91Ó°ÊÓ

Consider the inequality $x-6y+3>0$.

The boundary line equaton is $x-6y+3=0$.

Here the inequality is $>$, graph the boundary line as a dashed line.

Find the slope intercept form of $x-6y+3=0$:

$\begin{array}{l}x−6y+3=0\\ x−6y=−3\text{ â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„S³Ü²ú³Ù°ù²¹³¦³Ù3´Ú°ù´Ç³¾²ú´Ç³Ù³ó²õ¾±»å±ð²õ}\\ −6y=−x−3\text{ â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„S³Ü²ú³Ù°ù²¹³¦³Ù}x\text{frombothsides}\\ y=\frac{−x−3}{−6}\text{ â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„D¾±±¹¾±»å±ð²ú´Ç³Ù³ó²õ¾±»å±ð²õ²ú²â}−6\\ y=\frac{1}{6}x+\frac{1}{2}\end{array}$

The slope intercept form is $y=\frac{1}{6}x+\frac{1}{2}$with slope $\frac{1}{6}.$

Choose the test point as $\left(0,0\right).$

Substitute $x=0,y=0$in $x-6y+3>0$:

$\begin{array}{l}x−6y+3>0\\ 0−6\left(0\right)+3>0\text{ â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„s³Ü²ú²õ³Ù¾±³Ù³Ü³Ù±ð0´Ú´Ç°ù}x,0\text{for}y\\ 3>0\text{ â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶ÄÔò°ù³Ü±ð}\end{array}$

Since, $3>0$results a true inequality, so, shade the region that contains the point $\left(0,0\right).$

Graphical representation of $x-6y+3>0$: