6x + 5y = 39 ......(i)
3x + 5y = 27.....(ii)
Because of the 5y in both equations (i) and (ii), we subtract, so as to eliminate y
6x + 5y = 39 ......(i)
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3x + 5y = 27.....(ii)
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3x + 0 = 12
3x = 12
x = 12/3 = 4, x = 4.
Substituting into 3x + 5y =27, x = 4
3*4 + 5y = 27
12 + 5y = 27
5y = 27 - 12
5y = 15
y = 15/5
y = 3
So the solutions are x = 4, y = 3.
Hope this helped.