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I am trying to convert this LP into standard form and I am stuck on the part with $-1\leq x_1\leq 4$. I am unsure how to turn this into variables that are all greater than 0.

$\begin{array}{l@{}l}\text{minimize} & 4x_1+x_2+ 5x_3\\ \text{subject to}& x_1+x_3 \geq 10 \\ &2x_1+x_2+3x_3 \geq 15 \\ &-1 \leq x_1 \le 4, x_2 \leq 0, x_3 \geq 0 \\ \end{array}$

My approach so far:

Let $x_2'=-x_2$

$\begin{array}{l@{}l}\text{maximize} & -4x_1+x_2'- 5x_3\\ \text{subject to}& -x_1-x_3 \leq -10 \\ &-2x_1+x_2'-3x_3 \leq -15 \\ &-1 \leq x_1 \le 4,~x_2',~x_3 \geq 0 \\ \end{array}$

but then I don't know what to do with the $-1\leq x_1\leq 4$

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Substitute $x_1'=x_1+1\in[0,5]$ to get $$\begin{array}{l@{}l}\text{maximize} &Z-\color{red}4=-4x_1'+x_2'- 5x_3\\ \text{subject to}& -x_1'-x_3 \leq -11 \\ &-2x_1'+x_2'-3x_3 \leq -17 \\ &x_1' \le 5\\ &x_{1,2}',~x_3 \geq 0 \\ \end{array}$$and note that $Z$ is maximized iff $Z-4$ is maximized.

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