so first we solve for y in second equation
3x+y=4
y=4-3x
sub for y in first equiton
2x-(4-3x)^2=1
2x-(16-12x-12x+9x^2)=1
2x-(16-24x+9x^2)=1
2x-16+24x-9x^2=1
-9x^2+26x-16=1
times -1 both sides
9x^2-26x+16=-1
add 1 to both sides
9x^2-26x+17=0
factor
(x-1)(9x-17)=0
set each to zero
x-1=0
x=1
9x-17=0
9x=17
x=17/9
so sub back to find y
y=4-3x
y=4-3(1)
y=4-3
y=1
(1,1)
y=4-3x
y=4-3(17/9)
y=4-17/3
y=12/3-17/3
y=-5/3
(17/9,-5/3)
the 2 intersection points are
(1,1) and [tex]( \frac{17}{9} , \frac{-5}{3} )[/tex]
