Question

what is one of the solutions to the following system?\\(\

$$\begin{cases} y - 3 = x \\\\ x^2 - 6x + 13 = y \\end{cases}$$

\\)\\( (-5, 2) \\)\\( (-2, 1) \\)\\( (2, 5) \\)\\( (8, 5) \\)

Explanation:

Step1: Substitute \( y \) from first equation into second

From \( y - 3 = x \), we get \( y = x + 3 \). Substitute into \( x^2 - 6x + 13 = y \):
\( x^2 - 6x + 13 = x + 3 \)

Step2: Simplify the equation

Rearrange to standard quadratic form:
\( x^2 - 7x + 10 = 0 \)

Step3: Factor the quadratic

Factor: \( (x - 2)(x - 5) = 0 \)

Step4: Solve for \( x \)

Set each factor to zero: \( x - 2 = 0 \) or \( x - 5 = 0 \), so \( x = 2 \) or \( x = 5 \)

Step5: Find corresponding \( y \)

For \( x = 2 \), \( y = 2 + 3 = 5 \) (gives point \( (2, 5) \))
For \( x = 5 \), \( y = 5 + 3 = 8 \) (not among options, so check options with \( x = 2 \))

Answer:

(2, 5)