Solve:
3x+6y=7;9x+3y=11(x≠0,y≠0)
3x+6y=7 ....(1)
9x+3y=11 ....(2)
Let u=1x,v=1y the (1) and (2) becomes,
3u+6v=7 .....(3)
9u+3v=11 .....(4)
3×eqn(4)−9×eqn(3) gives
27u+9v−27u−54v=33−63
⇒−45v=−30
⇒v=3045=23
Substitute the value of v=23 in (3) we get
3u+6v=7⇒3u=7−6×23=7−4=3
∴u=33=1
∴u=1x=1⇒x=1
∴v=1y=23⇒y=32
Hence x=1,y=32