# Examples of Linear Inequalities in Two Variables

Example:

Graph the solution set of the system of linear inequalities
$\begin{gathered} 3{\text{x}} + 5{\text{y}} \leqslant 9 \\ {\text{x}} – 6{\text{y}} \leqslant 3 \\ \end{gathered}$

Solution:

We have

$\begin{gathered} 3{\text{x}} + 5{\text{y}} \leqslant 9\,\,\, – – – \left( A \right) \\ {\text{x}} – 6{\text{y}} \leqslant 3\,\,\, – – – \left( B \right) \\ \end{gathered}$

The corresponding equations of inequalities (A) and (B)

 $3{\text{x}} + 5{\text{y}} = 9\,\,\, – – – \left( 1 \right)$ ${\text{x}} – 6{\text{y}} = 3\,\,\, – – – \left( 2 \right)$ For x – intercept For x – intercept Put ${\text{y}} = 0$ in (1) we get Put ${\text{y}} = 0$ in (2) we get $\Rightarrow$$3{\text{x}} + 5\left( {\text{0}} \right) = 9$ $\Rightarrow$${\text{x}} – 6\left( {\text{0}} \right) = 3$ $\Rightarrow$$3{\text{x}} = 9$ $\Rightarrow$${\text{x}} = 3$ $\Rightarrow$${\text{x}} = 3$ $\therefore \left( {3,0} \right)$ $\therefore \left( {3,0} \right)$ For y – intercept For y – intercept Put ${\text{x}} = 0$ in (1) we get Put ${\text{x}} = 0$ in (2) we get $\Rightarrow$$3\left( {\text{0}} \right) + 5{\text{y}} = 9$ $\Rightarrow$${\text{0}} – 6{\text{y}} = 3$ $\Rightarrow$$5{\text{y}} = 9$ $\Rightarrow$$– 6{\text{y}} = 3$ $\Rightarrow$${\text{y}} = \frac{9}{5}$ $\Rightarrow$${\text{y}} = – \frac{1}{2}$ $\therefore \left( {0,\frac{9}{5}} \right)$ $\therefore \left( {0, – \frac{1}{2}} \right)$ Test Test Put origin $\left( {0,0} \right)$ in eq (A) Put origin $\left( {0,0} \right)$ in eq (B) $0 + 0 < 9$ $0 – 0 < 3$ $\Rightarrow 0 < 9$ which is the true solution of set origin side. $\Rightarrow 0 < 3$ which is the true solution of set origin side.