Question

solve the following system of equations algebraically: y= x² + 20x + 63 y= 3x - 9The solutions should be in points.

Answer 1

Given the following System of equations:

`\begin{cases}y=x^2+20x+63 \\ y=3x-9​\end{cases}`

You can apply the following procedure to solve it:

1. Make the equations equal to each other:

`x^(2)+20x+63=3x-9​`

2. Solve for the variable "x":

- Make the equation equal to zero:

`\begin{gathered} x^2+20x+63-3x+9=0 \\ x^2+17x+72=0 \end{gathered}`

- You need to use the Quadratic formula to find the values of "x". This is:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

In this case:

`\begin{gathered} a=1 \\ b=17 \\ c=72 \end{gathered}`

Substituting values and evaluating, you get:

`\begin{gathered} x=\frac{-17\pm\sqrt[]{17^2-4(1)(72)}}{2(1)} \\ \\ x=\frac{-17+\sqrt[]{1}}{2} \\ \\ x=\frac{-17-\sqrt[]{1}}{2} \\ \\ x=-8;x=-9 \end{gathered}`

3. Substitute each value of "x" into the second original equation and evaluate in order to find the corresponding values of "y":

- For:

`x=-8`

You get:

`\begin{gathered} y=3(-8)-9​ \\ y=-24-9 \\ y=-33 \end{gathered}`

- For:

`x=-9`

You get:

`\begin{gathered} y=3(-9)-9​ \\ y=-27-9 \\ y=-36 \end{gathered}`

The points are:

`(-8,-33);(-9,-36)`