Question

Solve the given inequalities graphically:

$x+2y\le 10,x+y\ge 1,x-y\le 0,x\ge 0,y\ge 0$

Answer 1

First we solve $x+2y\le 10$

Lets first draw graph of

$x+2y=10$..........(1)

Putting $x=0$ in (1)

$0+2y=10$

$2y=10$

$y=\frac{10}{2}$

$y=5$

$y=0$ in (1)

$x+2\left(0\right)=10$

$x+0=10$

$x=10$

Drawing graph

Refer fig.1

$x010$

$y50$

Points to be plotted are $(0,5),(10,0)$

Checking for $(0,0)$

Putting

$x=0,y=0$

$x+2y\le 10$

$0\le 10$

Which is true

So, we shade left of line.

Now we solve $x+y\ge 1$

Lets first draw graph of

$x+y=1$..........(2)

Putting $y=0$ in (1)

$x+0=1$

$x=1$

Putting $x=0$ in (1)

$0+y=1$

$y=1$

Drawing graph

Refer fig .2

$x01$

$y10$

Points to be plotted are $(0,1),(0,1)$

Checking for $(0,0)$

Putting

$x=0,y=0$

$x+y\ge 1$

$0+0\ge 1$

$0\ge 1$

Which is false

Hence origin does not lie in plane $x+y\ge 1$

So, we shade right upper side of line

Now we solve $x-y\le 0$

Lets first draw graph of

$x-y=0$..........(3)

Putting $x=0$ in (3)

$0-y=0$

$-y=0$

$y=0$

Putting $y=2$ in (2)

$x-2=0$

$x=0+2$

$x=2$

Drawing graph

Refer fig. 3

$x02$

$y02$

Points to be plotted are $(0,0),(2,2)$

Checking for $(10,0)$

Putting

$x=10,y=0$

$x-y\le 0$

$10-0\le 0$

$10\le 0$

Which is false

Hence $(10,0)$ does not lie in plane $x>y$

So, we shade left side of line.

Also, given

Refer fig. 4

$x\ge 0,y\ge 0$

So shaded region will lie in first quadrant.

Hence the shaded region represents the given inequality.