Question 1013927
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At a fast food restaurant, a family bought 4 hamburgers and 3 bags of french fries for 4.20. 
At the same time, a family traveling with them bought 5 hamburgers and 2 bags of french fries for 4.55. 
What was the cost of one hamburger and one bag of french fries? What was the cost if you ordered 5 hamburgers and 4 french fries?'
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From the condition, you have this system of two equations in two unknowns

4h + 3f = 420,     (1)
5h + 2f = 455.     (2)

Here h is the price of one hamburger and f is the price of one bag of french fries in cents.

To solve it, multiply equation (1) by 2 (both sides) and equation (2) by 3. You will get 

 8g + 6f =  840,   (1')
15h + 6f = 1365.   (2')

Now distract the equation (1') from (2') (both sides). In this way you eliminate the unknown f and obtain a single equation for unknown h:

15h - 8h = 1365 - 840,   or

7h = 525.

Hence, h = {{{525/7}}} = 75.

Substitute the found value of h = 75 into equation (1). You get

3f = 420 - 4h = 420 -4*75 = 420 - 300 = 120,

f = {{{120/3}}} = 40.

Thus the cost of one hamburger is $0.75; the cost of one bag of french fries is $0.40.

Now calculate the cost of 5 hamburgers and 4 french fries. It is

5*$0.75 + 4*$0.40 = $5.35.
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