SOLUTION: Hamburgers cost $2 each and pizzas cost $6 each. If the number of hamburgers was 2 more than twice the number of pizzas, and a total of $84 was spent, how many hamburgers and how m

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Question 64234This question is from textbook An Incremental Development
: Hamburgers cost $2 each and pizzas cost $6 each. If the number of hamburgers was 2 more than twice the number of pizzas, and a total of $84 was spent, how many hamburgers and how many pizzas were purchased?
thank you! This question is from textbook An Incremental Development

Found 2 solutions by 303795, ptaylor:
Answer by 303795(602) About Me  (Show Source):
You can put this solution on YOUR website!
Call the number of pizza x. The number of hamburgers must be 2x + 2.
The cost of pizzas at $6 will be 6x and the cost of hamburgers at $2 will be 2(2x + 2)
The cost should therefore be
6x + 2(2x + 2) = 84
6x + 4x + 4 = 84
10x + 4 = 84
10x + 4 - 4 = 84 - 4
10x = 80
x = 8
so 8 pizzas were purchased and 18 hamburgers

Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=number of pizzas
2x=twice the number of pizzas
2x+2=two more than twice the number of pizzas, therefore:
2x+2=number of hamburgers
We are told that the total cost of hamburgers ($2(2x+2)) plus total cost of pizzas ($6(x)) equals $84. So our equation to solve is:
2(2x+2)+6x=84 simplifyiing
4x+4+6x=84 collecting like terms
10x=80
x=8 number of pizzas
2x+2=2(8)+2=18 number of hamburgers
ck
$6(8)+$2(18)=$84
$48+$36=$84
$84=$84
Hope this helps. Happy holidays.-----ptaylor