which system of linear inequalities is represented by the graph?\no $y\\geq\\frac{1}{3}x + 3$ and $3x…

which system of linear inequalities is represented by the graph?\no $y\\geq\\frac{1}{3}x + 3$ and $3x - y>2$\no $y\\geq\\frac{1}{2}x + 3$ and $3x - y>2$\no $y\\geq\\frac{1}{3}x + 3$ and $3x + y>2$\no $y\\geq\\frac{1}{3}x + 3$ and $2x - y>2$

which system of linear inequalities is represented by the graph?\no $y\\geq\\frac{1}{3}x + 3$ and $3x - y>2$\no $y\\geq\\frac{1}{2}x + 3$ and $3x - y>2$\no $y\\geq\\frac{1}{3}x + 3$ and $3x + y>2$\no $y\\geq\\frac{1}{3}x + 3$ and $2x - y>2$

Answer

Explanation:

Step1: Identify the slope - intercept form of one line

The red line has a y - intercept of 3 and a slope of $\frac{1}{3}$. The shaded region is above the line, so the inequality is $y\geq\frac{1}{3}x + 3$.

Step2: Rewrite the second - type inequalities in slope - intercept form

For the form $ax+by>c$, rewrite it as $y>-\frac{a}{b}x+\frac{c}{b}$ or $y<-\frac{a}{b}x+\frac{c}{b}$. For $3x - y>2$, rewrite it as $y<3x - 2$. For $3x + y>2$, rewrite it as $y>-3x + 2$. For $2x - y>2$, rewrite it as $y<2x - 2$. The blue line has a negative slope. The line with a negative slope in the inequalities should have the shaded region above it. The line with a negative slope in the graph has a steeper negative slope compared to the line $y = - 3x+2$. The line for $3x - y>2$ (or $y<3x - 2$) has a positive slope which is not correct for the blue line in the graph. The line for $3x + y>2$ (or $y>-3x + 2$) has a negative slope and the shaded region above it which is consistent with the blue - shaded region in the graph.

Answer:

$y\geq\frac{1}{3}x + 3$ and $3x + y>2$