Chapter 6 - Systems of Equations and Inequalities - 6-2 Solving Systems Using Substitution - Practice and Problem-Solving Exercises - Page 377: 41

Therefore, the system has no solutions.

If we solve the first equation for y, $ y+x=x\qquad$ ... add $-x$ $y=0$ Substituting $0$ for $y$ in the second equation, the LHS is undefined, because there is a zero in the denominator. The RHS is defined, the number 4. There is no x that would make this equation true. Therefore, the system has no solutions.