Question

pretest: systems of equations inequalities
$y \leq 3x + 2$
$y > -2 \cdot x$
graphs a, b, c, d as shown
\\(\circ\\) a. graph a
\\(\circ\\) b. graph b
\\(\circ\\) c. graph c
\\(\circ\\) d. graph d

Explanation:

Step1: Analyze \( y \leq 3x + 2 \)

The line \( y = 3x + 2 \) has a slope of \( 3 \) and y - intercept \( 2 \). Since the inequality is \( \leq \), the line should be solid (because the inequality includes equality) and we shade below the line.

Step2: Analyze \( y > - 2x \)

The line \( y=-2x \) has a slope of \( - 2 \) and y - intercept \( 0 \). Since the inequality is \(> \), the line should be dashed (because the inequality does not include equality) and we shade above the line.

Step3: Match with graphs

• For the line \( y = 3x+2 \), a solid line with slope \( 3 \) and y - intercept \( 2 \) is present in Graph A and Graph B (Graphs C and D have dashed lines for \( y = 3x + 2 \), which is incorrect).
• For the line \( y=-2x \), a dashed line with slope \( - 2 \) and y - intercept \( 0 \) is present in Graph A and Graph B. Now, check the shading:
• For \( y\leq3x + 2 \), we shade below \( y = 3x+2 \), and for \( y > - 2x \), we shade above \( y=-2x \). In Graph A, the shading regions (the blue regions) match the required shadings. In Graph B, the shading does not match the correct regions for both inequalities.

Answer:

A. Graph A