To solve the system of equations using the substitution method, we need to set the two equations equal to each other.
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The equations given are: \[ y = x^2 - x - 40 \quad (1) \] \[ y = -3x - 5 \quad (2) \]
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Set the two equations equal to each other by substituting \(-3x - 5\) for \(y\) in equation (1): \[ -3x - 5 = x^2 - x - 40 \]
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Rearrange the equation to set it to zero: \[ x^2 - x - 40 + 3x + 5 = 0 \] Simplifying this gives: \[ x^2 + 2x - 35 = 0 \]
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Now, factor the quadratic equation: \[ (x + 7)(x - 5) = 0 \]
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Set each factor to zero to find the values of \(x\): \[ x + 7 = 0 \quad \Rightarrow \quad x = -7 \] \[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \]
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Now, substitute each value of \(x\) back into equation (2) to find the corresponding \(y\) values.
For \(x = -7\): \[ y = -3(-7) - 5 = 21 - 5 = 16 \quad \Rightarrow \quad (-7, 16) \]
For \(x = 5\): \[ y = -3(5) - 5 = -15 - 5 = -20 \quad \Rightarrow \quad (5, -20) \]
Thus, the solutions to the linear-quadratic system are: \[ (5, -20) \quad \text{and} \quad (-7, 16) \]
Since these points are present in your responses, the correct answer is: (5,−20) and (−7,16).