quad a + c = 16 \
  2. & \quad 4a + 2c = 56 \end{align*} \]

Step 1: Simplify the second equation

We can simplify the second equation by dividing everything by 2: \[ 2a + c = 28 \]

Step 2: Set up the system of equations

Now we have:

  1. \( a + c = 16 \)
  2. \( 2a + c = 28 \)

Step 3: Solve the system

Now we can solve this system using elimination or substitution. Here, we'll use elimination.

Subtract the first equation from the second equation: \[ (2a + c) - (a + c) = 28 - 16 \] \[ 2a + c - a - c = 12 \] \[ a = 12 \]

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

Now that we have \( a = 12 \), we can substitute this back into one of the original equations to find \( c \): \[ 12 + c = 16 \] \[ c = 16 - 12 \] \[ c = 4 \]

Conclusion

Thus, they bought:

  • 12 adult tickets
  • 4 child tickets

The answer is: D. 12 adult tickets, 4 child tickets.

3x + y = 3
x + y = 2

Solve the system of equations.
Responses
A x =




, y = 3x = 1/2 , y = 3
B x = 3, y =




x = 3, y = 1/2
C x =




, y =




x = 1/2 , y = 3/2
D x =




, y =




x = 3 /2 , y = 1/2
E x = 5/2, y = -9/2

To solve the system of equations given:

  1. \( 3x + y = 3 \)
  2. \( x + y = 2 \)

We can use substitution or elimination. Here, I will use the elimination method.

Step 1: Rewrite the second equation

From the second equation, we can express \(y\) in terms of \(x\): \[ y = 2 - x \]

Step 2: Substitute into the first equation

Now we can substitute \(y\) into the first equation: \[ 3x + (2 - x) = 3 \]

Step 3: Simplify and solve for \(x\)

\[ 3x + 2 - x = 3 \] Combine like terms: \[ 2x + 2 = 3 \]

Now, subtract 2 from both sides: \[ 2x = 1 \]

Finally, divide by 2: \[ x = \frac{1}{2} \]

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

Now substitute \(x = \frac{1}{2}\) back into the equation for \(y\): \[ y = 2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \]

Conclusion

The solution to the system of equations is: \[ x = \frac{1}{2}, \quad y = \frac{3}{2} \]

Thus, the correct option is: C. \(x = \frac{1}{2}, y = \frac{3}{2}\)

Suppose that the relation is defined as follows.
H = (-4, 6), (8, 2), (8, -6), (0, -4)
Give the domain and range of H.
Write your answers using set notation.

To find the domain and range of the relation \( H \) defined by the set of pairs:

\[ H = { (-4, 6), (8, 2), (8, -6), (0, -4) } \]

Domain

The domain of a relation is the set of all first elements (or x-coordinates) in the ordered pairs.

From the pairs in \( H \), the first elements are:

  • \( -4 \)
  • \( 8 \) (appears twice, but we only list it once)
  • \( 0 \)

Thus, the domain is: \[ \text{Domain}(H) = { -4, 8, 0 } \]

Range

The range of a relation is the set of all second elements (or y-coordinates) in the ordered pairs.

From the pairs in \( H \), the second elements are:

  • \( 6 \)
  • \( 2 \)
  • \( -6 \)
  • \( -4 \)

Thus, the range is: \[ \text{Range}(H) = { 6, 2, -6, -4 } \]

Final Answer

In set notation, the domain and range of the relation \( H \) are:

  • Domain: \( { -4, 8, 0 } \)
  • Range: \( { 6, 2, -6, -4 } \)