+0

help

0
125
1

At the fair, burgers are \$6, fries are \$3 and drinks are \$2. During lunch, they sold 53 items and made \$204. Everyone who bought a burger got fries as well except for two people. Nobody bought fries only or more than 1 burger. How many of each item was sold.

Mar 27, 2020

#1
+21957
0

Let       B =  number of burgers

B - 2  =  number of fries

D  =  number of drinks.

B + (B - 2) + D  =  53

B + B - 2 + D  =  53

2B + D  =  55

6·B + 3·(B - 2) + 2·D  =  204.00

6B + 3B - 6 + 2D  =  204.00

9b + 2D  =  210.00

Combining these two equations:

9B + 2D  =  210                                        --->      9B + 2D  =   210

2B +   D  =    55     --->   (multiply by -2)   --->     -4B - 2D  =  -110

(adding down the columns)                                   5B          =  100

B  =  20

Since  2B + D  =  55   --->   2(20) + D  =  55   --->              D  =  15

Since the number of burgers is 20, the number of fries is 18, and the number of drinks is 15.

Mar 27, 2020