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

Responses

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

(−2,−2) and (−2,2)

1 answer

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

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

Since the elimination method typically applies to linear equations, we can substitute the expression for \( y \) from the linear equation into the quadratic equation.

Now, set the second equation equal to the first:

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

Next, we will rearrange this equation to form a standard quadratic equation:

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

Now, we can factor this quadratic:

\[ 0 = (x - 2)(x - 2) \]

Thus, we have a double root:

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

Next, we can find the corresponding \( y \) value using the linear equation:

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

So, the solution to the system is:

\[ (2, -2) \]

Now, let's check our answer. We can plug \( x = 2 \) back into the quadratic equation:

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

This confirms that both equations yield the same point \((2, -2)\).

Thus, the correct response is:

\[ (2, -2) \]