First we have any one of the terms equal in signs(x or y).. I will take y but it's ur wish..
(5x+3y=27)×2= 10x+6y= 54______(equation 1)
(7x-2y=13)×3 = 21x-6y= 39______(equation 2)
6y and -6y get cancelled..
so that leaves 10x+21x=31x
and 54+39=93..
31x=93
x=3
on solving the 1st equation by putting in the value of x we get:- 5x+3y=27
5×3+3y=27
15+3y=27
3y=27-15
3y=12
y=4..
