Given,

x + 2y 10,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 5 and x = 10

The required points are (0, 5) and (10, 0)

Checking for the origin (0, 0)

0 ≤ 10, this is true.

Hence the solution area would be toward origin including the same. The solution area would be toward the left of the line`s graph.

x + y 1,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 1 and x = 1

The required points are (0, 1) and (1, 0)

Checking for the origin (0, 0)

0 ≥ 1, this is not true.

Hence the origin would not be included into the solution area. The required solution area would be toward right of the line`s graph.

x – y 0,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 0 and x = 0

Hence the origin would lie on the line.

To check which side of the line graph would be included in the solution area we would check for the (10, 0)

10 ≤ 0 which is not true hence the solution area would be on the left side of the line`s graph.

x 0, y 0

Since both x and y are greater than 0, the solution area would be in the 1^st quadrant.

Hence, the solution area for the given inequalities would be the shaded area of the graph satisfying all the given inequalities.