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Today, 03:44

Solve the system of equations. y = x + 3 y = x2 - 2x - 1 A. (1,4) and (-4,-1) B. (-1,7) and (4,2) C. (-1,4) and (4,-1) D. (-1,2) and (4,7)

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  1. Today, 04:04
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    Step-by-step explanation:

    Given the 2 equations

    y = x + 3 → (1)

    y = x² - 2x - 1 → (2)

    Substitute y = x² - 2x - 1 into (1)

    x² - 2x - 1 = x + 3 ← subtract x + 3 from both sides

    x² - 3x - 4 = 0 ← in standard form

    (x - 4) (x + 1) = 0 ← in factored form

    Equate each factor to zero and solve for x

    x - 4 = 0 ⇒ x = 4

    x + 1 = 0 ⇒ x = - 1

    Substitute these values into (1) for corresponding values of y

    x = 4 ⇒ y = 4 + 3 = 7 ⇒ (4, 7)

    x = - 1 ⇒ y = - 1 + 3 = 2 ⇒ ( - 1, 2)
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