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Use the substitution method to find the solutions to the linear-quadratic system y = x^2 - x - 40, y = -3x - 5? (1 point)
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Set x^2 - x - 40 = -3x - 5.
x^2 + 2x - 35 = 0 → (x + 7)(x - 5) = 0 → x = -7 or x = 5.
For x = -7: y = -3(-7) - 5 = 21 - 5 = 16.
For x = 5: y = -3(5) - 5 = -15 - 5 = -20.
Solutions: (-7, 16) and (5, -20).
x^2 + 2x - 35 = 0 → (x + 7)(x - 5) = 0 → x = -7 or x = 5.
For x = -7: y = -3(-7) - 5 = 21 - 5 = 16.
For x = 5: y = -3(5) - 5 = -15 - 5 = -20.
Solutions: (-7, 16) and (5, -20).
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