##### Solution to given system of equations

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These are the equations: y = -3x + 5 and y = 2x^2 + 8x - 16

A: No Solutions, B: All real numbers, C: (-7, 26), (3/2, 1/2), and D: (7/6, 3/2),(4,-7)

May 6th, 2015

$y=-3x+5---(1)\\ y=2x^2+8x-16---(2)\\$

From equations (1) and (2) we get

$-3x+5=2x^2+8x-16\\ 2x^2+8x-16+3x-5=0\\ 2x^2+11x-21=0\\ a=2\\ b=11\\ c=-21\\ We \hspace{5}can \hspace{5}solve\hspace{5}using \hspace{5}the \hspace{5}quadratic \hspace{5}formula\\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x=\frac{-11\pm\sqrt{11^2-4*2*(-21)}}{2*2}\\ x=\frac{-11\pm\sqrt{121+168}}{4}\\ x=\frac{-11\pm\sqrt{289}}{4}\\ x=\frac{-11\pm17}{4}\\ x=\frac{-11+17}{4}\hspace{5}or\hspace{5}x=\frac{-11-17}{4}\\ x=\frac{6}{4}\hspace{5}or\hspace{5}x=\frac{-28}{4}\\ x=1.5\hspace{5}or\hspace{5}x=-7\\$

Put x = 1.5 in equation (1)

y = -3*1.5 + 5

y = -4.5 + 5

y = 0.5

Put x = -7 in equation (1)

y = -3*(-7) + 5

y = 21 + 5

y = 26

So the line and the curve intersect at 2 points

(1.5, 0.5)  and (-7, 26)

ANSWER:  C) (-7, 26), (3/2, 1/2)

May 6th, 2015

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May 6th, 2015
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May 6th, 2015
Dec 8th, 2016
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