show graphically that the solution set of the following system of inequalities is empty
x-2y greater than = 0, 2x-y less than = -2 ,x greater than = 0, y greater than = 0
Given, x- 2y ≥ 0, 2x - y ≤ -2, x ≥ 0, y ≥ 0
On converting the given inequations into equations, we get
x- 2y = 0, 2x - y = -2, x = 0, y = 0
Now, consider the line x - 2y = 0. Its solution set is:
x | 0 | 2 |
y | 0 | 1 |
We find that (0,0) satisfies the inequation x- 2y ≥ 0. So, the portion containing the origin represents the solution set of the inequation x- 2y ≥ 0.
Again, consider the line 2x - y = -2. Its solution set is:
x | 0 | -1 |
y | 2 | 0 |
We find that (0,0) doesn't satisfy the inequation 2x - y ≤ -2. So, the portion not containing the origin represents the solution set of the inequation 2x - y ≤ -2.
Clearly, x ≥ 0 and y ≥ 0 represents the first quadrant.
As all the four lines doesn't possess any common region. So, the solution set of the given linear inequations is empty.