Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 662: 33

See below

Given: $y^2+x-3=0\\x^2-4x+3y+1=0$ The first equation becomes: $$x=3-y^2$$ Substitute $x$: $(3-y^2)^2-4(3-y^2)+3y+1=0\\9-6y^2+y^4-12+4y^2+3y+1=0\\y^4-2y^2+3y-2=0$ Rational root theorem gives us: $y=1\\y=-2$ Find x: $x_1=3-y^2=3-(1)^2=2\\x_2=3-(-2)^2=-1$