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susanwiederspan
(2, 5) and (5, 8) are the solutions to the system of equations.

We have [tex]y=x^2-6x+13[/tex] and [tex]x=3-y[/tex]. 

Let's substitute [tex]x+3[/tex] for [tex]y[/tex] in [tex]y=x^2-6x+13[/tex]. We get [tex](x+3)=x^2-6x+13[/tex]. Now, factor to solve for [tex]x[/tex].

[tex]0=x^2-7x+10 \\ 0=(x-5)(x-2) \\ x = 2, \ x = 5[/tex]

Now, substitute each value of [tex]x[/tex] into one of the original equations to solve for [tex]y[/tex]. 

[tex]y-3=2 \\ y = 5[/tex]

[tex]y-3=5 \\ y=8[/tex]
[tex](2,5),(5,8)[/tex]

Louli
Answer:
(2,5)

Explanation:
The solution of the system is the order pair that satisfies both equations.
The first given equation is:
x = y - 3
This can be written as:
y = x+3 .......> I

The second given equation is:
x^2 - 6x + 13 = y .........> II

Substitute with I in II and solve for x as follows:
x^2 - 6x + 13 = y
x^2 - 6x + 13 = x + 3
x^2 - 6x + 13 - x - 3 = 0
x^2 - 7x + 10 = 0
(x-5)(x-2) = 0
either x = 5
or x = 2

Substitute with x in equation I to get y as follows:
at x = 5 : y = 5+3 = 8...> first solution is (5,8)
at x = 2 : y = 2+3 = 5...> second solution is (2,5)

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