Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\{(x,y) \mid x=c-by/a\}$.

share|cite|improve this question

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.

share|cite|improve this answer


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.

share|cite|improve this answer
@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: – Douglas S. Stones Nov 8 '10 at 2:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.