Question

which system of linear inequalities is represented by the graph?
\\(\bigcirc\\) \\(y \geq \frac{1}{3}x + 3\\) and \\(3x - y > 2\\)
\\(\bigcirc\\) \\(y \geq \frac{1}{2}x + 3\\) and \\(3x - y > 2\\)
\\(\bigcirc\\) \\(y \geq \frac{1}{3}x + 3\\) and \\(3x + y > 2\\)
\\(\bigcirc\\) \\(y \geq \frac{1}{3}x + 3\\) and \\(2x - y > 2\\)

Explanation:

Step1: Find first inequality's slope

Identify 2 points on the solid line: $(-3, 2)$ and $(0, 3)$.
Slope $m = \frac{3-2}{0-(-3)} = \frac{1}{3}$.
Y-intercept $b=3$, so line is $y=\frac{1}{3}x+3$. Shading is above the solid line, so inequality: $y\geq\frac{1}{3}x+3$.

Step2: Find second inequality's line

Identify 2 points on the dashed line: $(0, -2)$ and $(1, 1)$.
Slope $m = \frac{1-(-2)}{1-0} = 3$.
Y-intercept $b=-2$, so line is $y=3x-2$. Rearrange to standard form: $3x-y=2$.

Step3: Determine second inequality sign

Shading is above the dashed line. Test point $(0,0)$: $0 > 3(0)-2$ → $0 > -2$ (true). So inequality: $3x-y>2$.

Answer:

$\boldsymbol{y\geq \frac{1}{3}x + 3}$ and $\boldsymbol{3x - y > 2}$ (first option)