+0  
 
0
52
1
avatar

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
avatar+20768 
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

32 Online Users

avatar
avatar
avatar