Question 87278
John is 12 years younger than Mary.  Mary's age in 8 years will exceeed twice
John's age 3 year's ago by 4 years. How old is each now?: 
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I can't tell if you are studying solving using 1 unknown or 2 unknowns.
This can be done either way.  I'll do it both ways:

Using only one unknown:

>>...John is 12 years younger than Mery...<<

John's age is defined in terms of Mary's age, and whenever possible we
like to define in terms of x, so let Mary's age = x, then

John's age now = Mary's age now minus 12, or

John's age = x - 12

>>...Mary's age in 8 years will exceeed twice John's age 3 year's ago by 4 years...<<

That is to say:

>>...Mary's age in 8 years equals twice John's age 3 years ago PLUS 4...<<

Mary's age in 8 years = Mary's age now + 8 
Mary's age in 8 years = x + 8

John's age 3 years ago = John's age now - 3
John's age 3 years ago = (x - 12) - 3 = x - 12 - 3 = x - 15
Twice John's age 3 years ago = 2(x - 15) = 2x - 30

x + 8 = (2x - 30) + 4

Solve that and get x = 34

So Mary is 34 and John is 12 years younger or 34-12 = 22

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Using two unknowns:

John is 12 years younger than Mary. 

J = M - 12

Mary's age in 8 years will exceeed twice John's
 age 3 year's ago by 4 years.

M + 8 = 2(J - 3) + 4

So you have the system of two equations in two unknowns.
Simplify the second and use substitution

J = M - 12
M + 8 = 2(J - 3) + 4

and you will get J = 22 and M = 34.

Edwin</pre>