Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 20

{$(1,0),(-1,0)$}

After adding the given two equations, we get $8x^2=8$ or, $x^2=1$; or $x=\pm 1$ From first equation $4x^2-y^2=4$ when $x=1$, we have $y=0$ From first equation $4x^2-y^2=4$ when $x=-1$, we have $y=0$ Thus, solution set is {$(1,0),(-1,0)$}