Answer:
[tex]y\geq \frac{1}{3}x+3[/tex]
[tex]3x-y>2[/tex]
Step-by-step explanation:
The complete question is
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?
step 1
Find the inequality of the first solid line
we have the points (0,3) and (3,4)
The slope is
[tex]m=(4-3)/(3-0)=\frac{1}{3}[/tex]
The y-intercept is the point (0,3)
so
The equation in slope intercept form is equal to
[tex]y=\frac{1}{3}x+3[/tex]
Remember that everything above the line is shaded and is a solid line
so
[tex]y\geq \frac{1}{3}x+3[/tex]
step 2
Find the inequality of the second line
we have the points (0,-2) and (1,1)
The slope is
[tex]m=(1+2)/(1-0)=3[/tex]
The y-intercept is the point (0,-2)
so
The equation in slope intercept form is equal to
[tex]y=3x-2[/tex]
Remember that everything to the right of the line is shaded and is a dashed line
so
[tex]y<3x-2[/tex]
Rewrite
[tex]2<3x-y[/tex] -----> [tex]3x-y>2[/tex]
therefore
The system of inequalities is
[tex]y\geq \frac{1}{3}x+3[/tex]
[tex]3x-y>2[/tex]
