1x 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?

gh+d = 300
12h+ d = 420

Sh+d = 420
12h+d - 300
C
jh+d = 300
n+ 2d = 420
D
jh+d = 720
12h+d = 300

3 answers

Let's denote the number of hamburgers sold as \( h \) and the number of hotdogs sold as \( d \).

From the problem, we have the following information:

  1. The total number of hamburgers and hotdogs sold is 300.
  2. The total revenue from hamburgers and hotdogs is $420.

We can set up the following system of equations based on this information:

  1. Total items sold: \[ h + d = 300 \]

  2. Total money made: \[ 2h + 1d = 420 \]

Now, let's match this system of equations with the given options:

  • The first equation \( h + d = 300 \) represents the total items sold.
  • The second equation \( 2h + d = 420 \) represents the total revenue made.

The correct system of equations matches the one we've formed:

\[ \begin{align*} h + d &= 300 \ 2h + d &= 420 \end{align*} \]

Looking at the options provided:

  1. Option A: \[ gh + d = 300 \ 12h + d = 420 \] This does not match.

  2. Option B: \[ Sh + d = 420 \ 12h + d = 300 \] This does not match.

  3. Option C: \[ jh + d = 300 \ n + 2d = 420 \] This does not match.

  4. Option D: \[ jh + d = 720 \ 12h + d = 300 \] This does not match.

None of the options provided exactly match the system of equations derived from the problem statement. The correct representation of the problem is:

\[ \begin{align*} h + d &= 300 \ 2h + d &= 420 \end{align*} \]

If jh corresponds to h, then choice C looks close but would need adjustments. Since no options are exactly correct, please verify provided options or explore additional details.

A What is the solution to the system of equations?
-2x +y=-5
3x - 2y = 12
』×
A
(3, 1)
4x

(6, 3)

C
(-2, -9)
《×
D
(-2, -1)

E
no solution

To solve the system of equations given:

  1. \(-2x + y = -5\)
  2. \(3x - 2y = 12\)

We'll solve it using the substitution or elimination method. Let's use substitution for this case.

From the first equation, we can express \(y\) in terms of \(x\):

\[ y = 2x - 5 \]

Now, we will substitute this expression for \(y\) into the second equation:

\[ 3x - 2(2x - 5) = 12 \]

Distribute the \(-2\):

\[ 3x - 4x + 10 = 12 \]

Combine like terms:

\[ -x + 10 = 12 \]

Now, isolate \(x\):

\[ -x = 12 - 10 \] \[ -x = 2 \implies x = -2 \]

Now that we have \(x\), we substitute back into the expression we found for \(y\):

\[ y = 2(-2) - 5 \] \[ y = -4 - 5 = -9 \]

Thus, the solution to the system of equations is:

\[ (x, y) = (-2, -9) \]

Now, let's look at the options provided:

\[ \text{A: } (3, 1) \ \text{B: } (6, 3) \ \text{C: } (-2, -9) \ \text{D: } (-2, -1) \ \text{E: } \text{no solution} \]

The correct option is:

C: (-2, -9).