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.

**
**