SOLUTION: help!!!! Problem: Joe is giving out some coupons for 3 free hours of canoe rental. melissa has 5 more coupons than joe to give out, and her coupons are for 4 free hours of canoe r

Algebra ->  Test -> SOLUTION: help!!!! Problem: Joe is giving out some coupons for 3 free hours of canoe rental. melissa has 5 more coupons than joe to give out, and her coupons are for 4 free hours of canoe r      Log On


   



Question 730849: help!!!! Problem:
Joe is giving out some coupons for 3 free hours of canoe rental. melissa has 5 more coupons than joe to give out, and her coupons are for 4 free hours of canoe rental. if their coupons represent 90 free hours of canoe rental in all, how many coupons do they each have?

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let
x = # of coupons Joe has

y = # of coupons Melissa has


We are told that "melissa has 5 more coupons than joe to give out", so we can say

y = x + 5

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Joe has 3 free hour coupons. So if he has x of them, then he has 3x hours free total.

Similarly for Melissa, she has y 4-hr free coupons. This gives her a total of 4y free hours.

In total, they have 3x + 4y free hours. This total equals 90 hours (given), so...

3x+4y = 90

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Now use these equations to find x and y

3x+4y = 90

3x+4(x+5) = 90 ... plug in y = x+5

3x+4x+20 = 90

7x+20 = 90

7x = 90-20

7x = 70

x = (70)/(7)

x = 10

y = x + 5

y = 10 + 5

y = 15

To sum things up, we found that

x = 10 and y = 15

So Joe has 10 coupons and Melissa has 15 coupons