Graphing the Solution Region of System of Linear Inequalities

Example:

Graph the solution set of the system of linear inequalities.

\[\begin{gathered} 4{\text{x}} – 5{\text{y}} \leqslant 20 \\ 3{\text{x}} + 2{\text{y}} \leqslant 6 \\ 3{\text{x}} + {\text{y}} \leqslant 2 \\ \end{gathered} \]

Solution:
We have

\[\begin{gathered} 4{\text{x}} – 5{\text{y}} \leqslant 20\,\,\, – – – \left( A \right) \\ 3{\text{x}} + 2{\text{y}} \leqslant 6\,\,\,\, – – – \left( B \right) \\ 3{\text{x}} + {\text{y}} \leqslant 2\,\,\,\, – – – \left( C \right) \\ \end{gathered} \]

For the corresponding equations of inequalities (A), (B) and (C), we get

\[\begin{gathered} 4{\text{x}} – 5{\text{y}} = 20\,\,\, – – – \left( 1 \right) \\ 3{\text{x}} + 2{\text{y}} = 6\,\,\,\, – – – \left( 2 \right) \\ 3{\text{x}} + {\text{y}} = 2\,\,\,\, – – – \left( 3 \right) \\ \end{gathered} \]

For x–Intercepts:
Put $${\text{y}} = 0$$in Eq (1), Eq (2) and Eq (3) and we get

$$ \Rightarrow 4{\text{x}} – 5\left( {\text{0}} \right) = 20$$	$$ \Rightarrow 3{\text{x}} + 2\left( {\text{0}} \right) = 6$$	$$ \Rightarrow 3{\text{x}} + {\text{0}} = 2$$
$$ \Rightarrow 4{\text{x}} = 20$$	$$ \Rightarrow 3{\text{x}} = 6$$	$$ \Rightarrow 3{\text{x}} = 2$$
$$ \Rightarrow {\text{x}} = 5$$	$$ \Rightarrow {\text{x}} = 2$$	$$ \Rightarrow {\text{x}} = \frac{2}{3}$$
$$\therefore \left( {5,0} \right)$$	$$\therefore \left( {2,0} \right)$$	$$\therefore \left( {\frac{2}{3},0} \right)$$

 

For y–Intercepts:
Put $${\text{x}} = 0$$in Eq (1), Eq (2) and Eq (3) and we get

$$ \Rightarrow 4\left( {\text{0}} \right) – 5{\text{y}} = 20$$	$$ \Rightarrow 3\left( {\text{0}} \right) + 2{\text{y}} = 6$$	$$ \Rightarrow 3\left( {\text{0}} \right) + {\text{y}} = 2$$
$$ \Rightarrow – 5{\text{y}} = 20$$	$$ \Rightarrow 2{\text{y}} = 6$$	$$ \Rightarrow {\text{y}} = 2$$
$$ \Rightarrow {\text{y}} = – 4$$	$$ \Rightarrow {\text{y}} = 3$$	$$ \Rightarrow {\text{y}} = 2$$
$$\therefore \left( {0, – 4} \right)$$	$$\therefore \left( {0,3} \right)$$	$$\therefore \left( {0,2} \right)$$

 

Test:
Put the origin as test point in inequalities (A), (B) and (C).

$$0 – 0 < 20$$	$$ \Rightarrow 0 + 0 < 6$$	$$ \Rightarrow 0 + {\text{0}} < 2$$
$$ \Rightarrow 0 < 20$$ (True)	$$ \Rightarrow 0 < 6$$ (True)	$$ \Rightarrow 0 < 2$$ (True)
$$\therefore $$ Solution set towards origin side.	$$\therefore $$ Solution set towards origin side.	$$\therefore $$ Solution set towards origin side.