There are different methods to solve an LP problem, such as graphical method, simplex method, or using software. Here we will use the simplex method.

First, we need to convert the problem to standard form, which means that all constraints are inequalities, all variables are non-negative, and the objective function is to be minimized. We introduce slack variables to convert the inequalities to equalities:

min z = 8x1 - 2x2

subject to:

-4x1 + 2x2 + x3 = 1
5x1 - 4x2 + x4 = 3
x1, x2, x3, x4 ≥ 0

Next, we set up the simplex tableau:

BV | x1 | x2 | x3 | x4 | RHS
----|----|----|----|----|----
x3 | -4| 2 | 1 | 0 | 1
x4 | 5| -4 | 0 | 1 | 3
z | 8| -2 | 0 | 0 | 0

The basic variables (BV) are x3 and x4, which have non-zero entries in their respective columns. The other variables are non-basic. The RHS column shows the right-hand side of each equation.

We select the most negative coefficient in the bottom row, which is -2. This corresponds to variable x2 in the objective function. We will use x2 as the entering variable. To determine the leaving variable, we calculate the ratios of the RHS to the coefficients of x2 in each equation:

1/2 = 0.5
3/(-4) = -0.75

The smallest ratio corresponds to the second equation, which means that x4 will leave the basis, and x2 will enter the basis. We perform row operations to make x2 the pivot element and to eliminate the other entries in its column:

BV | x1 | x2 | x3 | x4 | RHS
----|----|----|----|----|----
x3 | 0| 1 | 1 | 2 | 1
x2 | 5/4| 1 | 0 | -1/4| 3/4
z | -4 | 0 | 8 | 10| 6

Now x2 is a basic variable, and its coefficient in the objective function is zero, which means that the objective function cannot be further improved by increasing or decreasing x2. We select the most negative coefficient in the bottom row, which is -4. This corresponds to variable x1 in the objective function. We will use x1 as the entering variable. To determine the leaving variable, we calculate the ratios of the RHS to the coefficients of x1 in each equation:

1/5 = 0.2
3/5 = 0.6

The smallest ratio corresponds to the first equation, which means that x3 will leave the basis, and x1 will enter the basis. We perform row operations to make x1 the pivot element and to eliminate the other entries in its column:

BV | x1 | x2 | x3 | x4 | RHS
----|----|----

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