The given system of inequalities x−2y≤3...(i) 3x+4y≥12...(ii) x≥0...(iii) y≥1...(iv)
Step I Consider the inequations as strict equations i.e., x−2y=3,3x+4y=12x=0,y=1
Step II Find the point on the X-axis and Y-axis
Step III Plot the graph using the above tables.
(i) For x−2y=3 and 3x+4y=12 use the above tables.
(ii) Graph of x=0 will be Y-axis.
(iii) Graph of y=1 will be a line parallel to X-axis, intersecting Y-axis at 1 .
Step IV Take a point (0,0) and put it in the inequations (i) and (ii), 0−0≤3 (true)
So, the shaded region will be towards origin.
and 0+0≥12(false)
So, the shaded region will be away from the origin.
Again, take a point (1,0).
Put it in the inequations (iii), we get 1≥0
So, the shaded region will be towards origin.
And a point (1,0) put it in the inequations (iv), we get 0≥1 (false)
So, the shaded region will be away from the origin.
Thus, common shaded region shows the solution of the inequalities.