Question

solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
$y \geq \frac{1}{2}x + 2$
$y > -\frac{3}{2}x - 2$
graph of a coordinate plane with x-axis from -10 to 10 and y-axis from -10 to 10
answer attempt 1 out of 3
point: (\boxed{ }, \boxed{ })
submit answer

Explanation:

Step1: Analyze the first inequality \( y \geq \frac{1}{2}x + 2 \)

The line \( y=\frac{1}{2}x + 2 \) has a slope of \( \frac{1}{2} \) and a y - intercept of 2. Since the inequality is \( \geq \), we draw a solid line for this equation and shade the region above the line.

Step2: Analyze the second inequality \( y>-\frac{3}{2}x - 2 \)

The line \( y =-\frac{3}{2}x-2 \) has a slope of \( -\frac{3}{2} \) and a y - intercept of - 2. Since the inequality is \( > \), we draw a dashed line for this equation and shade the region above the line.

Step3: Find the intersection region

The solution set of 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 and check if it satisfies both inequalities. Let's try the point \( (0,3) \).

• For the first inequality \( y\geq\frac{1}{2}x + 2 \): Substitute \( x = 0 \) and \( y=3 \). We get \( 3\geq\frac{1}{2}(0)+2\), which simplifies to \( 3\geq2 \), and this is true.
• For the second inequality \( y>-\frac{3}{2}x - 2 \): Substitute \( x = 0 \) and \( y = 3 \). We get \( 3>-\frac{3}{2}(0)-2\), which simplifies to \( 3>- 2 \), and this is true.

Answer:

The point \( (0,3) \) (other valid points like \( (2,4) \) etc. are also correct as long as they satisfy both inequalities)

Point: \( (0, 3) \)