Final answer:
None of the provided options (A, B, C, or D) solve the system of equations graphically or algebraically. By solving the equations algebraically, the solution to the system is (-1, -1).
Step-by-step explanation:
To solve the system of equations graphically, we need to express both equations in the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.
For the first equation, 2x - y = -1, we rearrange to get y = 2x + 1. For the second equation, x + y = -2, we rearrange to get y = -x -2.
Plotting these two lines on a graph, we look for the point where they intersect. This point represents the solution to the system of equations. Checking each option by substituting into both equations, we find that:
- A) (1, -1): 2(1) - (-1) = 2 + 1 = 3 (not -1), disqualifies.
- B) (2, -2): 2(2) - (-2) = 4 + 2 = 6 (not -1), disqualifies.
- C) (3, -3): 2(3) - (-3) = 6 + 3 = 9 (not -1), disqualifies.
- D) (4, -4): 2(4) - (-4) = 8 + 4 = 12 (not -1), disqualifies.
None of the options A, B, C, or D is the correct solution when checked against the original equations. We must find the point of intersection by plotting the lines correctly or solving the equations algebraically.
Algebraically, adding the two equations (2x - y) + (x + y) = -1 + (-2) gives us 3x = -3, which means x = -1. Substituting x into the first equation, we get y = 2(-1) + 1 = -1. So the solution is (-1, -1), which is not one of the options provided.