Since the decision variables x₁,x₂ are non-negative, the solution lies in the I quadrant.
Consider the equations
\(3{ x }_{ 1 }+5{ x }_{ 2 }=15\)                                             

\(5{ x }_{ 1 }+2{ x }_{ 2 }=10\)

The feasible region is OABC and its co-ordinates are O(0, 0) A(2, 0) C(0, 3) and B is the point of intersection of the lines
\(3{ x }_{ 1 }+5{ x }_{ 2 }=15\) ----(1)
and \(5{ x }_{ 1 }+2{ x }_{ 2 }=10\)   ... (2)
Verification of B:
\((1)\times 5\Rightarrow 15{ x }_{ 1 }+25{ x }_{ 2 }=75\\ \quad \quad (-)\quad (-)\quad \quad (-) \\ (2)\times 3\Rightarrow 15{ x }_{ 1 }+6{ x }_{ 2 }=30\\ --------------\\ 19{ x }_{ 2 }=45 \Rightarrow { x }_{ 2 }=\frac { 45 }{ 19 } \)
\(From(1), 3{ x }_{ 1 }+5\left( \frac { 45 }{ 19 } \right) =15\)
\(\Rightarrow 3{ x }_{ 1 }=15-\frac { 225 }{ 19 } \)
\(\Rightarrow { x }_{ 1 }=\frac { 20 }{ 19 } \)
\(\therefore \ B \ is\ \left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \)

Maximum of Z occurs at B \(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \)
Hence, the solution is \({ x }_{ 1 }=\frac { 20 }{ 19 } ,{ x }_{ 2 }=\frac { 45 }{ 19 } \ and\ { Z }_{ max }=\frac { 345 }{ 19 } \) 

Since the decision variables x₁ ,x₂ are non-negative, the solution lies in the I quadrant of the plane.
Consider the equations
\(-{ x }_{ 1 }+3{ x }_{ 2 }=10\)

\({ x }_{ 1 }+{ x }_{ 2 }=6\)

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

The feasible region is OABCD and its co-ordinates are O(0, 0)A(2, 0) B(4, 2) C(2, 4) and D(0, 10/3)

Maximum of Z occurs at\(D\left( 0,\frac { 10 }{ 3 } \right) \). Hence, the solution is \({ x }_{ 1 }=0,{ x }_{ 2 }=\frac { 10 }{ 3 } \quad and\quad { Z }_{ max }=\frac { 20 }{ 3 } \)

Since the decision variables are non-negative, the solution lies in the I quadrant of the plane.
Consider the equations
\({ x }_{ 1 }-{ x }_{ 2 }= 10\) 

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

x₁ = 3 is a line parallel to x₂-axis at a distance of 3 units from it.
Also, x₂ = 4 is a line parallel to x₁ - axis at a distance of 4 units from it.

The feasible region is ABCD and its co-ordinates are A is the point of intersection of x₁ = 3 and −x₁ + 2x₂ = 2
Verification:
\(-3+{ 2x }_{ 2 }=2 \Rightarrow { 2x }_{ 2 }=5 \Rightarrow { x }_{ 2 }=\frac { 5 }{ 2 } \)
\( \therefore A\left( 3,\frac { 5 }{ 2 } \right) \)
B is the point of intersection of x₂ = 4 and −x₁ + 2x₂ = 2
\(\Rightarrow -{ x }_{ 1 }+8=2\Rightarrow -{ x }_{ 1 }=-6 \Rightarrow { x }_{ 1 }=6\)
\(\therefore B\quad is\quad \left( 6,4 \right) \)
C is the point of intersection of x₂ = 4 and \({ x }_{ 1 }{ -x }_{ 2 }=2\)
\(\Rightarrow { x }_{ 1 }=4\)
\(\therefore \text {C is (4,4)}\)
D is the point of intersection of x₁= 3 and \({ x }_{ 1 }{ -x }_{ 2 }=2\)
\(\Rightarrow { x }_{ 2 }=3\)
∴D is (3,3)   

Minimum of Z occurs at C(4, 4)
Hence, the solution is x₁ = 4, x₂ = 4 and Z_min = 4.

Since the decision variables are non-negative, the solution lies in the I quadrant of the plane.
Consider the equation 
\({ x }_{ 1 }+{ x }_{ 2 }= 4\)

The feasible region is OAB and its co-ordinates are O(0, 0), A(4, 0) and B(0, 4)

Maximum of Z occurs at B(0, 4)
Hence, the solution is x₁= 0, x₂ = 4 and Z_max = 16

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.