How do you solve the system $x^2-2x+2y+2=0$ and $-x^2+2x-y+3=0$ ?

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How do you solve the system $x^2-2x+2y+2=0$ and $-x^2+2x-y+3=0$ ?

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Answer 1 · 2018-06-03T10:57:12

$(-2,-5),(4,5)$

Explanation:

$x^2-2x+2y+2=0to(1)$

$-x^2+2x-y+3=0to(2)$

$"add equations "(1)" and "(2)" term by term"$

$(x^2-x^2)+(-2x+2x)+(2y-y)+5=0$

$y+5=0rArry=-5$

$"substitute "y=-5" into either "(1)" or "(2)" and"$
$"solve for x"$

$(1)tox^2-2x-10+2=0$

$x^2-2x-8=0larrcolor(blue)"in standard form"$

$(x-4)(x+2)=0$

$x-4=0rArrx=4$

$x+2=0rArrx=-2$

$"solutions are "(-2,-5)" or "(4,-5)$
graph{(x^2-2x+2y+2)(x^2-2x+y-3)=0 [-10, 10, -5, 5]}

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How do you solve the system $x^2-2x+2y+2=0$ and $-x^2+2x-y+3=0$ ?

1 Answer

Explanation:

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Humanities

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How do you solve the system x^2-2x+2y+2=0x^2-2x+2y+2=0 and -x^2+2x-y+3=0-x^2+2x-y+3=0?

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Explanation:

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How do you solve the system $x^2-2x+2y+2=0$ and $-x^2+2x-y+3=0$ ?