Question 654507: i need help with problems like this. i know it says only one problem per submission but i cant figure out how to ask another question. i have tried to solve these two problems but they are extremely difficult for me and i cant find any good examples anywhere else. please help :( thanks
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a restaurant manager bought 80 lb of hamburger and 30 lb of steak for a total cost of 163 a second purchase at the same prices includes 150 lb of hamburger and 50 lb of steak and costs 290 find price of 1 lb of hamburger
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a coin bank contains only dimes and quarters the total value is 3.70. if the dimes were quarters and the quarters were dimes the total value would be 2.95 find the number of quarters
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Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! a restaurant manager bought 80 lb of hamburger and 30 lb of steak for a total
cost of 163 a second purchase at the same prices includes 150 lb of hamburger
and 50 lb of steak and costs 290 find price of 1 lb of hamburger
:
let h = cost per pound for hamburger
let s = cost per pound for steak
:
Write a total cost equation for each order
80h + 30s = 163
150h + 50s = 290
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We are going to us elimination; mult the 1st eq by 5, the 2nd eq by 3, results:
400h + 150s = 815
450h + 150s = 870
------------------subtraction eliminates s, find h
-50h + 0 = -55
h = -55/-50
h = $1.10 a pound for hamburger
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a coin bank contains only dimes and quarters the total value is 3.70.
if the dimes were quarters and the quarters were dimes the total value
would be 2.95 find the number of quarters
:
Let x = original number of dimes
Let y = original number of quarters
:
Write an equation for each scenario
.10x + .25y = 3.70
.25x + .10y = 2.95
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Using elimination again, mult the 1st equation by 5, the 2nd eq by 2
.50x + 1.25y = 18.50
.50x + .20y = 5.90
-------------------subtraction eliminates x, find y (quarters)
0 + 1.05y = 12.60
y = 12.6/1.05
y = 12 quarters
:
You should check both these problems by finding the other variable value.
Then substitute the values in the original equations, to see if it is true.
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