91Ó°ÊÓ

Consider the inequality $4x-5y-10≤0$.

The boundary line equaton is $4x-5y-10=0$.

Here the inequality is $≤$, graph the boundary line as a solid line.

Find the slope intercept form of $4x-5y-10=0$:

$\begin{array}{l}4x−5y−10=0\\ 4x−5y=10\text{ â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„A»å»å10´Ç²Ô²ú´Ç³Ù³ó²õ¾±»å±ð²õ}\\ −5y=−4x+10\text{ â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„S³Ü²ú³Ù°ù²¹³¦³Ù}4x\text{frombothsides}\\ y=\frac{−4x+10}{−5}\text{ â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„D¾±±¹¾±»å±ð²ú´Ç³Ù³ó²õ¾±»å±ð²õ²ú²â}−5\\ y=\frac{4}{5}x−2\end{array}$

The slope intercept form is $y=\frac{4}{5}x-2$with slope $\frac{4}{5}.$

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

Substitute $x=0,y=0$in $4x-5y-10≤0$:

$\begin{array}{l}4x−5y−10≤0\\ 4\left(0\right)−5\left(0\right)−10≤0\text{ â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„s³Ü²ú²õ³Ù¾±³Ù³Ü³Ù±ð0´Ú´Ç°ù}x,0\text{for}y\\ −10≤0\text{ â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶Ä„â¶ÄÔò°ù³Ü±ð}\end{array}$

Since, $-10≤0$results a true inequality, so, shade the region that contains the point $\left(0,0\right).$

Graphical representation of $4x-5y-10≤0$: