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I have a pretty straightforward linear programming problem here:

$$ maximize \hskip 5mm -x_1 + 2x_2 -3x_3 $$

subject to

$$ 5x_1 - 6x_2 - 2x_3 \leq 2 $$ $$ 5x_1 - 2x_3 = 6 $$ $$ x_1 - 3x_2 + 5x_3 \geq -3 $$ $$ 1 \leq x_1 \leq 4 $$ $$ x_3 \leq 3 $$

Convert to standard form.

what boggles me is how to substitute $x_1$ since it’s restricted from both sides and I can’t move forward in the problem until I figure it out...

I’m not asking for the whole standard form, just how to approach this one variable. :)

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    $\begingroup$ It is like having two separate inequalities $x_1\geq 1$ and $x_1\leq 4$. Treat them as individual constraints and proceed with your method. $\endgroup$ – Michal Adamaszek Dec 2 at 8:42
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Answer to your problem: You can always split a two sided inequality to take two inequalities like how you take 2 equalities as 3 different ones.

Split $a ≥ b ≥ c$ into $a ≥ c$ , $b ≥ c$ and $a ≥ b$

Solution: $x_1=\frac{6+2x_3}{5}$ Plug in x everywhere

$6−4x_2−2x_3≤2$

$27x_3−15x_2≥−21$

$x_3≤3$

$\frac{-1}{2}≤x_3$

Take one on x and one on y.

Desmos

To Max. $\frac{-6+10x-17y}{5}$, We see that x has to as high as possible and y as low as possible and priority goes to y as it contributes more to the value of the given expression.

So we take this vertex, enter image description here

Which will be Intersection of $27y-15x=-21$ and $6-4x-2y=2$. I leave the rest to you.

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The standard form should use the inequalities of the form $h(\mathbf{x}) \le 0$, where $\mathbf{x} = (x_1, x_2, x_3)^\top$. Hence, e.g. $5x_1 - 2x_3 = 6$ should be rewritten as follows: $$ 5x_1 - 2x_3 - 6 \le 0, \\ -5x_1 + 2x_3 + 6 \le 0. $$

Hope it is clear how to proceed with other constraints.

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