Question 895219
Listed under "Simultaneous equation of which one is linear and one is quadratic"
y=x^2+2x-15 and x=2y+10
sub (2y+10)for x in 1st equation
y=(2y+10)^2+2(2y+10)-15
y=4y^2+40y+100+4y+20-15
4y^2+43y+105=0
(y+7)(4y+15)=0
y=-7
x=-4
or
y=-15/4=-3.75
x=2.5
points of intersection: (-4,-7) and (2.5,-3.75)