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Is it possible to solve for two unknowns using only one equation?

For example:

$x+3y=32$

Where $x$ and $y$ are integers.

Thanks :)

Inceptio
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BittersweetNostalgia
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  • $x,y \epsilon$ Natural Numbers? – Inceptio Mar 21 '13 at 09:30
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    You labeled the question as 'Diophantine-equations'. If $x,y$ don't have to be integers, this label makes no sense. – Dennis Gulko Mar 21 '13 at 09:33
  • The set of solutions here is trivially ${(32-3k,k) : k \in \Bbb Z}$. For a more interesting example, you can consider the equation $2x+3y=32$. Note that such equations need not have solutions, for instance $2x + 6y = 31$ has none. – Siméon Mar 21 '13 at 09:57
  • @ANerd: You should try squaring $x$ or $y$, that will be a challenging one. Something like $x^2+3y=32$ or $x+3y^2=32$. – Inceptio Mar 21 '13 at 09:59

2 Answers2

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The answer to the original question 'Is it possible to solve for two variables related by single equation' is, YES!

Here is an example. Solve for two unknown real numbers $x$ and $y$ given a single equation $$x+\sqrt{y-1}=\sqrt[4]{1-y}$$ The only real solutions are $x=0,y=1$

Solution As $x,y$ are real numbers, either $y-1\ge0$ or $1-y \le0.$

Assume that $y$ is strictly greater than $1$, then the LHS of the equation is a real number, whereas, the RHS will have non-zero complex part. A similar argument is also true when the value of $y$ is strictly less than $1$. Therefore, $y = 1$ is the only possible solution. Indeed, $x = 1$ and $y = 0$, is a solution, and the only one.

This may be considered as a hack, since apart from the equation, we have two constraints, i.e., $x$ and $y$ are Real. There are complex solutions to the equation.

hrkrshnn
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Not uniquely. You can solve for $x$ with $x = 32 - 3y$, and if you choose any $y$, you'll have an $x$ so the corresponding $(x, y)$ pair is a solution. Conversely, you can solve for y in terms of $x$, and then use (any? or just integer $x$ so that $y$ is an integer -- you filed this under "diophantine equations") $x$ choices to get $y$s for $(x, y)$ solution pairs.

The_Sympathizer
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  • Ah ha. So there can be like any possible number of solutions? – BittersweetNostalgia Mar 21 '13 at 09:33
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    There are an infinite number of solutions, in fact. One for each $y$ (or conversely, for each $x$ -- I see the comment you made.). When graphed, the solution set is a line in the plane (unless you really do interpret it as a Diophantine equation, in which case it's just disconnected, isolated points.). – The_Sympathizer Mar 21 '13 at 09:35