Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 662: 25

$(0.5,0.5)$

Adding the first and the second equation we get: $-4x-4y+4=0\\-x-y+1=0\\x=1-y$ Plugging this into the first equation we get: $(1-y)^2-y^2-4(1-y)+2=0\\1+y^2-2y-y^2-4+4y+2=0\\2y-1=0\\y=0.5$ Thus $x=1-0.5=0.5$. Thus the solutions are: $(0.5,0.5)$