Question

solve system by substitution

y=3x-20

y=-x^(2)+34

Answer 1

`\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)=(-b-√(\Delta ))/(2a) =(-3-√(225))/(2)=(-3-15)/(2)= (-18)/(2)=-9`


`x_(2)=(-b+√(\Delta ))/(2a) =(-3+√(225))/(2)=(-3+15)/(2)= (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}`