Question

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

Answer 1

Answer:

The solutions of the system of equations are the points

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

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

Step-by-step 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

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

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$  -----> $x=\frac{1-\sqrt{33}} {2}$  

$x=\frac{-1-\sqrt{33}} {-2}$  -----> $x=\frac{1+\sqrt{33}} {2}$  

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$  

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$  ---->$y=\frac{5-\sqrt{33}} {2}$  

For $x=\frac{1+\sqrt{33}} {2}$  

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$  ---->$y=\frac{5+\sqrt{33}} {2}$  

therefore

The solutions of the system of equations are the points

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

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