If $a, b, c \in R$, then $a x+b y+c=0$ is called linear equation in two variables $x$ and $y$ whereas the inequalities $ax+b y \leqslant c, a x+b y \geqslant c$ $a x+b yc$are called linear inequation in two variables$x$and$y$. An ordered pair$(x, y)$is a solution of linear inequation if the inequality is satisfied by$(x, y)$

Checking Solutions of inequalities:

Check whether the given ordered pair is a solution of $2 x+3 y \geqslant 5$

(a) $(0,1)$

(b) $(4,-1)$

(c) $(2,1)$

$2 x+3 y>1$

$(1,2)$

$2 \times 1+3 \times 2=8>1$

$x-y<0$

$(-1,3)$

$-1-3=-4<0$

$(x, y)$

1. $x+y=5$
2. $x+y>5$
3. $x+y<5$

Consider the graph. The graph of the line $x+y=5$ divides the points in the plane into three sets.

(i) Closed Half Plane:

$a x+b y \geqslant c / a x+b y \leqslant c$

(ii) Open Half Plane:

$a x+b y>c / a x+b y

$2 x+3 y \leqslant 1 \text { or } x+y \geqslant 3$

$x+y<2 \text { or } 3 x-y>1$

Solution set:

The set of all ordered pair $(\alpha, \beta)$ of a real number which satisfy a given inequality is called the solution set of the given inequality.

Solution region (or Graph):

The region of the plane containing all the points whose co-ordinates satisfy a given inequality is called the solution region (or graph) of the inequality.

$(\alpha,\beta)$

$3 \alpha-\beta<2 \quad(\alpha, \beta) \in \text{solution set}$

$3 \alpha -\beta\nless 2 \quad(\alpha,\beta) \not\in\text{solution set}$

$\text{solution set}$ =$\{(\alpha,\beta)| (\alpha,\beta) \text{satisfy the inequality} ax+by \leqslant c ~ or ~ ax+by \geqslant c ~ or ~ ax+by < c ~ or ~ ax+by > c\}$

In order to find the solution set of a linear inequation in two variables, we follow the following algorithm.

Step I: Write the associated equation ax+by=c for the given inequation.

Step II: put y=0⇒x=c/a, put x=0⇒y=c/b.

Step III: Join point (c/a,0) & (0,c/b) to obtain the graph of line from the given inequation.

Step IV: Shading of Solution Region by origin test or arbitrary point test.

Step V: Shaded Region $\Rightarrow$ Solution Region.

Step I: $\quad 2 x-y \geqslant 1$

Associated equation is

$2 x-y=1$

Step II:

put $y=0 \Rightarrow x=1 / 2$

that is $\quad(\frac{1}{2}, 0)$

put

$x=0 \Rightarrow-y=1 \Rightarrow y=-1$

that is $(0,-1)$

Step III : Draw the graph of the line,

$2 x-y=1$

Step IV : $\quad 2 x-y \geqslant 1$

$(1,2)$

$2 \times 1-2=0 \geqslant 1$

$\therefore$ point $(1,2)$ does’nt lies in the solution region of $2 x-y \geqslant 1$

Origin test:

$O(0,0) \quad 2 x-y \geqslant 1$
$2 \times 0-0=0 \cancel{\geqslant} 1$

$\therefore O(0,0)$ does’nt lies in the solution region.

$2 x+3 y \leqslant 6$,

Solve it graphically.

Solution: Given inequation

$2 x+3 y \leqslant 6$

$\therefore$ Associated equation

$2 x+3 y=6$ put $y=0 \Rightarrow x=3$

ie. $\quad (3,0)$

put $x=0 \quad \Rightarrow y = 2$

$\therefore (0,2)$

Arbitrary point test: .

$(-1,2)\quad 2 x+3 y \leqslant 6$

$2 \times(-1)+3 \times 2= 4<6$

$\therefore(-1,2)$ satisfy the inequation

$2 x+3 y \leqslant 6$

$\therefore$ Half plane II will be the solution Region as $(-1,2)$ belongs to this region.

$\therefore$ Shaded region will be solution region.

Solve graphically: $2 y+x \geqslant 0$

Solution: Associated equation for the given inequation will be

$2 y+x=0$

put $y=0$ $\Rightarrow x=0 \quad(0,0)$

put $x=1$ $\Rightarrow y=-1 / 2 \quad(1,-1 / 2)$

$(1,2) \quad 2 y+x \geqslant 0$

$2 \times 2+1=5>0$

$\therefore(1,2)$ satisfy the inequation $2 y+x \geqslant 0$

$\therefore(1,2) \in$ Solution Region.

$\therefore$ Half plane I will be the solution region.

$|x| \leqslant 3$

$\Rightarrow\quad -3 \leqslant x \leqslant 3$

$\Rightarrow\quad x \geqslant-3$ & $x \leqslant 3$

Solve Graphically: $|y-x| \leqslant 3$

Solution: Given, $|y-x| \leqslant 3$

$\Rightarrow \quad-3 \leqslant y-x \leqslant 3$

$\Rightarrow y-x \geqslant-3,$
$\quad y-x \leqslant 3$

$\Rightarrow x-y \leqslant 3, \quad x-y \geqslant-3$

$x-y \leqslant 3\cdots(i)$

$x-y \geqslant-3\cdots(ii)$

For (i):

$\therefore \quad x-y \leqslant 3$

$\Rightarrow \quad \frac{x}{3}-\frac{y}{3} \leqslant 1$

$\Rightarrow \quad \frac{x}{3}+\frac{y}{-3}\leqslant 1$

Associated equation $\frac{x}{3}+\frac{y}{-3}=1$

$\left[\frac{x}{a}+\frac{y}{b}=1\right]$

$x-y \leqslant 3$

$(0,0)$

$0-0=0<3$

$\therefore$ origin $(0,0)$ belongs to solution region of $x-y \leqslant 3$

Again for (ii), $\quad x-y \geqslant-3$

$\Rightarrow \quad \frac{x}{-3}+\frac{y}{3}=1$
$\ \ \text { (Associated equation) }$

$\therefore(-3,0)$ & $(0,3)$

origin test :-

${(0,0)}$

$0-0=0>-3$

$\therefore(0,0)$ will lie in solution region of

$x-y \geqslant-3$

$\therefore$ Common shaded region will be the solution region for

$|y-x| \leqslant 3$

Solve graphically: $|x-y| \geqslant 1$

Solution: $\quad|x-y| \geqslant 1$

$\Rightarrow \quad x-y \leqslant-1 \quad$ & $\quad x-y \geqslant 1$

$x-y=-1$

$\Rightarrow \frac{x}{-1}+\frac{y}{1}=1$

$x-y=1$

$\Rightarrow \frac{x}{1}+\frac{y}{-1}=1$

$x-y \leqslant-1$

$0-0=0 \leqslant-1 \text { (False) }$

$x-y \geqslant 1$

$0-0=0 \geqslant 1 \text { (False) }$

$\therefore$ Shaded portion will be the solution region for $|x-y| \geqslant 1$.