Question

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

Substitution
Please show your work
Need ASAP

Answer 1

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})`