Question

Solve the system of inequalities graphically:

$x-2y\le 3,3x+4y\ge 12,x\ge 0,y\ge 1$

Answer 1

First we solve $x-2y\le 3$

Lets first draw graph of

$x-2y=3$

Putting $x=0$ in (1)

$0-2y=3$

$-2y=3$

$y=\frac{3}{-2}$

$y=-1.5$

Putting $y=0$ in (1)

$x-2\left(0\right)=3$

$x-0=3$

$x=3$

$\begin{array}{ccc}x& 0& 3\\ y& -1.5& 0\end{array}$

Points to be plotted are

$(0,-1.5)$ and $(3,0)$ to get figure $1$.

Checking for $(0,0)$

Putting $x=0$, $y=0$

$x-2y\le 3$

$0-2\left(0\right)\le 3$

$0\le 3$

Which is true. Hence origin lies in plane $x-2y\le 3$

So, we shade left of line.

Now we solve $3x+4y\ge 12$

Lets first draw graph of

$3x+4y=12$

Putting $x=0$ in (1)

$3\left(0\right)+4y=12$

$0+4y=12$

$4y=12$

$y=\frac{12}{4}$

$y=3$

Putting $y=0$ in (1)

$3x+4\left(0\right)=12$

$3x+0=12$

$3x=12$

$x=\frac{12}{3}$

$x=4$

$\begin{array}{ccc}x& 0& 4\\ y& 3& 0\end{array}$

Points to be plotted are

$(0,3),(4,0)$ to get figure $2$.

checking for $(0,0)$

Putting $x=0,y=0$

$3+4y\ge 12$

$3\left(0\right)+2\left(0\right)\ge 12$

$0\ge 12$

which is false.

Hence, origin does not lie in plane $3x+2y>6$

So, we shade rights side of line.

Also $y\ge 1$

So, for all values $x$

$\begin{array}{cccc}x& 0& -1& 4\\ y& 1& 1& 1\end{array}$

Points to be plotted are

$(0,1),(-1,1),(4,1)$ to get figure $3$.

Also $x\ge 0$

So, the shaded region will lie on the right side of y-axis

Hence the shaded region represents the given inequality.