Use the elimination method to solve the linear-quadratic system y=x2+x−42

Question

, y=−x+6
.(1 point)
Responses

(6,0)
left parenthesis 6 comma 0 right parenthesis

(8,−2)
and (−6,12)
left parenthesis 8 comma negative 2 right parenthesis and left parenthesis negative 6 comma 12 right parenthesis

(−8,14)
left parenthesis negative 8 comma 14 right parenthesis

(−8,14)
and (6,0)

Answer 1

To solve the system of equations using the elimination method, we have the following equations:

1. $y=x^2+x-42$
2. $y=-x+6$

First, we will set them equal to each other because both equations equal $y$:

$$x^2+x-42=-x+6$$

Now, let's move all terms to one side of the equation:

$$x^2+x+x-42-6=0$$

This simplifies to:

$$x^2+2x-48=0$$

Next, we can factor the quadratic:

$$(x+8)(x-6)=0$$

Setting each factor to zero gives us the $x$ values:

1. $x+8=0$ → $x=-8$
2. $x-6=0$ → $x=6$

Now, we will find the corresponding $y$ values for each $x$.

For $x=-8$: Substituting into the second equation $y=-x+6$: $$y=-(-8)+6=8+6=14$$ So one solution is $(-8,14)$.

For $x=6$: Substituting into the second equation $y=-x+6$: $$y=-6+6=0$$ So the other solution is $(6,0)$.

Thus, the solutions to the system are:

$(-8,14)$ and $(6,0)$.

Therefore, the answer is:

(−8,14) and (6,0).