Question

solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in
y > \\(\frac{3}{2}x - 6\\)
y \\(\leq -x - 1\\)
answer attempt 1 out of 2
point:

Explanation:

Step1: Graph \( y > \frac{3}{2}x - 6 \)

The boundary line is \( y = \frac{3}{2}x - 6 \), which has a slope of \( \frac{3}{2} \) and a y - intercept of - 6. Since the inequality is \( y>\frac{3}{2}x - 6 \), we draw a dashed line (because the inequality is strict, \( y\) is not equal to \( \frac{3}{2}x - 6 \)) and shade the region above the line.

Step2: Graph \( y \leq -x - 1 \)

The boundary line is \( y=-x - 1 \), with a slope of - 1 and a y - intercept of - 1. Since the inequality is \( y\leq -x - 1 \), we draw a solid line (because \( y\) can be equal to \( -x - 1 \)) and shade the region below the line.

Step3: Find the intersection region

The solution to the system of inequalities is the region where the two shaded regions overlap. To find a point in this region, we can find the intersection of the two lines \( y=\frac{3}{2}x - 6 \) and \( y=-x - 1 \) first. Set \( \frac{3}{2}x-6=-x - 1 \).
Add \( x \) to both sides: \( \frac{3}{2}x+x-6=-1 \), which is \( \frac{3x + 2x}{2}-6=-1 \), or \( \frac{5x}{2}-6=-1 \).
Add 6 to both sides: \( \frac{5x}{2}=5 \).
Multiply both sides by \( \frac{2}{5} \): \( x = 2 \).
Substitute \( x = 2 \) into \( y=-x - 1 \), we get \( y=-2 - 1=-3 \). So the intersection point of the two lines is \( (2,-3) \). But we can also pick a test point in the overlapping region. Let's try \( x = 0 \).
For \( y>\frac{3}{2}(0)-6=-6 \) and \( y\leq - 0 - 1=-1 \). So we can pick \( (0,-2) \). Let's check:
For \( y>\frac{3}{2}x - 6 \): \( -2>\frac{3}{2}(0)-6=-6 \), which is true.
For \( y\leq -x - 1 \): \( -2\leq - 0 - 1=-1 \), which is true. Another point, let's try \( x = 0,y=-2 \) is valid, or we can also try \( x = 2,y=-3 \) (but the first line is dashed, but the second is solid, and at \( x = 2,y=-3 \), for \( y>\frac{3}{2}x-6 \), \( -3>\frac{3}{2}(2)-6=3 - 6=-3 \)? No, \( -3=-3 \), so it's not in the first region. Let's try \( x = 0,y=-2 \): as above, it works. Or \( x = 1 \).
For \( y>\frac{3}{2}(1)-6=\frac{3}{2}-6=\frac{3 - 12}{2}=-\frac{9}{2}=-4.5 \) and \( y\leq -1 - 1=-2 \). Let's pick \( y=-2 \) when \( x = 1 \). Check \( y>\frac{3}{2}(1)-6 \): \( -2>-4.5 \), true. Check \( y\leq -1 - 1 \): \( -2\leq -2 \), true. So \( (1,-2) \) is a valid point.

Answer:

\((1, - 2)\) (Other valid points like \((0,-2)\) etc. are also acceptable)