Question 1099540
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3x + 2y = 48,   (1)
6x + 6y = 64.   (2)


Divide eq(2) by 2 (both sides).  You will get the modified system

3x + 2y = 48,   (1')
3x + 3y = 32.   (2')


Subtract eq(1) from eq(2)  (both sides).  You will get

y = 32 - 48 = -16.


Thus you just found the solution for y.

Now substitute it into either equation (1) or (2). I will substitute into eq(1). You will get

3x + 2*(-16) = 48  ====>  3x = 48 + 32 = 80  ====>  x = {{{80/3}}}.


<U>Answer</U>.  The solution is (x,y) = ({{{80/3}}},-16).
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