Question

y > x + 2

y < -\frac{1}{3}x - 3

a)

graph a

b)

graph b

c)

graph c

d)

graph d

Explanation:

Step1: Analyze \( y > x + 2 \)

The line \( y = x + 2 \) has a slope of \( 1 \) and a y - intercept of \( 2 \). Since the inequality is \( y>x + 2 \), the line should be dashed (because the inequality is strict, \( y
eq x + 2 \)) and we shade the region above the line.

Step2: Analyze \( y<-\frac{1}{3}x - 3 \)

The line \( y=-\frac{1}{3}x - 3 \) has a slope of \( -\frac{1}{3} \) and a y - intercept of \( - 3 \). Since the inequality is \( y<-\frac{1}{3}x - 3 \), the line should be dashed and we shade the region below the line.

Step3: Check the intersection of the two shaded regions

• For option A: The shading does not match the regions defined by the two inequalities. The region for \( y > x+2 \) and \( y<-\frac{1}{3}x - 3 \) should be the area that is above \( y=x + 2 \) and below \( y =-\frac{1}{3}x-3 \). The intersection of these two regions (the area that satisfies both inequalities) is in the region where both conditions are met. Looking at the graphs, option B has the region that is above \( y=x + 2 \) (the dashed line with slope 1) and below \( y=-\frac{1}{3}x - 3 \) (the dashed line with slope \( -\frac{1}{3} \)). The other options (A, C, D) do not have the correct intersection of the shaded regions as per the two inequalities.

Answer:

B)