1. Solving simultaneous linear inequation means finding the set of points $(x, ~y)$ for which all the constraints are satisfied.

Solution set may be

(i) An empty set $(3 x+2 y \geqslant 24,3 x+y \leqslant 15, x \geqslant 4)$

(ii) A bounded region $(2 x+3 y \leqslant 12, x \geqslant 2, y \geqslant 1)$

(iii) An unbounded region $(x+2 y \leqslant 8,2 x+y \leqslant 8, x \geqslant 0, y \geqslant 0)$

2. To solve a system of Linear inequation in two variables graphically

(i) Draw the graphs of all the given linear inequalities.

(ii) Mark the feasible region of each line.

(iii) Find the common region which satisfies all the given linear inequalities.

(iv) This common region is the required solution of the given system of linear inequalities.

Solve system of linear inequalities graphically.

$3 x+2 y \geqslant 24 - \text{(i) }$

$3 x+y \leqslant 15 - \text{(ii) }$

$x \geqslant 4 - \text{(iii)}$

Solution: Associated equation for (i), (ii) & (iii) are

$3x + 2y = 24$

$3x + y = 15$ & $x = 4$

for (i) $3 x+2 y=24$

put $y=0 \Rightarrow x=8$

$x=0 \Rightarrow y=12$

$\therefore$ points are $(8,0)$ & $(0,12)$

for (ii) $\quad 3 x+y=15$

put $y=0 \Rightarrow x=5$

$x=0 \Rightarrow y=15$

for (iii) $x=4$ is a line $\parallel$ to $y-$ axis passing through $(4,0)$

Now for (i) i.e. $3 x+2 y \geqslant 24$

origin test: put $x=0$ & $y=0$

$3 \times 0+2 \times 0=0 \ngeq 24$

$\therefore(0,0) \notin \text{solution region}.$

for (ii) i.e. $3 x+y \leqslant 15$

put $x=0 \quad$ &$~y=0$

$3 \times 0+0=0 \leqslant 15$ (True)

$\therefore(0,0) \in$ solution region.

for (iii) i.e. $x \geqslant 4$

According to defined region for different given inequalities no region will be common region.

$\therefore \text{solution region}=\phi$

Solve graphically

$2 x+3 y \leqslant 12 -\text { (i) }$

$x \geqslant 2 - \text { (ii) }$

$y \geqslant 1 -\text { (iii) }$

Solution: Associated equation for (i), (ii), & (iii)

$2 x+3 y=12$

$x=2$

$y=1$

for (i) $2 x+3 y=12$

put $y=0 \Rightarrow x=6$

$x=0 \Rightarrow y=4$

$\therefore$ points are $(6,0)$ & $(0,4)$

for(ii) $x=2$ is a line || to $y-\text{axis}$ and passing through $(2,0)$

for (iii) $y=1$ is a line || to $x-\text{axis}$ and passing through $(0,1)$

$\underline{\text{Origin test:}}$

for (i) $2 x+3 y \leqslant 12$

put $x=0$ & $y=0$

$2 \times 0+3 \times 0=0 \leqslant 12$ (True)

$\therefore(0,0)$ lies in the solution region for $2 x+3 y \leqslant 12$

for (ii) $x\geqslant 2 \ \ \Rightarrow \$ solution region lies in right hand side of line $x=2$.

for (iii) $y \geqslant 1 \ \ \Rightarrow \$ solution region lies in upper part of the line $y = 1$

Shaded region $ABC$ will satisfy all the three given constraints. Hence, solution region will be the shaded region $ABC.$

Solve graphically

$x+2 y \leqslant 8 -\text {(i) }$

$2 x+y \leqslant 8- \text { (ii) }$

$x \geqslant 0, y \geqslant 0 -\text {(iii) }$

Solution: Associated equation for (i) & (ii)

$x+2 y=8$

$2 x+y=8$

for (i) $x+2 y=8$

put $y=0 \Rightarrow x=8$

$x=0 \Rightarrow y=4$

$\therefore$ points are $(8,0)$ & $(0,4)$

for (ii) $2 x+y=8$

$\text { put } y=0 \Rightarrow x=4$

$x=0 \Rightarrow y=8$

$\therefore$ points are $(4,0)$ & $(0,8)$

$x \geqslant 0$ & $y \geqslant 0$

$\rArr I^{st} \text{quadrant}$

$\underline{\text{Origin test:}}$

for (i) $x+2 y \leqslant 8$

put $x=0, y=0 \Rightarrow 0+2 \times 0=0 \leqslant 8$ (True)

$\therefore(0,0) \in$ Solution Region.

for (ii) $2 x+y \leqslant 8$

put $x=0, y=0 \Rightarrow 2 \times 0+0=0 \leqslant 8$ (True)

$(0,0) \in$ Solution Region

Common solution Region for the given system of inequalities (i), (ii) & (iii) will be shaded region $O A B C.$

Solve graphically

$x+y<5 -\text { (i) }$

$4 x+y \geqslant 4 -\text { (ii) }$

$x+5 y \geqslant 5 -\text { (iii) }$

$x \leqslant 4 -\text{(iv)}$

$y \leqslant 3 -\text { (v) }$

Associated equation for given inequation (i), (ii), (iii), (iv) & (v) are

$x+y=5$

$4 x+y=4$

$x+5 y=5$

$x=4$

$y=3$

for (i) $x+y=5$

put $y=0 \Rightarrow x=5$

$x=0 \Rightarrow y=5$

$\therefore$ points are $(5,0)$ & $(0,5)$

for (ii) $4 x+y=4$

$\text { put } y=0 \Rightarrow x=1$

$x=0 \Rightarrow y=4$

points are $(1,0)$ & $(0,4)$

for (iii)

$x+5 y=5$

$\text { put } y=0 \Rightarrow x=5$

$x=0 \Rightarrow y=1$

points are $(5,0)$ & $(0,1)$

for (iv) $x=4$ is a line $|| ~to ~y-\text{axis}$ and passing through $(4,0)$

for (v) $y=3$ is a line $|| ~to ~x-\text{axis}$ and passing through $(0,3)$

Common region for all the given inequalities (i), (ii), (iii), (iv) & (v) will be the shaded region $ABCDE.$

Hence solution region will be region $ABCDE.$