Question

Solve $3x+4y\le 12$ graphically in two-dimensional plane.

Answer 1

Solving inequality for graph.
Given $3x+4y\le 12$
Let first draw graph of $3x+4y=12$ $.....\left(1\right)$

Putting $x=0$ in equation $\left(1\right)$	Putting $y=0$ in equation $\left(1\right)$
$\Rightarrow 3\left(0\right)+4y=12$	$\Rightarrow 3x+4\left(0\right)=12$
$\Rightarrow 4y=12$	$\Rightarrow 3x=12$
$\Rightarrow y=\frac{12}{4}$	$\Rightarrow x=\frac{12}{3}$
$\Rightarrow y=3$	$\Rightarrow x=4$

Hence,

Drawing graph.
Points to be plotted are $(0,3),(4,0)$
Solve for required region
Substituting $x=0,y=0$ in the given inequality,

$3x+4y\le 12$
$3\left(0\right)+4\left(0\right)\le 12$
$0\le 12$
Which is true.

Hence origin lies in plane.
So, we shade left side of line $3x+4y=12$
which contains origin.

Final answer :