Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 9

$(x,y)=${$(-\dfrac{3}{2},-4),(2,3)$}

Re-write the second equation as: $y=2x-1$ First equation yields: $x(2x-1)=6$ or, $2x^2-x-6=0$ or, $(2x+3)(x-2)=0$ or, $x=${$-\dfrac{3}{2},2$} From first equation when $x=-\dfrac{3}{2}$, we have $y=-4$ From first equation when $x=2$, we have $y=3$ Thus, solution set is $(x,y)=${$(-\dfrac{3}{2},-4),(2,3)$}