Solve the following systems of inequalities graphically 5x + 4y ≤ 20, x ≥ 1, y ≥ 2.

x ≥ 1 , ...(ii)

y ≥ 2 . ' ...(iii)

Graph of inequality (i): Let us draw the graph of the line

5x + 4y = 20 ...(iv)

At y = 0 ⇒ 5x + 4(0) = 20 ⇒ x=4

At x = 0 ⇒ 5(0) + 4y = 20 ⇒ x=5

(4,0) and (0, 5) are on the line (iv).
Putting x = y = 0 in (i) we get 5(0) + 4(0) ≤ 20 i.e., 0 ≤ 20 which is true.

Hence, half plane region containing the origin is the solution region of the given inequality.

Graph of inequality (ii). Let us draw the graph of the line x = 1 which is parallel to the y-axis and is at a unit distance from it.

Putting x = 0 in (ii), we have 0 ≥ 1 which is not true. Hence, half-plane region not containing the origin is the solution region of the given inequality.

Graph of inequality (iii). Let us draw the graph of the line y = 2 which is parallel to x-axis and is at a distance of 2 units from it.

Putting y = 0 in (iii), we have 0 > 2 which is not true.

Hence, half plane region not containing the origin is the solution region of the given inequality.

The common region of the above three regions represents the solution set of the given linear system.