Given the following System of equations:
[tex]\begin{cases}y=x^{2}-15x+60 \\ y=x-3\end{cases}[/tex]You can solve it as following:
1. You can make the equations equal to each other:
[tex]x^{2}-15x+60=x-3[/tex]2. Now you must solve for the variable "x". See the procedure below:
[tex]\begin{gathered} x^2-15x+60-x+3=0 \\ x^2-16x+63=0 \end{gathered}[/tex]Notice that you need to find two numbers whose sum is -16 and whose product is 63. These numbers are -7 and -9, because:
[tex]\begin{gathered} -7-9=-16 \\ \\ (-7)(-9)=63 \end{gathered}[/tex]Then, you get:
[tex]\begin{gathered} (x-7)(x-9)=0 \\ \\ x=7;x=9 \end{gathered}[/tex]3. Substitute each value of "x" into the second equation and then evaluate, in order to find the correponding values of "y".
- For:
[tex]x=7[/tex]You get:
[tex]\begin{gathered} y=7-3 \\ y=4 \end{gathered}[/tex]- For:
[tex]x=9[/tex]You get:
[tex]\begin{gathered} y=9-3 \\ y=6 \end{gathered}[/tex]Therefore, the points are:
[tex](7,4);(9,6)[/tex]
