Graph the solution sets of the following inequations: $x-y_{s} 3$

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Graph the solution sets of the following inequations: $x-y_{s} 3$

Given $x-y \leq 3$
$\Rightarrow-y \leq 3-x$
Multiplying by minus on both the sides, we’ll get
$\begin{array}{l} y \geq-3+x \\ y \geq x-3 \end{array}$
Consider the equation $y=x-3$ .
Finding points on the coordinate axes:
If $x=0$ , the $y$ value is – 3 i.e, $y=-3$
$\Rightarrow$ the point on the $Y$ axis is $A(0,-3)$
If $y=0,0=x-3$ $\Rightarrow x=3$
The point on the $X$ axis is $B(3,0)$
Plotting the points on the graph: fig. $2 a$

Now consider the inequality $y \geq x-3$
Here we need the y value greater than or equal to $x-3$
$\Rightarrow$ the required region is above point $A$ .
Therefore the graph of the inequation $x+y \geq 4$ is fig. $2 b$