Solve the following system of linear inequation graphically: `2x+3ylt=6,\ 3x+22 ylt=6,\ xgeq0,\ ygeq0`

Answer

N/a

Explanation

`x ge + 0, y ge0`

`rArr ` The solution of these inequations lies in first quadrant
For the inequation `2x + 3y lre 6 :`
Corresponding equation `2x +3y = 6 ` meets the X-axis and Y-axis at A (3, 0) and B (0, 0) respectively. Draw a thick line through AB. Put (0, 0) in inequation, we get
`0 le 6`
which is ture i.e., inequation satisfies with (0, 0) .
`:'` AB divides XY-plane into two parts, we shade that region in which the point (0, 0) lies.
For the inequation `3x + 2y le 6 :`
Corresponding equation `3x + 2y = 6` meets the X-axis and Y-axis at C (2,0) and D (0, 3) respectively Draw a thick line through CD.
Put (0, 0) in the inequation,we get
`0le6`
which is true i.e., the inquatioin satisfies with (0, 0).
`:'`CD divides XY-plane into two parts, we shade that region in which (0, 0) lies.
Now the common shaded region of given four inequation is the required solution.