So, there's this problem:

maximize 

> 
> $$6x\_1 + 8x\_2 + 5x\_3 + 9x\_4$$
> 
> 
> 

subject to

$$x\_1+x\_2+x\_3+x\_4 = 1 \;\text{ knowing that }\;
x\_1, x\_2, x\_3, x\_4\geq0$$

The solution is obvious: since the sum of all variables must be 1, $x\_4 = 1$ gives the highest value to the objective function

BUT
the simplex method goes wrong when I use it. There's a negative value on the axuliary equations, which means it is an infeasible dictionary, therefore I should use two phases method. But this one goes wrong as well, as there are too many variables to replace the original objective function with slack variables. 

But the problem HAS a solution, it's obvious, I just can't find it through simplex method.
Can anyone help me?

Here's how the 2 phases method gave:

first dictionary:

\begin{align\*} z&= -x\_0\\
w\_1 &= 1 -x\_1 -x\_2 -x\_3 -x\_4 +x\_0\\
w\_2 &= -1 +x\_1 +x\_2 +x\_3 +x\_4 +x\_0
\end{align\*}

second dictionary:

\begin{align\*} z&= -1 +x\_1 +x\_2 +x\_3 +x\_4 -w\_2\\
w\_1 &= 2 -2x\_1 -2x\_2 -2x\_3 -2x\_4 +w\_2\\
x\_0 &= 1 -x\_1 -x\_2 -x\_3 -x\_4 +w\_2
\end{align\*}

third dictionary:

\begin{align\*} z&= -w\_1 -w\_2\\
x\_1 &= 1 -x\_2 -x\_3 -x\_4 +\frac{w\_2}{2} - \frac{w\_1}{2}\\
x\_0 &= \frac{w\_1}{2} +\frac{w\_2}{2}
\end{align\*}

this is supposed to be optimal but there's no way I'll be able to convert $6x\_1 + 8x\_2 + 5x\_3 + 9x\_4$ using those auxiliary equations