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

The boundary line is \( y = \frac{1}{2}x + 2 \), which has a slope of \( \frac{1}{2} \) and a y - intercept of 2. Since the inequality is \( y < \frac{1}{2}x + 2 \), the line should be dashed (because the inequality is strict, \( y
eq\frac{1}{2}x + 2 \)) and we shade the region below the line. To graph the line, we can plot the y - intercept (0, 2) and then use the slope to find another point. For example, from (0, 2), moving up 1 unit and right 2 units gives (2, 3).

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

The boundary line is \( y=-\frac{5}{4}x - 5 \), with a slope of \( -\frac{5}{4} \) and a y - intercept of - 5. Since the inequality is \( y>-\frac{5}{4}x - 5 \), the line is dashed (because \( y
eq-\frac{5}{4}x - 5 \)) and we shade the region above the line. To graph the line, plot the y - intercept (0, - 5). Then, from (0, - 5), moving down 5 units and right 4 units (or up 5 units and left 4 units) gives another point, for example, (4, 0) (since when \( x = 4 \), \( y=-\frac{5}{4}(4)-5=-5 - 5=-10\)? Wait, no. Let's recalculate: \( y =-\frac{5}{4}x-5 \), when \( x = 0 \), \( y=-5 \); when \( x = 4 \), \( y=-\frac{5}{4}\times4-5=-5 - 5=-10 \); when \( x=-4 \), \( y =-\frac{5}{4}\times(-4)-5 = 5 - 5=0 \). So a point on the line is (- 4, 0).

Step3: Find the solution region

The solution to the system of inequalities is the region that is shaded by both inequalities. To find a point in the solution set, we can pick a point that is in the overlapping shaded region. Let's test the point (0, 0).

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

For \( y>-\frac{5}{4}x - 5 \): Substitute \( x = 0 \), \( y = 0 \). We get \( 0>-\frac{5}{4}(0)-5=-5 \), 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, such as (2, 0), ( - 2, 0) etc., as long as they satisfy both inequalities). The graphing involves dashed lines for both inequalities, shading below \( y=\frac{1}{2}x + 2 \) and above \( y =-\frac{5}{4}x-5 \), and the overlapping region is the solution set. The final answer for a point in the solution set is \(\boldsymbol{(0,0)}\) (or any other point in the overlapping shaded region).