# Prove that $ax + by = c$ has same solution set as $ax+dy=e$

Prove that, where $a,b, \ldots, e$ are real numbers and $a \neq 0$, if $ax + by = c$ has the same solution set as $ax+dy=e$ then they are the same equation. What if $a=0$?

Note: If $a \ne 0$ then the solution set of the first equation is $\{(x,y) \mid x=c-by/a\}$.

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What other possible tags are appropriate here apart from homework? –  Guess who it is. Nov 7 '10 at 13:29

## 2 Answers

Follow the supplied hint: evaluate $\rm\ x = (c-b\: y)/a = (e-d\:y)/a\$ at $\rm\ y = 0\$ and $\rm\ y = -1$

More generally, see also this closely related recent question.

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Hint:

If $ax + by = c$ has the same solution set as $ax + dy = e$ then any pair $(x,y)$ solving the first equation solves the second too. Let $(x,y)$ and $(x',y')$ with $y \neq y'$ be solutions. Subtracting the two we have $(b - d)y = c - e$ and $(b - d)y' = c - e$ but the only way for $(b - d)y = (b - d)y'$ to hold is when $b = d$, because $y$ is not equal to $y'$. Clearly then, $0 = c - e$ and this proves that $c = e$.

The reason this proof does not go through when $a = 0$ is because there are no solutions with $y$ distinct.

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@muad: To avoid a circular proof you need to explicitly say why there are solutions with different values of y. –  Bill Dubuque Nov 7 '10 at 17:49
I don't think there's an official response regarding homework questions yet, but there seems to be a consensus to not give answers that could be "copied verbatim and submitted as a solution". See: meta.math.stackexchange.com/questions/106 –  Douglas S. Stones Nov 8 '10 at 2:13