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21 June, 07:13

What is the solution of the linear-quadratic system of equations?

y=x^2+5x-3

y-x=2

+3
Answers (1)
  1. 21 June, 07:16
    0
    -x=-y+2

    x=y-2

    now substitute iin the first one

    y = (y-2) ^2+5 (y-2) - 3

    y=y^2-4y+4 + 5y-10-3

    y=y^2+y-9

    y^2-9=0.

    y^2=9

    y=3

    3-x=2

    -x=2-3

    -x=-1

    x=1

    the solution are x=1 and y=3

    3=1^2+5 (1) - 3

    3=1+5-3

    3=6-3

    3=3
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