Question

Use the elimination method to solve the linear-quadratic system y=x2−2x−2, y=2x−6.(1 point)%0D%0AResponses%0D%0A%0D%0A(2,−2)%0D%0Aleft parenthesis 2 comma negative 2 right parenthesis%0D%0A%0D%0A(−2,−2) and (−2,2)%0D%0A left parenthesis negative 2 comma negative 2 right parenthesis and left parenthesis negative 2 comma 2 right parenthesis%0D%0A%0D%0A(2,−2) and (−2,2)%0D%0Aleft parenthesis 2 comma negative 2 right parenthesis and left parenthesis negative 2 comma 2 right parenthesis%0D%0A%0D%0A(−2,−10)

Answer 1

To solve the linear-quadratic system using the elimination method, we have the following equations:

1. $y=x^2-2x-2$ (quadratic equation)
2. $y=2x-6$ (linear equation)

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

$$x^2-2x-2=2x-6$$

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

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

This simplifies to:

$$x^2-4x+4=0$$

Next, factor the quadratic:

$$(x-2{)}^2=0$$

This implies:

$$x-2=0\Rightarrow x=2$$

Now we will substitute $x=2$ into the linear equation to find $y$:

$$y=2(2)-6=4-6=-2$$

Thus, we have one solution:

$$(2,-2)$$

Now, let's check if there are any other solutions by substituting $y=2x-6$ into the quadratic equation again. Since we found $(2,-2)$, we can check for a possible repeat:

1. Substitute $y$ into the quadratic: $2x-6=x^2-2x-2$
   • Rearranging gives us: $$x^2-4x+4=0⟹(x-2{)}^2=0$$

We already established that the quadratic equation has a double root at $x=2$ with $y=-2$. This means the only solution for the system is:

$$(2,-2)$$

Hence, the answer is:

$$\text{(2, −2)}$$