To determine which system of equations has the point (2, 5) as a solution, we can substitute \( x = 2 \) and \( y = 5 \) into each of the equations in the given options:
A.
- \( y = x - 8 \) → \( 5 = 2 - 8 \) → \( 5 \neq -6 \) (not a solution)
- \( 2x + y = 7 \) → \( 2(2) + 5 = 7 \) → \( 4 + 5 = 7 \) → \( 9 \neq 7 \) (not a solution)
B.
- \( y = -\frac{1}{2}x + 6 \) → \( 5 = -\frac{1}{2}(2) + 6 \) → \( 5 = -1 + 6 \) → \( 5 = 5 \) (a solution)
- \( y = 3x - 1 \) → \( 5 = 3(2) - 1 \) → \( 5 = 6 - 1 \) → \( 5 = 5 \) (a solution)
Since both equations are satisfied, (2, 5) is a solution to system B.
C.
- \( y = x + 2 \) → \( 5 = 2 + 2 \) → \( 5 = 4 \) (not a solution)
- \( y = x + 5 \) → \( 5 = 2 + 5 \) → \( 5 = 7 \) (not a solution)
D.
- \( y = \frac{2}{3}x + 6 \) → \( 5 = \frac{2}{3}(2) + 6 \) → \( 5 = \frac{4}{3} + 6 \) → \( 5 \neq 7.33 \) (not a solution)
- \( 3y + 6x - 18 = 0 \) → \( 3(5) + 6(2) - 18 = 0 \) → \( 15 + 12 - 18 = 0 \) → \( 9 \neq 0 \) (not a solution)
Based on this analysis, the correct answer is:
B. \( y = -\frac{1}{2}x + 6 \) and \( y = 3x - 1 \)