The concession stand is selling hot dogs and hamburgers during a game. At halftime, they sold a total of 78 hot dogs and hamburgers and brought in $105.50. how many of each item did they sell if hamburgers sold for $1.50 and hot dogs sold for $1.25?

1 Answer
Jul 5, 2016

Answer:

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

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.