Answer:
The solutions of the system of equations are the points
[tex](\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})[/tex]
[tex](\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})[/tex]
Step-by-step explanation:
we have
[tex]y=-x^{2} +2x+10[/tex] ----> equation A
[tex]y=x+2[/tex] ----> equation B
Solve the system by substitution
substitute equation B in equation A
[tex]x+2=-x^{2} +2x+10[/tex]
solve for x
[tex]-x^{2} +2x+10-x-2=0[/tex]
[tex]-x^{2} +x+8=0[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]-x^{2} +x+8=0[/tex]
so
[tex]a=-1\\b=1\\c=8[/tex]
substitute in the formula
[tex]x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}[/tex]
[tex]x=\frac{-1\pm\sqrt{33}} {-2}[/tex]
[tex]x=\frac{-1+\sqrt{33}} {-2}[/tex] -----> [tex]x=\frac{1-\sqrt{33}} {2}[/tex]
[tex]x=\frac{-1-\sqrt{33}} {-2}[/tex] -----> [tex]x=\frac{1+\sqrt{33}} {2}[/tex]
Find the values of y
For [tex]x=\frac{1-\sqrt{33}} {2}[/tex]
[tex]y=x+2[/tex]
[tex]y=\frac{1-\sqrt{33}} {2}+2[/tex] ---->[tex]y=\frac{5-\sqrt{33}} {2}[/tex]
For [tex]x=\frac{1+\sqrt{33}} {2}[/tex]
[tex]y=x+2[/tex]
[tex]y=\frac{1+\sqrt{33}} {2}+2[/tex] ---->[tex]y=\frac{5+\sqrt{33}} {2}[/tex]
therefore
The solutions of the system of equations are the points
[tex](\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})[/tex]
[tex](\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})[/tex]
