To solve the system of equations given by the linear equation \(y = x + 2\) and the quadratic equation \(y = x^2 - x - 6\), we can set them equal to each other since they both equal \(y\):
\[ x + 2 = x^2 - x - 6 \]
Now, we can rearrange this equation to form a standard quadratic equation:
\[ x^2 - x - 6 - x - 2 = 0 \] \[ x^2 - 2x - 8 = 0 \]
Next, we can factor the quadratic:
\[ (x - 4)(x + 2) = 0 \]
This gives us two solutions for \(x\):
\[ x - 4 = 0 \implies x = 4 \] \[ x + 2 = 0 \implies x = -2 \]
Now, we can find the corresponding \(y\) values using the linear equation \(y = x + 2\):
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For \(x = 4\): \[ y = 4 + 2 = 6 \] So one solution is \((4, 6)\).
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For \(x = -2\): \[ y = -2 + 2 = 0 \] So the other solution is \((-2, 0)\).
Thus, the solutions to the system are:
\[ (-2, 0) \quad \text{and} \quad (4, 6) \]
Among the options given, the correct response is:
(−2, 0) and (4, 6).