Question 529656
can you help me solve the system algebraically
y=2(x+3)^2-5
y=14x+17


Since the 2 equations are equal to y, then we can say that:
{{{2(x + 3)^2 - 5 = 14x + 17}}}


{{{2(x^2 + 6x + 9) - 5 = 14x + 17}}}


{{{2x^2 + 12x + 18 - 5 = 14x + 17}}}


{{{2x^2 + 12x + 13 = 14x + 17}}}


{{{2x^2 + 12x - 14x + 13 - 17 = 0}}}


{{{2x^2 - 2x - 4 = 0}}}


{{{2(x^2 - x - 2) = 2(0)}}}


{{{x^2 - x - 2 = 0}}}


(x - 2)(x + 1) = 0


x = 2 or – 1


When x = 2, then:
y = 14(2) + 17, or 28 + 17, or {{{y = 45}}}. Solution: {{{highlight_green(x = 2_when_y =45)}}}


When x = - 1, then:
y = 14(- 1) + 17, or - 14 + 17, or {{{y = 3}}}. Solution: {{{highlight_green(x =- 1_when_y =3)}}}


You should be able to do the check to make sure that the solution pairs make the 2 equations true.


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