Algebra - concession word problem

Question

A basketball concession stand sells a soda for $1.25 , a bag of popcorn for $0.50 , and a candy bar for $.75 At one game, 10 fewer candy bars were sold than bags of popcorn. That same night, the number of sodas sold was 3 more than the number of candy bars and bags of popcorn sold altogether. If the concession stand made $130 that night, how many of each item were sold?

Answer 1

1.25s +.5b + .75c = 130

c= b-10, or b= c+10

s = b+c+3

3 equations 3 unknowns, solve by substitution elimination, matrix algebra or graphing

to find the number of sodas (s) bags of popcorn (b) and candy bars (c) were sold

1.25a + .5b + .75c= 130

1.25(b +c+3) + .5(b) + .75c =130 substitute to eliminate a and b then solve for c

1.25((c+10) + (c) + 3)) +.5(c+10) + .75c = 130

3.75c = 130 - 1.25(13) - 5 = 130 -16.25 -5 = 113.75 -5 = 108.75

c = 108.75/3.75 = 29 candy

b = c+10 = 39 bags of popcorn

s = b+c+3 = 71 sodas

check the answers

1.25(71) + .5(39) + .75(29) = 130. It works