To find the solutions to the system given by the equations \( y = x^2 + 3x - 13 \) and \( y = 5 \), we can set them equal to each other.
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Substitute \( y = 5 \) into the first equation: \[ 5 = x^2 + 3x - 13 \]
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Rearranging the equation gives: \[ x^2 + 3x - 13 - 5 = 0 \implies x^2 + 3x - 18 = 0 \]
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Next, we can factor the quadratic: \[ (x + 6)(x - 3) = 0 \] This gives us the solutions: \[ x + 6 = 0 \implies x = -6 \] \[ x - 3 = 0 \implies x = 3 \]
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Now, we have two \( x \) values: \( x = -6 \) and \( x = 3 \). Since both solutions correspond to \( y = 5 \), the points of intersection (solutions to the system) are: \[ (-6, 5) \text{ and } (3, 5) \]
Looking at your response options, the answer matches with (3, 5) and (-6, 5).
Thus, the correct answer is: (3, 5) and (-6, 5).