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Y=-x^2+2x+10
y=x+2

Substitution
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1 Answer

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Answer:

The solutions of the system of equations are the points


(\frac{1-√(33)} {2},\frac{5-√(33)} {2})


(\frac{1+√(33)} {2},\frac{5+√(33)} {2})

Explanation:

we have


y=-x^(2) +2x+10 ----> equation A


y=x+2 ----> equation B

Solve the system by substitution

substitute equation B in equation A


x+2=-x^(2) +2x+10

solve for x


-x^(2) +2x+10-x-2=0


-x^(2) +x+8=0

The formula to solve a quadratic equation of the form


ax^(2) +bx+c=0

is equal to


x=\frac{-b\pm\sqrt{b^(2)-4ac}} {2a}

in this problem we have


-x^(2) +x+8=0

so


a=-1\\b=1\\c=8

substitute in the formula


x=\frac{-1\pm\sqrt{1^(2)-4(-1)(8)}} {2(-1)}


x=\frac{-1\pm√(33)} {-2}


x=\frac{-1+√(33)} {-2} ----->
x=\frac{1-√(33)} {2}


x=\frac{-1-√(33)} {-2} ----->
x=\frac{1+√(33)} {2}

Find the values of y

For
x=\frac{1-√(33)} {2}


y=x+2


y=\frac{1-√(33)} {2}+2 ---->
y=\frac{5-√(33)} {2}

For
x=\frac{1+√(33)} {2}


y=x+2


y=\frac{1+√(33)} {2}+2 ---->
y=\frac{5+√(33)} {2}

therefore

The solutions of the system of equations are the points


(\frac{1-√(33)} {2},\frac{5-√(33)} {2})


(\frac{1+√(33)} {2},\frac{5+√(33)} {2})

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