Question

solve the following system of equations algebraically:y= x² - 15x + 60y= x - 3 What are the points for the final solution?

Answer 1

Given the following System of equations:

`\begin{cases}y=x^(2)-15x+60 \\ y=x-3​\end{cases}`

You can solve it as following:

1. You can make the equations equal to each other:

`x^(2)-15x+60=x-3​`

2. Now you must solve for the variable "x". See the procedure below:

`\begin{gathered} x^2-15x+60-x+3=0 \\ x^2-16x+63=0 \end{gathered}`

Notice that you need to find two numbers whose sum is -16 and whose product is 63. These numbers are -7 and -9, because:

`\begin{gathered} -7-9=-16 \\ \\ (-7)(-9)=63 \end{gathered}`

Then, you get:

`\begin{gathered} (x-7)(x-9)=0 \\ \\ x=7;x=9 \end{gathered}`

3. Substitute each value of "x" into the second equation and then evaluate, in order to find the correponding values of "y".

- For:

`x=7`

You get:

`\begin{gathered} y=7-3​ \\ y=4 \end{gathered}`

- For:

`x=9`

You get:

`\begin{gathered} y=9-3​ \\ y=6 \end{gathered}`

Therefore, the points are:

`(7,4);(9,6)`