Given the system of equations:
[tex]\begin{gathered} y=x^2+2x+6 \\ y=-x+4 \end{gathered}[/tex]
Using the substitution :
So,
[tex]x^2+2x+6=-x+4[/tex]
Solve for x:
[tex]\begin{gathered} x^2+2x+6=-x+4 \\ x^2+2x+x+6-4=0 \\ x^2+3x+2=0 \end{gathered}[/tex]
Factor the last equation :
[tex]\begin{gathered} x^2+3x+2=0 \\ (x+1)(x+2)=0 \\ x+1=0,x+2=0 \\ x=-1\text{or x = -2} \end{gathered}[/tex]
So, the answer is option A
The solutions are: x = -1 , -2
y = 5 , 6
As order pairs , the solution will be : ( -1 , 5 ) and ( -2 , 6 )
The following image represents the graphical solution of the system
