which is the graph of the linear inequality $2x - 3y < 12$?

Answer

Explanation:

Step1: Rewrite the inequality

Start with (2x - 3y < 12). Solve for (y):
Subtract (2x) from both sides: (-3y < -2x + 12).
Divide by (-3) (remember to reverse the inequality sign): (y > \frac{2}{3}x - 4).

Step2: Analyze the boundary line

The boundary line is (y = \frac{2}{3}x - 4). Since the inequality is (y > \dots), the line should be dashed (not solid, because the inequality is strict: (<), so (y >) means the line is not included).

Find intercepts:

(x)-intercept: Set (y = 0): (0 = \frac{2}{3}x - 4 \implies \frac{2}{3}x = 4 \implies x = 6). So the line passes through ((6, 0)).
(y)-intercept: Set (x = 0): (y = -4). So the line passes through ((0, -4)).

Step3: Determine the shaded region

Test a point not on the line, e.g., ((0, 0)):
Plug into (y > \frac{2}{3}x - 4): (0 > \frac{2}{3}(0) - 4 \implies 0 > -4), which is true. So the region containing ((0, 0)) should be shaded.

Now, check the graphs:

The boundary line is dashed (since (y > \dots), not (y \geq)).
The line passes through ((6, 0)) and ((0, -4)).
The shaded region includes ((0, 0)) (above the line, since (y > \dots)).

Looking at the options, the third graph (rightmost) has a dashed line (wait, no—wait, let's recheck. Wait, the inequality is (2x - 3y < 12) → (y > \frac{2}{3}x - 4). So the shaded region is above the line. Let's check the graphs:

First graph: Shaded below? No. Second: Shaded below? Wait, no—wait, the third graph (rightmost) has the shaded region above the line (since (y > \dots)), with dashed line? Wait, no, maybe I mixed up. Wait, let's re-express:

Wait, the line (y = \frac{2}{3}x - 4) has a positive slope ((\frac{2}{3})), passes through ((6, 0)) and ((0, -4)). The inequality (y > \frac{2}{3}x - 4) means the region above the line. So the graph with the shaded area above the dashed line (since (<) in the original inequality, so the line is dashed) and including ((0, 0)) (which is above (y = \frac{2}{3}(0) - 4 = -4), so (0 > -4) is true).

Looking at the options, the third graph (rightmost) has the shaded region above the line (the line goes from ((6, 0)) down to ((0, -4)), and the shaded area is above it, including ((0, 0))). The other graphs have shaded below. So the correct graph is the third one (rightmost) with the dashed line? Wait, no, the original inequality is (2x - 3y < 12), so (y > \frac{2}{3}x - 4), so the shaded region is above the line. The third graph (rightmost) shows the shaded area above the line (since the line is going up from ((0, -4)) to ((6, 0)), and the shaded area is above it, including ((0, 0))).