# Real solutions to equation set

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.

• You have two equations in four unknowns, and thus you have an underdetermined system. – J. M. is a poor mathematician Sep 2 '11 at 16:50
• P.S. The diophantine-equations tag is intended for equations that require integer solutions. – J. M. is a poor mathematician Sep 2 '11 at 16:52
• 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. – josh Sep 2 '11 at 16:52
• 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. – josh Sep 2 '11 at 16:55
• 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? – J. M. is a poor mathematician Sep 2 '11 at 16:55

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?

• You are terrific, thanks a lot! – josh Sep 2 '11 at 17:47
• @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. – Srivatsan Sep 2 '11 at 17:52
• "Do the algebra and find the equation" - I like that. – davidlowryduda Sep 2 '11 at 23:46
• @mixedmath Thanks. :-) – Srivatsan Sep 3 '11 at 0:38