I believe the last equation (3x-y=10) does not apply to the system.
Given:
x+4y=16 .....................(1)
3x+5y=27 ...................(2)
3(1)-(2)
3x-3x+12y-5y=48-27 => 7y=21 => y=3 ........................(Y)
Substitute y=3 in (1)
x+4(3)=16 => x=16-12=4 => x=4 ...............................(X)
So solution is {x=4, y=3}
