Question

below are two inequalities and the graphs of their lines without the shading. by imagining where the shading should be, identify which point would satisfy both inequalities.
$y > -\frac{5}{2}x - 2$
$y > \frac{1}{3}x + 1$

Explanation:

Step1: Recall Inequality Shading Rules

For \( y > mx + b \), shading is above the dashed line. So for \( y > -\frac{5}{2}x - 2 \), shade above its line; for \( y > \frac{1}{3}x + 1 \), shade above its line. The solution is the intersection of both shaded regions.

Step2: Analyze Key Points (e.g., Test a Point)

Let's test a point, say \( (0, 3) \):

• For \( y > -\frac{5}{2}x - 2 \): \( 3 > -\frac{5}{2}(0) - 2 \Rightarrow 3 > -2 \), true.
• For \( y > \frac{1}{3}x + 1 \): \( 3 > \frac{1}{3}(0) + 1 \Rightarrow 3 > 1 \), true.

Check another point, say \( (-2, 0) \):

• For \( y > -\frac{5}{2}x - 2 \): \( 0 > -\frac{5}{2}(-2) - 2 = 5 - 2 = 3 \)? \( 0 > 3 \), false.

Check \( (3, 4) \):

• \( y > -\frac{5}{2}(3) - 2 = -\frac{15}{2} - 2 = -\frac{19}{2} \), \( 4 > -\frac{19}{2} \), true.
• \( y > \frac{1}{3}(3) + 1 = 1 + 1 = 2 \), \( 4 > 2 \), true. Wait, but let's check the graph's line positions. The steeper line is \( y = -\frac{5}{2}x - 2 \) (dashed, negative slope), the shallower is \( y = \frac{1}{3}x + 1 \) (dashed, positive slope). The intersection of their upper regions: points above both lines. Let's confirm with \( (0, 3) \) (on the graph, above both dashed lines). Another way: the region above both lines is where \( y \) is greater than both linear expressions. So a point like \( (0, 3) \) or \( (3, 4) \) works. But let's check the graph's grid. The steeper line crosses \( y \)-axis at \( -2 \), the other at \( 1 \). The overlapping shaded area is above both, so points with \( y \) above both lines. Let's take \( (0, 3) \): satisfies both. Let's verify algebraically:

For \( y > -\frac{5}{2}x - 2 \) and \( y > \frac{1}{3}x + 1 \), the solution set is all \( (x, y) \) where \( y \) is greater than both. So any point in the intersection of the two upper half-planes.

Answer:

A point like \( (0, 3) \) (or \( (3, 4) \), etc.) satisfies both. To confirm, test \( (0, 3) \): \( 3 > -2 \) (first inequality) and \( 3 > 1 \) (second inequality), both true. So a valid point is \( (0, 3) \) (or other points in the overlapping region above both dashed lines).