Question

which graph shows the solution to this system of inequalities
$y \leq -\frac{1}{3}x + 1$
$y > 2x - 3$
a. graph with text description
b. graph

Explanation:

Step1: Analyze \( y \leq -\frac{1}{3}x + 1 \)

The inequality \( y \leq -\frac{1}{3}x + 1 \) represents the region below (including the line) the line with slope \( -\frac{1}{3} \) and y - intercept 1. The line should be solid (because of the "≤" sign).

Step2: Analyze \( y > 2x - 3 \)

The inequality \( y > 2x - 3 \) represents the region above (not including the line) the line with slope 2 and y - intercept - 3. The line should be dashed (because of the ">" sign).

Step3: Find the intersection region

We need to find the region that satisfies both inequalities. Let's also find the intersection point of the two lines \( y=-\frac{1}{3}x + 1 \) and \( y = 2x-3 \).

Set \( -\frac{1}{3}x + 1=2x - 3 \)

Multiply through by 3 to get rid of the fraction: \( -x + 3 = 6x-9 \)

Add \( x \) to both sides: \( 3=7x - 9 \)

Add 9 to both sides: \( 12 = 7x \)

So \( x=\frac{12}{7}\approx1.71 \)

Substitute \( x = \frac{12}{7} \) into \( y = 2x-3 \): \( y=2\times\frac{12}{7}-3=\frac{24}{7}-\frac{21}{7}=\frac{3}{7}\approx0.43 \)

Now, let's analyze the options:

• For the line \( y\leq-\frac{1}{3}x + 1 \), the region is below the solid line. For \( y > 2x - 3 \), the region is above the dashed line.

Looking at the graphs, in option A, the shaded region is below the solid line \( y = -\frac{1}{3}x + 1 \) and above the dashed line \( y=2x - 3 \), which matches the solution of the system. In option B, the region does not match the required regions for both inequalities.

Answer:

A