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I have two equations: $|c x_1 + d x_2 | = u $ and $|c x_2 - d x_1 | = v $. I know c, d, v and u. Is it possible to find out if $c x_1 + d x_2$ and $c x_2 - d x_1$ is bigger than or smaller than $0$? I have tried googling and also to solve it with wolfram alpha, but I am stuck.

Thanks in advance.

miracle173
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QCQCQC
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2 Answers2

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We can't. $|c x_1 + d x_2 | = u \implies c x_1 + d x_2 = \pm u$ (and similarly for the $2nd$). Without further restrictions on $x_1$ and $x_2$, we can get valid solutions for $x_1$ and $x_2$ by choosing any one case from each of the $2$ sets of equations.

Ishan Deo
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  • Thanks for the answer. To be clear, I don't need to solve for x1 and x2. I am only interested in if u is positive or negative and the same for v. You are saying this is not possible? – QCQCQC Feb 13 '20 at 13:46
  • Correct, that's not possible. As Ishan explained, any combinations of the signs for $cx_1+dx_2$ and $cx_2-dx_1$ is possible generally. Of course, if $u=0$ or $v=0$ then the the sign of the respective term will always be zero. – Ingix Feb 13 '20 at 13:51
  • Thanks so much! In my case, c = cos(theta/2) and d = sin(theta/2). I can change theta to my liking and evaluate on of the two (or both) equations again (but its costly). Given that I have the u and v from earlier, can I change theta to get enough information to retrieve the sign of u and v? If so, any ideas how? – QCQCQC Feb 14 '20 at 08:09
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The linear equation $$\begin{eqnarray} c x_1 + d x_2 = U \\ - d x_1 +c x_2= V \end{eqnarray}$$ has exactly one solution if and only if its determinant $c^2+d^2\neq 0$ . For real $c$ and $d$ this is equivalent to $(c,d)\neq (0,0)$. It does not depend on $U$ and $V$. If $(c,d)\neq (0,0)$ then there is a solution for every $(U,V)\in\{(u,v),(u,-v),(-u,v),(-u,-v)\}$ .

miracle173
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