6x+5y=39
3x+5y=27
solveing systems by elimination

X=4 and Y=3 If you need to the work shown I can put a picture

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)
-
3x + 5y = 27(ii)

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.