Question

39 choose all solutions of the following system of linear inequalities listed below. a (0,-1) b (6,2) c (0, 0) d (-8,2) e (-2,3) f (1,-6)

Explanation:

Step1: Recall solution region

A solution to a system of linear inequalities is a point in the overlapping shaded region (feasible region) of the graphs of the inequalities.

Step2: Analyze each point

- **Point A: (0, -1)**
Check if (0, -1) is in the shaded region. From the graph, the shaded region is above the solid line and left of the dashed line. (0, -1) is below the solid line (since the solid line passes through (0, -1)? Wait, no—wait, the solid line: let's see the slope. Wait, the solid line goes from, say, (-8, 0) to (4, -4)? Wait, maybe better to check if the point is in the shaded area. Wait, (0, -1): the shaded region is the blue area. Looking at the grid, (0, -1) is below the solid line (since the solid line at x=0 is at y=0? Wait, no, the solid line passes through (0, -1)? Wait, maybe I misread. Wait, the solid line: let's find its equation. Let's take two points: when x = -8, y = 0; when x = 4, y = -4. So slope $m = \frac{-4 - 0}{4 - (-8)} = \frac{-4}{12} = -\frac{1}{3}$. Equation: $y - 0 = -\frac{1}{3}(x + 8) \implies y = -\frac{1}{3}x - \frac{8}{3}$. At x=0, $y = -\frac{8}{3} \approx -2.67$. Wait, maybe my initial analysis is wrong. Wait, the shaded region is the blue area. Let's check each point:
- **Point A: (0, -1)**
Is (0, -1) in the shaded region? The shaded region is above the solid line (since the solid line is part of the solution, and the shading is above it) and left of the dashed line. Let's plug into the inequalities. Assume the solid line is $y \geq -\frac{1}{3}x - \frac{8}{3}$ (since shading is above) and the dashed line is $x < -2$ (or some vertical/non-vertical line). Wait, the dashed line is steep, maybe vertical? Wait, the dashed line is on the right, maybe x = -2? Wait, no, the dashed line has a negative slope? Wait, maybe better to visually inspect:
- (0, -1): Let's see the shaded area. The blue area is left of the dashed line and above the solid line. (0, -1): is it in the blue? Maybe. Wait, the options: A is (0, -1), C is (0,0), D is (-8,2), E is (-2,3), B is (6,2), F is (1, -6).
- **Point B: (6, 2)**
(6, 2) is to the right of the dashed line (since dashed line is on the left), so not in the shaded region. Eliminate B.
- **Point C: (0, 0)**
(0, 0): above the solid line (since solid line at x=0 is y ≈ -2.67, so 0 > -2.67) and left of the dashed line. In shaded region.
- **Point D: (-8, 2)**
(-8, 2): left of dashed line, above solid line (solid line at x=-8 is y=0, so 2 > 0). In shaded region.
- **Point E: (-2, 3)**
(-2, 3): left of dashed line (dashed line is at x > -2? Wait, no, dashed line is on the right, so x < dashed line. (-2, 3): check if in shaded. Above solid line (solid line at x=-2: $y = -\frac{1}{3}(-2) - \frac{8}{3} = \frac{2}{3} - \frac{8}{3} = -2$, so 3 > -2) and left of dashed line. In shaded region.
- **Point F: (1, -6)**
(1, -6): below solid line (solid line at x=1: $y = -\frac{1}{3}(1) - \frac{8}{3} = -3$, so -6 < -3). Not in shaded region. Eliminate F.
- **Point A: (0, -1)**
Solid line at x=0: $y = -\frac{8}{3} \approx -2.67$. So -1 > -2.67, so above solid line. Left of dashed line? Yes. So (0, -1) is in shaded region? Wait, maybe my earlier slope calculation was wrong. Alternatively, visually: the shaded region includes points where x is small (left) and y is above the solid line. Let's recheck:
- (0, -1): in blue? Maybe.
- (0, 0): in blue (yes, above solid line, left of dashed).
- (-8, 2): in blue (left of dashed, above solid line: solid line at x=-8 is y=0, 2 > 0).
- (-2, 3): in blue (left of dashed, above solid line: solid line at x=-2 is y=-2, 3 > -2).
- (6, 2): right of dashed line, out.
- (1, -6): below solid line, out.

Answer:

A. (0, -1), C. (0, 0), D. (-8, 2), E. (-2, 3)