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I am trying to find all real solutions to the following set of equations: $$\begin{align*} b(a-c)-ad&=0\\ 2ab+cb+ad-2cd&=0 \end{align*}$$

My algebra is a bit rusty, and I really have no idea where to begin (aside from maybe a long sequence of 'brute force' by substitution). I would appreciate help with this.

Thanks in advance!

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    $\begingroup$ You have two equations in four unknowns, and thus you have an underdetermined system. $\endgroup$ – J. M. is a poor mathematician Sep 2 '11 at 16:50
  • $\begingroup$ P.S. The diophantine-equations tag is intended for equations that require integer solutions. $\endgroup$ – J. M. is a poor mathematician Sep 2 '11 at 16:52
  • $\begingroup$ J.M. thank you. I'm aware this has infinite solutions, and I am looking for answers of the form: a=0,b=0,c!=0,d!=0, and the such. $\endgroup$ – josh Sep 2 '11 at 16:52
  • $\begingroup$ I've asked Wolfram Alpha to solve this system: wolframalpha.com/input/…*c%3D0. This is the kind of answer I'm looking for, but I have no idea how it arrived at this answer. $\endgroup$ – josh Sep 2 '11 at 16:55
  • $\begingroup$ Okay... if you know that $a$ and $b$ are zero, why not substitute those in into your equations? Did you really need integer solutions (in which case I'll restore the former tag) or are you allowing real solutions? $\endgroup$ – J. M. is a poor mathematician Sep 2 '11 at 16:55
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HINT Your question can be equivalently written as: $$ \begin{align*} 3ab=2cd \tag{1} \\ bc+ad = ab \tag{2} \end{align*} $$

You can eliminate $d$ from the system by taking $a \times (1) + 2c \times (2)$, when you will get the equation: $$ \ldots (\text{do the algebra and find the equation}) $$ Collect together all terms on one side and factor the equations. You will find that either $b = 0$ or $\ldots$. (Fill in the blank.)

  • Case 1: Suppose $b = 0$. Can you handle this case?

  • Case 2: Suppose $\ldots$ holds. Then in this case, $a = c = 0$. (Can you prove this? It is slightly nontrivial to show this.) In this case, what can you say about $b$ and $d$?

So, to conclude, what are all the solutions of the equation?

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  • $\begingroup$ You are terrific, thanks a lot! $\endgroup$ – josh Sep 2 '11 at 17:47
  • $\begingroup$ @josh, Were you able to go all the way through? In that case, if you want to, you can edit the question, showing your full work and the final solution; we can check if everything's ok. $\endgroup$ – Srivatsan Sep 2 '11 at 17:52
  • $\begingroup$ "Do the algebra and find the equation" - I like that. $\endgroup$ – davidlowryduda Sep 2 '11 at 23:46
  • $\begingroup$ @mixedmath Thanks. :-) $\endgroup$ – Srivatsan Sep 3 '11 at 0:38

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