Question

The shaded region representing the solution set of the system of inequalities $x-2 y \leq 3,3 x+4 y \geq 12$, $x \geq 0, y \geq 1$ in the following figure is

• A. $S_1$
• B. $S_2$
• C. $S_3$
• D. $S_4$

Answer 1

Answer: (A)

The given system of inequalities
$x-2 y \leq 3$...(i)
$3 x+4 y \geq 12 $...(ii)
$x \geq 0 $...(iii)
$y \geq 1$...(iv)
Step I Consider the inequations as strict equations
$\text { i.e., } x-2 y=3,3 x+4 y=12 x=0, y=1$
Step II Find the point on the $X$-axis and $Y$-axis

Step III Plot the graph using the above tables.
(i) For $x-2 y=3$ and $3 x+4 y=12$ use the above tables.
(ii) Graph of $x=0$ will be $Y$-axis.
(iii) Graph of $y=1$ will be a line parallel to $X$-axis, intersecting $Y$-axis at 1 .
Step IV Take a point $(0,0)$ and put it in the inequations (i) and (ii),
$0-0 \leq 3 \text { (true) }$
So, the shaded region will be towards origin.
and
$0+0 \geq 12$(false)
So, the shaded region will be away from the origin.
Again, take a point $(1,0)$.
Put it in the inequations (iii), we get
$1 \geq 0$
So, the shaded region will be towards origin.

And a point $(1,0)$ put it in the inequations (iv), we get
$0 \geq 1$ (false)
So, the shaded region will be away from the origin.
Thus, common shaded region shows the solution of the inequalities.