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Maximise $8x_1+4x_2+5x_3$
subject to $x_1+2x_2+x_3\le10\\2x_1-3x_2\le8\\x_1,x_2\ge0;x_3\text{free}$

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First you have to replace $x_3$ by $x_3^+-x_3^-$, where $x_3^+,x_3^- \geq 0$.

The coefficients of the objective function have to be muliplied by $(-1)$, because the objective function has to be maximized.

$s_1$ and $s_2$ are the slack variables.

The initial table is

$\begin{array}{|c|c|c|c|}\hline x_1&x_2&x_3^+&x_3^-&s_1&s_2&RHS \\ \hline \color{blue}{-8}&-4&-5&5&0&0&0 \\ \hline \color{green}{1}&2&1&-1&1&0&\color{green}{10} \\ \hline \color{red}{\boxed{2}}&-3&0&0&0&1&\color{blue}{8 }\\ \hline \end{array}$

The first column your pivot column, because $\color{blue}{-8}$ is the smallest value in the first row (objective function).

And the fraction of $\color{green}{10}$ and $\color{green}{1}$ is bigger than the fraction of $\color{blue}{8 }$ and $\color{red}{\boxed{2}}$. Therefore $\color{red}{\boxed{2}}$ is your first pivot element.

I hope you can start from here. If you have any questions, feel free to ask.

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