Given
system of equations:
[tex]\begin{gathered} y=-\left(x+2\right)^2+1 \\ y=3x+7 \end{gathered}[/tex]
Find
Solutions of the system of equations
Explanation
as we have given
[tex]\begin{gathered} y=-\left(x+2\right)^2+1 \\ y=3x+7 \end{gathered}[/tex]
so , from both equations , we obtain
[tex]\begin{gathered} -(x+2)^2+1=3x+7 \\ -(x^2+4+4x)+1=3x+7 \\ -x^2-4x-4+1=3x+7 \\ -x^2-4x-3x-3-7=0 \\ -x^2-7x-10=0 \\ x^2+7x+10=0 \end{gathered}[/tex]
now , solve the quadratic equation ,
[tex]\begin{gathered} x^2+7x+10=0 \\ x^2+5x+2x+10=0 \\ x(x+5)+2(x+5)=0 \\ (x+2)(x+5)=0 \\ x=-2,-5 \end{gathered}[/tex]
so , when x = -2 then ,
[tex]\begin{gathered} y=3x+7 \\ y=3(-2)+7=1 \end{gathered}[/tex]
when x = -5 then
[tex]\begin{gathered} y=3x+7 \\ y=3(-5)+7=-8 \end{gathered}[/tex]
Final Answer
Hence , the solutions of the system of equations are (-2 , 1) and (-5 , -8). So , the correct option is B
