Answer
Infinitely many solutions with $x+y-1=2(y-x) $ or $y=3x-1$
Work Step by Step
To solve the given problem we will have to plug the simplest equation into the other equation and then solve for the other variable.
$x+y-1=2(y-x) $ or $y=3x-1$
Now, $x+(3x-1)-1=2((3x-1)-x)$
or, $x+3x-1-1=4x-2$
or, $4x-1-1=4x-2$
or, $4x-2=4x-2$ or, $0=0$
Hence, the given equation has Infinitely many solutions with $x+y-1=2(y-x) $ or $y=3x-1$
