First, we draw the graph line 5x + 4y = 40
Now , `5x + 4y = 40 rArr x/8+y/10 = 1 `
This line meets the axes at A(8, 0) and B(0, 10).
Clearly, the line AB represents 5x + 4y = 40
Thus, the line AB and part of the plane containing 0(0, 0) and separated by AB represent the solution set of the inequation, `5x+4y le 40 `
(ii) Clearly, x = 2 is the line CD parallel to the y-axis at a distance of 2 units to its right .
Clearly, (0, 0) does not satisfy the inequation, `x ge 2 `
. So, `x ge 2 ` consists of the line CD and part of the plane on right of CD
(iii) Also, y = 3 is the line EF parallel to the x-axis lying above it at a distance of 3 units from it.
Clearly (0,0) does not satisfy the inequation , ` y ge 3 `
So, ` y ge 3 ` consists of the line EF and part of the plane above it. The intersection of al thse planes is the required shaded part, representing of all these planes is the required shaded part, representing the solution of the given system,as shown below.
(`##RSA_MATH_XI_C07_SLV_009_S01.png" width="80%">
