Question

x

Answer 1

\(\bf \begin{cases}
{\color{brown}{ y}}=-x^2+36\\
{\color{brown}{ y}}=2x+21
\end{cases}\qquad well\qquad
\begin{array}{cccllll}
{\color{brown}{ y}}&=&{\color{brown}{ y}}
\\ \quad \\
-x^2+36&=&2x+21
\end{array}\qquad thus
\\ \quad \\
-x^2+36=2x+21\implies 36=x^2+2x+21
\\ \quad \\
0=x^2+2x+21-36\implies0=x^2+2x-15\qquad
\\ \quad \\
notice\qquad
\begin{array}{cccllll}
x^2&+2x&-15\\
&\uparrow &\uparrow \\
&5-3&5\cdot -3
\end{array}
\\ \quad \\
0=(x+5)(x-3)\)

solve for "x"
once you found "x", get "y" by substitution