## Intermediate Algebra for College Students (7th Edition)

Infinitely many solutions with $x+y-1=2(y-x)$ or $y=3x-1$
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 $y=3x-1$