Answer:
1. [tex]x^2-6x+8=0[/tex]
2. [tex](x-4)(x-2)=0[/tex]
3. [tex]x=4 \text{ and } x=2[/tex]
4. [tex]x=4 \text{ and } x=2[/tex]
Step-by-step explanation:
1. When using substitution all we do would be is substituse the y for a 3.
This leaves us with the equation:
[tex]3=-x^2+6x-5[/tex]
Rewriting it we get:
[tex]0=-x^2+6x-8[/tex]
or if we shift the 0 to the other side:
[tex]x^2-6x+8=0[/tex]
2. In order to factor the equation we can use the butterfly method:
[tex]\left[\begin{array}{ccc}1&-4\\1&-2\end{array}\right][/tex]
So it factors out to:
[tex](x-4)(x-2)=0[/tex]
You can also use the quadratic formula.
3. To find the solutions we just set each factor to 0
[tex]x-4=0\\\text{and}\\x-2=0[/tex]
So the x-values would be:
[tex]x=4 \text{ and } x=2[/tex]
4. To find the solution to the system we just plug in the values and it turns out to be the same numbers as before.
