Answer:
1.
![x^2-6x+8=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/qil45skc6anc86x6gtsauvjnobgy443vlm.png)
2.
![(x-4)(x-2)=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/e1zyd366peeh6gy1uxpsq75vlhbcxkzxaa.png)
3.
![x=4 \text{ and } x=2](https://img.qammunity.org/2023/formulas/mathematics/high-school/7cxytd49huapslsz3y81sx8yegfchp747m.png)
4.
![x=4 \text{ and } x=2](https://img.qammunity.org/2023/formulas/mathematics/high-school/7cxytd49huapslsz3y81sx8yegfchp747m.png)
Explanation:
1. When using substitution all we do would be is substituse the y for a 3.
This leaves us with the equation:
![3=-x^2+6x-5](https://img.qammunity.org/2023/formulas/mathematics/high-school/nywvu4ywu983c3u47zfvc4evy32wgnpd3c.png)
Rewriting it we get:
![0=-x^2+6x-8](https://img.qammunity.org/2023/formulas/mathematics/high-school/dmiuisko60tjcxfcpfmkbj9unrti5jyx54.png)
or if we shift the 0 to the other side:
![x^2-6x+8=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/qil45skc6anc86x6gtsauvjnobgy443vlm.png)
2. In order to factor the equation we can use the butterfly method:
![\left[\begin{array}{ccc}1&-4\\1&-2\end{array}\right]](https://img.qammunity.org/2023/formulas/mathematics/high-school/teskhcconnowaybh8a1x16asuj1s4uwu1s.png)
So it factors out to:
![(x-4)(x-2)=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/e1zyd366peeh6gy1uxpsq75vlhbcxkzxaa.png)
You can also use the quadratic formula.
3. To find the solutions we just set each factor to 0
![x-4=0\\\text{and}\\x-2=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/w2xvlo76xngdmh86xzysfidahepyhupbe8.png)
So the x-values would be:
![x=4 \text{ and } x=2](https://img.qammunity.org/2023/formulas/mathematics/high-school/7cxytd49huapslsz3y81sx8yegfchp747m.png)
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.