Answer:
(2,2) and (3,1)
Step-by-step explanation:
Since we're given the value of y in the second equation, we can replace the y on the left side of the first equation with 4-x, giving us
[tex]4-x=x^2-6x+10\\0=x^2-5x+6[/tex]
We can factor the expression on the right to get us
[tex]0=x^2-5x+6\\0=x^2-3x-2x+6\\0=x(x-3)-2(x-3)\\0=(x-2)(x-3)[/tex]
Solving [tex]x-2=0[/tex] and [tex]x-3=0[/tex] gets us the solutions [tex]x=2,3[/tex], which we can plug into the second equation to get us
[tex]y=4-2=2\\y=4-3=1[/tex]
So, our solution set is the pair of points (2,2) and (3,1)
