Five years back, michael's age was 2/3 the age of james. after ten years he will be 5/6 of the age of james. what are the equations of this problem? x-5=2/3(y-5) and x+10=5/6(y+10) x-=2/3(y-5) and x+10=5/6(y+10) x-5=(y-5) and x+10=5/6(y+10) x-5=2/3(y-5) and x+10=5(y+10)

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

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.