I find it easier to start from scratch and write out the equations, then compare them with the given equations.
Let m and j represent the ages of the two boys. Then m-5=(2/3)(j-5)
Also, in ten years: m+10 = (5/6)j.
Let's solve this system: Mult the first eqn by 3 to remove the fraction:
3(m-5) = 2(j-5), or 3m - 15 = 2j -10. Next, mult. the 2nd eqn by 6 to remove the fraction: 6m+60 = 5j.
Our two equations are (at this point) 3 m - 15 = 2j - 10 and
6m +60 = 5j
Let's mult. the first equation by -2, so as to obtain the coefficient -6 for m:
-6m + 30 = -4j + 10
6m + 60 = 5j
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90 = j + 10, so j = 80 (wow!)
Let's now find m: 6m + 60 = 5(80), or 6m = 400+60 = 460
Then m = 76 2/3.
We must check our solution (m = 76 2/3, j = 80):
Let's subst. these values into m-5=(2/3)(j-5)
Does 76 2/3 - 5 = (2/3)(80) - 10/3?
Does 76 2/3 - 15/3 = 160/3 - 10/3?
Does 76 - 15 = 160 - 10 No. So something's wrong here.
Going back to the problem statement:
x-5=2/3(y-5) and x+10=5/6(y+10) seems correct!
Mike's age 5 years ago was x-5, and james' age 5 years ago was y-5. The multiplier (2/3) is also correct.
Mikes age 10 years from now will be x+10, and james' y+10.
The first answer set is the correct one.
