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I know how to solve system of linear equations: theorem of Rouché-Frobenius to know if it has solutions and Cramer for find them. But I need some theoretic material to know how to solve linear inequalities.

Specifically, I need to know how to find numbers $x_1, x_2, \ldots, x_r$ such that they satisfy all of these inequalities:

\begin{equation} \left\{ \begin{split} a_{11}x_1 + \ldots + a_{1r}x_r & \neq b_1\\ a_{21}x + \ldots + a_{2r}x_r & \neq b_2\\ a_{s1}x + \ldots + a_{sr}x_r & \neq b_s\\ \end{split} \right. \end{equation}

Is there any general way (theorem or so) to find these numbers?

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    $\begingroup$ This set is equal to $\mathbb{R}^4 \setminus \{(x,y,z,t) ; -x+z+t=1 \ or \ x-y-t=1\}$ $\endgroup$
    – PAM1499
    Commented Sep 4, 2020 at 17:31
  • $\begingroup$ "do not satisfy inequality" means to satisfy the corresponding equation, so probably you would say "such that they satisfy all of these inequalities:" $\endgroup$ Commented Sep 13, 2020 at 15:22
  • $\begingroup$ @enzotib solved $\endgroup$ Commented Sep 15, 2020 at 13:08

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Let $$A= \begin{bmatrix} a_{11}+a_{12}+ &\cdots & + a_{1r}\\ a_{21}+a_{22}+& \cdots &+ a_{2r}\\ \vdots\ \qquad \vdots & \ddots & \vdots\\ a_{s1}+a_{s2}+ &\cdots & + a_{sr} \end{bmatrix}$$and $$X=\begin{bmatrix}x_1\\ x_2\\ \vdots \\ x_s\end{bmatrix},\qquad B=\begin{bmatrix}b_1\\ b_2\\ \vdots \\ b_s\end{bmatrix}$$ and Hence the given condition is $$AX\neq B $$ Let $A_L^{-1} $ is the left inverse of the matrix $A $ such that $A_L^{-1}A=I_r $ Therefore \begin{align*} & AX\neq B\\ \implies & A_L^{-1}AX\neq A_L^{-1}B\\ \implies & \Big(A_L^{-1}A\Big)X\neq A_L^{-1}B\\ \implies & X\neq A_L^{-1}B \end{align*} In this way you can find out which values $x_i $ can not take

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    $\begingroup$ $AX\neq B$ means that at least one inequality is satisfied, while the OP ask for all inequality to be satisfied $\endgroup$ Commented Sep 13, 2020 at 15:20
  • $\begingroup$ $AX\neq B $ means both meaaning-1) all the some of the values can be equal and rest are not equal and 2) all the values are not equal..that is why by multiplying with the left inverse we are taking that all of the $x $'s are not equal to those values $\endgroup$ Commented Sep 13, 2020 at 15:31
  • $\begingroup$ @SohamChatterjee "means both meaaning-1) all the some of the values can be equal and rest are not equal and 2) all the values are not equal" This is false. This means that "at least one inequality is satisfy" as enzotib said $\endgroup$ Commented Nov 10, 2020 at 19:26

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