Question

On a coordinate plane, 2 straight lines are shown. The first solid line has a positive slope and goes through (0, 3) and (3, 4). Everything above the line is shaded. The second dashed line has a positive slope and goes through (0, negative 2) and (1, 1). Everything to the right of the line is shaded.
Which system of linear inequalities is represented by the graph?

y > One-thirdx + 3 and 3x – y > 2
y > One-halfx + 3 and 3x – y > 2
y > One-thirdx + 3 and 3x + y > 2
y > One-thirdx + 3 and 2x – y > 2

Answer 1

To solve this, we need to analyze the two lines and the shaded areas represented in the given inequalities.

1. **First Line (Solid Line)**:
- The line passes through the points (0, 3) and (3, 4).
- We can calculate the slope (m) as follows:
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 3}{3 - 0} = \frac{1}{3}
\]
- The equation of the line can be expressed in the slope-intercept form (y = mx + b):
\[
y = \frac{1}{3}x + 3
\]
- Since everything above this line is shaded, it represents the inequality:
\[
y > \frac{1}{3}x + 3
\]

2. **Second Line (Dashed Line)**:
- The dashed line passes through (0, -2) and (1, 1).
- We can calculate the slope here as well:
\[
m = \frac{1 - (-2)}{1 - 0} = \frac{3}{1} = 3
\]
- The equation for this line can also be derived:
\[
y = 3x - 2
\]
- The inequality corresponding to being shaded to the right of this dashed line is:
\[
y < 3x - 2
\]
- However, in terms of the inequalities given, we need to rearrange it. The representation of the shaded area on the right of the line can be rewritten as:
\[
3x - y > 2
\]

Combining these two inequalities, we have:
1. \( y > \frac{1}{3}x + 3 \)
2. \( 3x - y > 2 \)

So, the correct system of inequalities that represents the graph is:

\[
y > \frac{1}{3}x + 3 \quad \text{and} \quad 3x - y > 2
\]

Thus, the correct answer is:

**y > One-thirdx + 3 and 3x – y > 2**.