# Graphing the Solution Region of System of Linear Inequalities

Example:

Graph the solution set of the system of linear inequalities.

Solution:
We have

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

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.