7.7 Graphing linear inequalities in two variables

Point $A(1,-1)$ is a solution since

$\begin{array}{l}2x+3y\le 6\hfill \\ 2(1)+3(-1)\le 6?\hfill \\ 2-3\le 6?\hfill \\ -1\le 6.\text{True}\hfill \end{array}$

Point $B(2,5)$ is not a solution since

$\begin{array}{l}2x+3y\le 6\hfill \\ 2(2)+3(5)\le 6?\hfill \\ 4+15\le 6?\hfill \\ 19\le 6.\mathrm{False}\hfill \end{array}$

Graph $3x-2y\ge -4$ .

1. Graph the boundary line. The inequality is $\ge$ so we’ll draw the line solid . Consider the inequality as an equation.

2. Choose a test point. The easiest one is $\left(0,0\right)$ . Substitute $\left(0,0\right)$ into the original inequality.

Graph $x+y-3<0$ .

1. Graph the boundary line: $x+y-3=0$ . The inequality is $<$ so we’ll draw the line dotted .

2. Choose a test point, say $(0,0)$ .

Graph $y\le 2x$ .

1. Graph the boundary line $y=2x$ . The inequality is $\le$ , so we’ll draw the line solid .
   
   
2. Choose a test point, say $\left(0,0\right)$ .

   $\begin{array}{l}y\le 2x\hfill \\ 0\le 2\left(0\right)?\hfill \\ 0\le 0.\mathrm{True}\hfill \end{array}$

   Shade the half-plane containing $\left(0,0\right)$ . We can’t! $\left(0,0\right)$ is right on the line! Pick another test point, say $\left(1,6\right)$ .

   $\begin{array}{l}y\le 2x\hfill \\ 6\le 2\left(1\right)?\hfill \\ 6\le 2.\mathrm{False}\hfill \end{array}$

   Shade the half-plane on the opposite side of the boundary line.
   

Graph $y>2$ .

1. Graph the boundary line $y=2$ . The inequality is $>$ so we’ll draw the line dotted .

2. We don’t really need a test point. Where is $y>2?$ Above the line $y=2!$ Any point above the line clearly has a $y\text{-coordinate}$ greater than 2.