Question

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

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

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

Explanation:

Step1: Substitute y from first equation

From \( y - 3 = x \), we get \( y = x + 3 \). Substitute this into the second equation \( x^2 - 6x + 13 = y \), so \( x^2 - 6x + 13 = x + 3 \).

Step2: Simplify the equation

Rearrange the equation: \( x^2 - 6x - x + 13 - 3 = 0 \), which becomes \( x^2 - 7x + 10 = 0 \).

Step3: Factor the quadratic

Factor \( x^2 - 7x + 10 \) as \( (x - 2)(x - 5) = 0 \).

Step4: Solve for x

Set each factor to zero: \( x - 2 = 0 \) gives \( x = 2 \); \( x - 5 = 0 \) gives \( x = 5 \).

Step5: Find corresponding y

For \( x = 2 \), \( y = 2 + 3 = 5 \), so the point is \( (2, 5) \). For \( x = 5 \), \( y = 5 + 3 = 8 \) (not in options). Check the options, \( (2, 5) \) is a solution. Also, we can check each option in the original equations:

• For \( (-5, 2) \): \( 2 - 3 = -1

eq -5 \), so no.

• For \( (-2, 1) \): \( 1 - 3 = -2 \), but \( (-2)^2 - 6(-2) + 13 = 4 + 12 + 13 = 29

eq 1 \), so no.

• For \( (2, 5) \): \( 5 - 3 = 2 \) (satisfies first equation), \( 2^2 - 6(2) + 13 = 4 - 12 + 13 = 5 \) (satisfies second equation).
• For \( (8, 5) \): \( 5 - 3 = 2

eq 8 \), so no.

Answer:

C. (2, 5) (assuming the options are labeled as A. (-5, 2), B. (-2, 1), C. (2, 5), D. (8, 5))