solve the following system of inequation by graphical method : `5x +4y le 40, x ge 2, y ge 3`

Answer

N/a

Explanation

For the inequation `5x + 4y le 40.`

Corresponding equation `5x + 4y = 40` meets the X-axis and Y-axis at A (8, 0) and (0, 10) reapectively. Draw a thick line passing through A and B. Now put (0, 0) in the inequation, we get
`0 le 40`
which is ture, i.e., inequation satisfies with (0, 0). Now AB divides the XY-plane into two parts, we shade that region in which (0, 0) lies.
For inequation `xge 2 :`
Corresponding equation is x = 2 which is a line parallel to Y-axis and passes through (2, 0) .
The inequation does not satisfy with the point (0, 0). The line x = 2 divide the XY-plane into two parts. We shade that region in which (0, 0) does not lie.
For inequation `yge 3 :`
Corresponding equation is y = 3, which is a line parallel to X-axis and passes throuth (0 ,3).
The inequation does not satisfy with the point (0, 0) . The line y = 3 divide the XY-plane into two parts.
Now the common shaded region of the given three inequation is the required solution.