A concession stand sell hot dogs and hamburgers. At a football game, 84 hot dogs and 36 hamburgers were sold for $276. At another football game, 60 hotdogs and 18 hamburgers were sold for $174. Wrote a system of equations to represent this situation. Then solve the system of equations.

The concession stand sold 46 hot dogs and 32 hamburgers.

Step-by-step explanation:

The first thing to do in algebraic problems is assign variables to things we don't know, so let's start there:

We don't know how many hot dogs the concession stand sold, so we will call that number d.

We don't know how many hamburgers the concession stand sold, so we will call that number h.

Now we translate the statements into algebraic equations:

The number of hot dogs and hamburgers that were sold is 78, so d+h=78.

If each hot dog is sold for 1.25, then the total revenue from hot dogs is given by 1.25d. In the same way, the total revenue from hamburgers is 1.50h. The total revenue from both hot dogs and hamburgers should be the sum of these, and since we are told the total revenue is 105.50, we can say 1.25d+1.5h=105.5.

We now have a system of two linear equations:

d+h=78

1.25d+1.5h=105.5

We can solve it using several methods, though I'm going to go with substitution. Use the first equation to solve for d:

d+h=78

→d=78−h

Now plug this in for d in the second equation:

1.25d+1.5h=105.5

→1.25(78−h)+1.5h=105.5

Solving for h, we have:

97.5−1.25h+1.5h=105.5

0.25h=8

h=8.25→h=32

Since h+d=78,

32+d=78→d=46

The concession stand therefore sold 46 hot dogs and 32 hamburgers.

i think the answer is b.

step-by-step explanation:

not 100% sure though.

explanation:

step-by-step explanation: