Solve the following systems of inequalities graphically 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0.

y ≥ 2x …………… (2)

x ≥ 3 …………… (3)

x ≥ 0 …………… (4)

y ≥ 0 …………… (5)

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

4x + 3y = 60

y = 0 ⇒ 4x + 3(0) = 60 ⇒ x = 15.

x = 0 ⇒ 4(0) + 3y = 60 ⇒ y =20.

(15, 0) and (0, 20) are on the line 4x + 3y = 60.

Putting x = y = 0 in (i) we get 0 ≤ 60 which is true.

Hence, half-plane region containing the origin is the solution region.

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

y = 2x

y = 0 ⇒ 0 = 2x ⇒ x = 0.

x = 0 ⇒ y = 2(1) ⇒ y = 2.

∴ (0, 0) and (1, 2) are the points on the line y = 2x.

To determine the region represented by the given inequality (ii) consider the point not lying on the line y = 2x, say (2, 0) and it lies in the half plane of (ii) if 0 ≥ 4, which is not true. Therefore, the portion not containing (2, 0) represents the solution set of the given inequality.

Graph of inequality (iii): Clearly, x ≥ 0 represents the region lying on the right side of y-axis.

Graph of inequality (iv): Let us draw the graph of the line x = 3 which is parallel to the y – axis and is at a distance of 3 unit from it.

Putting x = 0 in (iii), we have 0 ≥ 3 which is not true.

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

Graph of inequality (iv): Clearly x ≥ 0 represents the region lying on the right side of y-axis.

Graph of inequality (v): Clearly y ≥ 0 represents the region lying above the x-axis.

The common region of the above five regions represent the solution set of the given linear system.

Triple shaded triangular area is the solution area in the solution region.