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stira [4]
3 years ago
9

The concession stand at the high school football stadium sells hot dogs and hamburgers to raise money for the high school athlet

ic programs. Each hot dog sold earns the programs $0.50, and each hamburger sold earns $0.75. This week, the concession stand sold a combination of 230 hot dogs and hamburgers and earned $138.50 for the athletic programs. If x represents the number of hot dogs sold and y represents the number of hamburgers sold, which system of equations represents this situation?
Mathematics
2 answers:
dimaraw [331]3 years ago
8 0
0.50x+0.75x=138.50

hd+hb=230

These are the system of equations.

please vote my answer brainliest. thanks!
mixer [17]3 years ago
3 0

Answer:

Corresponding system of equations is, x+y=230 and 0.5x+0.75y=138.5

Step-by-step explanation:

Let, the number of hot dogs = x and the number of hamburgers = y

It is given that, the concession sold total 230 hot dogs and hamburgers.

Thus, we have, x+y=230.

Also, the price of each hot dog and hamburger are $0.50 and $0.75 each.

Since, the total earnings is $138.50.

We get, 0.5x+0.75y=138.5

Thus, the corresponding system of equations is,

x+y=230

0.5x+0.75y=138.5

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Given the equation 5x − 4 = –2(3x + 2), solve for the variable. Explain each step and justify your process.
blagie [28]
A.
5x-4=-2(3x+2) \ \ \ \ \ \ \ \ \ |\hbox{expand the bracket} \\
5x-4=-2 \times 3x-2 \times 2 \\
5x-4=-6x-4 \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 6x to both sides} \\
11x-4=-4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 4 to both sides} \\
11x=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{divide both sides by 11} \\
x=0

B.
3(2x-4)=5x-1 \\
6x-12=5x-1 \\
\boxed{11x-12=-1} \Leftarrow \hbox{the first mistake} \\
11x=11 \\
\boxed{x=11} \Leftarrow \hbox{the second mistake}

Megan's solution isn't correct.
The first mistake: she subtracted 5x from the right-hand side of the equation, but added 5x to the left-hand side.
The second mistake: she divided the right-hand side of the equation by 11, but didn't divide the left-hand side.

The correct solution:
3(2x-4)=5x-1 \ \ \ \ \ \ \ \ \ \ \ |\hbox{expand the bracket} \\
3 \times 2x+3 \times (-4)=5x-1 \\ 6x-12=5x-1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{subtract 5x from both sides} \\
x-12=-1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  |\hbox{add 12 to both sides} \\
x=11
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3 years ago
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