Question

solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
$y < \frac{1}{2}x + 2$
$y > -\frac{5}{4}x - 5$

Explanation:

Step1: Graph \( y < \frac{1}{2}x + 2 \)

First, graph the line \( y = \frac{1}{2}x + 2 \). The slope is \( \frac{1}{2} \) and the y-intercept is 2. Since the inequality is \( y < \frac{1}{2}x + 2 \), the line should be dashed (because it's a strict inequality) and we shade below the line.

Step2: Graph \( y > -\frac{5}{4}x - 5 \)

Next, graph the line \( y = -\frac{5}{4}x - 5 \). The slope is \( -\frac{5}{4} \) and the y-intercept is -5. Since the inequality is \( y > -\frac{5}{4}x - 5 \), the line should be dashed and we shade above the line.

Step3: Find the Intersection Region

The solution set is the region where the two shaded areas overlap. To find a point in the solution set, we can pick a point that satisfies both inequalities. Let's test the origin (0, 0):

For \( y < \frac{1}{2}x + 2 \): \( 0 < \frac{1}{2}(0) + 2 \) → \( 0 < 2 \), which is true.

For \( y > -\frac{5}{4}x - 5 \): \( 0 > -\frac{5}{4}(0) - 5 \) → \( 0 > -5 \), which is also true. So (0, 0) is in the solution set.

Answer:

A point in the solution set is \((0, 0)\) (other valid points could also be used, e.g., (4, 3), (2, 1), etc.)