Solving Simultaneous Equations Using Elimination Quick Check

Question

5 of 55 of 5 Items

Question
Use the elimination method to solve the linear-quadratic system y=x2−2x−2, y=2x−6.(1 point)
Responses

(2,−2) and (−2,2)
left parenthesis 2 comma negative 2 right parenthesis and left parenthesis negative 2 comma 2 right parenthesis

(−2,−10)
left parenthesis negative 2 comma negative 10 right parenthesis

(2,−2)
left parenthesis 2 comma negative 2 right parenthesis

(−2,−2) and (−2,2)

Answer 1

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

1. \( y = x^2 - 2x - 2 \) (Quadratic)
2. \( y = 2x - 6 \) (Linear)

Since both expressions are equal to \( y \), we can set them equal to each other to eliminate \( y \):

\[ x^2 - 2x - 2 = 2x - 6 \]

Now, let's rearrange and simplify the equation:

\[ x^2 - 2x - 2 - 2x + 6 = 0 \] \[ x^2 - 4x + 4 = 0 \] \[ (x - 2)^2 = 0 \]

This gives:

\[ x - 2 = 0 \implies x = 2 \]

Now, substituting \( x = 2 \) back into one of the original equations to find \( y \). We'll use the linear equation:

\[ y = 2(2) - 6 = 4 - 6 = -2 \]

Thus, we have one solution:

\[ (2, -2) \]

Now, to determine if there are any other solutions, we note that the equation \( (x - 2)^2 = 0 \) indicates that the only solution for \( x \) is \( 2 \), which means there is only one pair \((2, -2)\).

The answer is:

\[ (2, -2) \]

So, the correct response is:

(2, -2)