liyaredmon
6

# solve system by substitution y=3x-20y=-x^(2)+34

$\begin{cases} y=3x-20 \\ y=-x^2+34 \end{cases}\\ \\ \begin{cases} y=3x-20 \\3x-20=-x^2+34 \end{cases}\\ \\\begin{cases} y=3x-20 \\x^2+3x-20-34=0 \end{cases}\\ \\\begin{cases} y=3x-20 \\x^2+3x-54=0 \end{cases}$ $\begin{cases} y=3x-20 \\x^2+3x-54=0 \end{cases} \\ \\x^2+3x-54=0 \\ \\a=1, \ \ b=3, \ \ c= -54 \\ \\ \Delta =b^2 -4ac =3^2-4*1*(-54)=9+216=225 \\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-3-\sqrt{225}}{2}=\frac{-3-15}{2}= \frac{-18}{2}=-9$ $x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{-3+\sqrt{225}}{2}=\frac{-3+15}{2}= \frac{12}{2}=6$ $\begin{cases} y=3x-20 \\ x=-9 \end{cases}\ \ \vee \ \ \begin{cases} y=3x-20 \\x=6 \end{cases}\\ \\ \begin{cases} y=3*(-9)-20 \\ x=-9 \end{cases}\ \ \vee \ \ \begin{cases} y=3*6-20 \\x=6 \end{cases}\\ \\ \begin{cases} y=-27-20 \\ x=-9 \end{cases}\ \ \vee \ \ \begin{cases} y=18-20 \\x=6 \end{cases}\\ \\ \begin{cases} y=-47 \\ x=-9 \end{cases}\ \ \vee \ \ \begin{cases} y=-2\\x=6 \end{cases}$