Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 18

{$(\dfrac{1}{5},\dfrac{18}{5}),(1,2)$}

Re-write the first equation as: $y=-2x+4$ First equation yields: $(x+1)^2+(-2x+4-2)^2=4$ or, $5x^2-6x+1=0$ or,$(5x-1)(x-1)=0$ or, $x=${$\dfrac{1}{5},1$} From first equation when $x=\dfrac{1}{5}$, we have $y=\dfrac{18}{5}$ From first equation when $x=1$, we have $y=2$ Thus, solution set is {$(\dfrac{1}{5},\dfrac{18}{5}),(1,2)$}