EAMCET Engineering 2006 Solved Paper

The equation of lines are
x−y−2‌=0 . . . (i)
x+y−4‌=0 . . . (ii)
‌ and ‌‌‌x+3y‌=6 . . . (iii)
On solving Eqs. (i) and (ii), we get
x=3,y=1
On putting these value in Eq. (iii)
3+3=6
which is satisfied. Thus the required point is (3,1).