SOLUTION: Yesterday, John bought 5 deluxe hamburgers and 2 orders of cheese fries from the Hamburger Palace and paid $23.65. Today John purchased 3 deluxe hamburgers and 1 order of cheese fr

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Question 1155672: Yesterday, John bought 5 deluxe hamburgers and 2 orders of cheese fries from the Hamburger Palace and paid $23.65. Today John purchased 3 deluxe hamburgers and 1 order of cheese fries and paid $13.80. What is the price of one deluxe hamburger? What is the price of one order of cheese fries?
Found 3 solutions by josgarithmetic, Boreal, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Very simple application for two-variable linear equations system
prices for hamburgers, h
price for "fries", f
Points (h,f) described are (5,2) and (3,1).

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
5H+2C=23.65
3H+1C=13.80
-6H-2C=-27.60, multiplying the bottom equation by -2. Add them.
-H=-3.95
Hamburgers at $3.95
so 3 are 11.85 and 1 C has to be $1.95 to add to $13.80
Cheese is $1.95 ANSWER
Check in the top
$19.75+$3.90=$23.65 checks.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Suppose that today, instead of 3 burgers and 1 fries for $13.80, he had bought twice as much -- 6 burgers and 2 fries for $27.60.

Then the difference between yesterday's purchase and today's would be 1 more burger for $3.95.

So the price of a burger is $3.95; from there it is easy to find the price of an order of fries.

That informal solution using logical reasoning follows exactly the same path as a formal algebraic solution using elimination:

5h%2B2f+=+23.65 -- (1) yesterday
3h%2Bf+=+13.60 -- (2) today
6h%2B2f+=+27.60 -- (3) today's, doubled
h+=+3.95 -- the difference between (1) and (3)