How do you solve $y = x^{2}, y = - 2 x - 1$ $y=x^2, y=-2x-1$ using substitution?

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Answer 1 · 2017-09-10T19:10:21

$(x, y) = (- 1, 1) .$ $(x,y)=(-1,1).$ graph{(y-x^2)(y+2x+1)=0 [-10, 10, -5, 5]}

Let, $y=x^2......(1), and, y=-2x-1......(2).$

Sub.ing $y$ from $(1)$ in $(2),$ we get,

$x^2=-2x-1, or, x^2+2x+1=0.$

$:. (x+1)^2=0.$

$:. x=-1, and,"by "(1)," then, "y=x^2=(-1)^2=1.$

Hence, the Soln. is, $(x,y)=(-1,1).$

N.B. : Geometrically, the Line $y=-2x-1$ is tangent to the

Parabola $y=x^2.$

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How do you solve y=x^2, y=-2x-1y=x2,y=−2x−1y=x^2, y=-2x-1 using substitution?