Since the decision variables are non-negative, the solution lies in the I-quadrant of the plane. Consider the equations
\({ x }_{ 1 }+2{ x }_{ 2 }= 8\)

\({ 3x }_{ 1 }+{ 2x }_{ 2 }=12\)

The feasible region is OABC and its co-ordinates are 0(0, 0) A( 4, 0) qo, 4) and B is the point of intersection of the lines
\({ x }_{ 1 }+2{ x }_{ 2 }=8\)   ... (1) \(and\quad { 3x }_{ 1 }+{ 2x }_{ 2 }=12\)   ...(2)
Verification of B:
\((1)\Rightarrow { x }_{ 1 }+2{ x }_{ 2 }=8\\ \quad \quad (-)\quad (-)\quad \quad (-)\\ (2)\Rightarrow 3{ x }_{ 1 }+2{ x }_{ 2 }=12\\ -----------\\ -2{ x }_{ 1 }=-4 \Rightarrow { x }_{ 1 }=2\)
\(From(1), 2+2{ x }_{ 2 }=8\Rightarrow 2{ x }_{ 2 }=6\Rightarrow { x }_{ 2 }=3\)

∴ B is (2,3) 

Minimum of Z occurs at A(4, 0).
Hence, the solution is x₁= 4, x₂ = 0 and Z_min = - 12.

Since the decision variables are non-negative, the solution lies in the I-quadrant of the plane.Consider the equations
\({ x }_{ 1 }+3{ x }_{ 2 }=3\)

\({ x }_{ 1 }+{ x }_{ 2 }=2\)

The feasible region is ABC and its co-ordinates are A(3, 0) C(0,2) and B its the point of intersection of the lines
\({ x }_{ 1 }+3{ x }_{ 2 }=3\)  ...(1) and  \({ x }_{ 1 }+{ x }_{ 2 }=2\)   ...(2)
Verification of B:
\((1)\Rightarrow { x }_{ 1 }+3{ x }_{ 2 }=3\)
\( \quad \quad (-)\quad (-)\quad \quad (-)\)
\((2)\Rightarrow { x }_{ 1 }+{ x }_{ 2 }=2\)
\( \quad -----------\)
\(2{ x }_{ 2 }=1\Rightarrow { x }_{ 2 }=\frac { 1 }{ 2 } \)
\( From(2),\ { x }_{ 1 }+\frac { 1 }{ 2 } =2\ \Rightarrow { x }_{ 1 }=2-\frac { 1 }{ 2 } \Rightarrow \frac { 3 }{ 2 } \therefore B\quad is\quad \left( \frac { 3 }{ 2 } ,\frac { 1 }{ 2 } \right) \)

Minimum of Z occurs at B(3/2, 1/2)
Hence, the solution is x₁= 3/2, x₂ = 1/2 and Z_min = 7

Since the decision variables are non-negative, the solution lies in the I quadrant of the plane.
Consider the equations 
\({ x }_{ 1 }-{ x }_{ 2 }=-1\)

\({ -x }_{ 1 }+{ x }_{ 2 }=0\)

The feasible region is not common. Thus, there is no maximum value of Z.