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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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    You have two equations in four unknowns, and thus you have an underdetermined system.2011-09-02
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    P.S. The [tag:diophantine-equations] tag is intended for equations that require **integer** solutions.2011-09-02
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    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.2011-09-02
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    I've asked Wolfram Alpha to solve this system: http://www.wolframalpha.com/input/?i=solve+ba-bc-ad%3D0%3B+2ab%2Bcb%2Bad-2d*c%3D0. This is the kind of answer I'm looking for, but I have no idea how it arrived at this answer.2011-09-02
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    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?2011-09-02
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    @J.M. It's a homogeneous system, so there are "effectively" only 3 variables :). (Actually, I am able to solve the equation to get a sensible-looking answer.)2011-09-02
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    J.M.: I did not need integer solutions; that's a result of me not understanding what a diophantine equation is. Sorry"! About substituting: it is easy to find reals that satisfy this equation but where, for example, b!=0...2011-09-02
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    @Sri, the problem is that josh didn't say the magic words "___ in terms of ___"... do I solve for a, for c, or for what?2011-09-02
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    J.M.: My bad. I'm expecting answers of the form "variables a,b,c are 0, d is any real", which is why I didn't consider this. You can solve for a, in example.2011-09-02
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    Yes, that is a problem. One guess is to say something like: The solution set is $\{ (a, b, 5a, 2b-a^2) \,:\, a,b \in \mathbb R \}$. Of course, we would think of this as solving in terms of $a$ and $b$. And of course, ultimately, what he wants, only he can tell.2011-09-02
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    Srivatsan: that's exactly the kind of answer I am looking for, yes. Everyone: I'm sorry for so many vague details, not used to posting questions here!2011-09-02

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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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    You are terrific, thanks a lot!2011-09-02
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    @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.2011-09-02
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    "Do the algebra and find the equation" - I like that.2011-09-02
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    @mixedmath Thanks. :-)2011-09-03