Question

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Answer 1

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

We have
`y=x^2-6x+13` and
`x=3-y`.

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


`0=x^2-7x+10 \\ 0=(x-5)(x-2) \\ x = 2, \ x = 5`

Now, substitute each value of
`x` into one of the original equations to solve for
`y`.


`y-3=2 \\ y = 5`


`y-3=5 \\ y=8`

`(2,5),(5,8)`

Answer 2

Answer:
(2,5)

Step-by-step 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)