Use the elimination method to solve the linear-quadratic system y = x^2 + 10, y = -7x -2

Question

A. (-3,19)
B. (-4,26)
C. (3,-23) and (4,-30)
D. (-3,19) and (-4,26)

Answer 1

To solve the system of equations using the elimination method, we start with the given equations:

1. $y=x^2+10$ (Equation 1)
2. $y=-7x-2$ (Equation 2)

Since both equations are equal to $y$, we can set them equal to each other:

$$x^2+10=-7x-2$$

Now, we'll rearrange this equation to one side:

$$x^2+7x+12=0$$

Next, we can factor this quadratic equation:

$$(x+3)(x+4)=0$$

Setting each factor equal to zero gives us the solutions for $x$:

1. $x+3=0$ → $x=-3$
2. $x+4=0$ → $x=-4$

Now we can substitute these values back into either original equation to find the corresponding values of $y$.

For $x=-3$:

Using Equation 2:

$$y=-7(-3)-2=21-2=19$$

So one point is $(-3,19)$.

For $x=-4$:

Using Equation 2 again:

$$y=-7(-4)-2=28-2=26$$

So another point is $(-4,26)$.

Thus, the solutions for the system are the points $(-3,19)$ and $(-4,26)$.

Therefore, the correct answer is:

D. (-3,19) and (-4,26).