FactChecker said:
Usually some artificial variables are added which are driven out (using artificially high cost values) to arrive at a feasible solution that only uses the original variables. I am not familiar with any other way to guarantee that a feasible solution can be found.
Modern industrial-scale computer implementations of the simplex method avoid artificial variables, but instead consider starting at infeasible bases and trying to get rid of infeasibilites first, before optimizing. For example, consider the problem
$$\begin{array}{ccl}
\max & Z = & 4 x_1 + 6 x_2 - x_3 \\
\text{s.t.}&&5 x_1 + 2 x_2 + 8 x_3 = 80\\
&&4 x_1 + 9 x_2 +3 x_3 \leq 60\\
&&6 x_1 + 4 x_2 + 5 x_3 \geq 60\\
&& x_1, x_2, x_3 \geq 0
\end{array}$$
If we introduce a slack variable ##s_1## for the first inequality and a surplus variable ##s_2## for the second one, our problem can be written as that of maximizing ##Z## over the non-negative variables ##x_1, x_2, x_3, s_1, s_2## that are related by the equations
$$\begin{array}{ccc}
Z- 4 x_1 - 6x_2 + x_3 &=&0\\
5 x_1 + 2 x_2 + 8 x_3 &= &80\\
4x_1 + 9x_2 + 3x_3 + s_1 &=& 60\\
6x_1 + 4 x_2 + 5 x_3 - s_2 &=& 60
\end{array}$$
We don't even know if the problem is feasible, so determining that is the first order of business. We do not yet have an LP basis.
Without worrying about feasibility, let us choose an initial basis in which the basic variables are ##x_1, s_2, s_3##. The equations can be re-written by putting basic variables on the left and non-basic variables on the right:
$$\begin{array}{ccl}
Z&=& 64 +(22/5) x_2 - (37/5) x_3 \\
x_1 &=& 16 -(2/5) x_2 - (8/5) x_3 \\
s_1 &=& -4 -(27/5) x_2 + (17/5) x_3 \\
s_2&=& 36 +(8/5) x_2 - (2/5) x_3
\end{array}$$
This basis is infeasible because it has ##s_1 = -4 < 0##. Temporarily, we can think of this as a problem of trying to maximize ##s_1## (so as to drive it up towards zero), and for that reason we choose to increase ##x_3##. Using the usual minimum-ratio rules to maintain feasibility of ##x_1, s_2## and to try to achieve feasibility of ##s_1##, the ratios are
$$\begin{array}{l}
s_1 \;\text{ratio} = (-4)/(-17/5) = 1.176\\
s_2 \;\text{ratio} = 36/(23/5) = 7.826\\
x_1 \; \text{ratio} = 16/(8/5) = 10
\end{array}$$
The minimum ratio is for ##s_1##, so ##s_1## leaves the basis and ##x_3## enters. The new equations are
$$\begin{array}{ccl}
Z &=& 940/17 -(199/17) x_2 - (37/17) s_1\\
x_1 &=&240/17 -(66/17) x_2+-(8/17) s_1\\
x_3 &=& 20/17+(37/17) x_2+(5/17)s_1 \\
s_2 &=& 520/17-(143/17) x_2 - (23/17) s_1
\end{array}$$
This gives us a feasible solution ##Z = 940/17, x_1 = 240/17, x_3 = 20/17, s_2 = 520/17.## The current ##Z##-equation also shows that this solution is optimal
By the way: what would happen if we had a truly infeasible problem but did not use artificial variables? In this case the above method would eventually contain an equation something like $$\text{basic variable}\; x = -2 - 4 u_1 - 3 u_2 - \cdots$$ where ##u_1, u_2, u_3 \ldots ## are the current non-basic variables. Note that the right-had-side is negative, and all the coefficients on the right are non-positive. That is, we have a negative basic variable, and increasing any of the non-basic variables up from 0 will not help (and may even make matters worse). This says that the basic variable ##x## cannot be larger than -2, so certainly cannot be non-negative. Our problem would be detected as being infeasible.