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3 August, 20:32

Solve the system of equations below algebraically.

5x^2 + y^2 - x + 20y - 48 = 0

x-2y+3=0

A) (0,0) and (1, 2)

B) (0, 2)

C) (1, - 3)

D) (-3,0) and (1, 2)

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Answers (1)
  1. 3 August, 23:17
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    D) (-3, 0) and (1, 2).

    Step-by-step explanation:

    5x^2 + y^2 - x + 20y - 48 = 0

    x - 2y + 3 = 0

    From the second equation:

    x = 2y - 3, so we substitute this in the first equation:

    5 (2y - 3) ^2 + y^2 - (2y - 3) + 20y - 48 = 0

    5 (4y^2 - 12y + 9) + y^2 - 2y + 3 + 20y - 48 = 0

    20y^2 - 60y + 45 + y^2 - 2y + 3 + 20y - 48 = 0

    21y^2 - 42y = 0

    21y (y - 2) = 0

    y = 0, 2.

    So when y = 0 x = 2 (0) - 3 = - 3

    and when y = 2, x = 2 (2) - 3 = 1.
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