SOLUTION: I am stuck. The problems is....The sum of 3 times John's age and 2 times Jill's age is 44. Jill is 8 years less than twice as old as John is. Find each of their ages. John=x

Question 151788: I am stuck. The problems is....The sum of 3 times John's age and 2 times Jill's age is 44. Jill is 8 years less than twice as old as John is. Find each of their ages.
John=x
Jill=y
I have tried 3x + 2y = 44 and y = 2x - 8. Together I have 3x + 2(2x-8)=44
the answer would be a fraction so it must not be correct. I must be setting up the two equations wrong. Please help
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Your initial equations look good!
3x+2y = 44
y = 2x-8 Substitute this into the first equation.
3x+2(2x-8) = 44 Simplify.
3x+4x-16 = 44
7x-16 = 44
7x = 60
x = 60/7
y = 2x-8
y = 2(60/7)-8
y = 120/7 - 56/7
y = 64/7
Well, mathematically, these are the solutions.
Was there anything in the instructions that led you to believe that the answers should be integers, besides the fact that you would normally expect integer answers in problems like this?
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