If I want to show that $ax + by = e$ for some $x$ and $y$, can I just substitute and algebraically derive $0 = 0$?

Question

If I want to show that $ax + by = e$ for some $x$ and $y$, can I just substitute and algebraically derive $0 = 0$?

$ax + by = e$

$cx + dy = f$

Let $x = \dfrac{de - bf}{ad - bc}$ where $ad - bc \not = 0$ and $y = \dfrac{af - ce}{ad - bc}$ where $ad - bc \not = 0$.

Show that, for these values of x and y, ax + by = e and cx + dy = f.

$\therefore a\left( \dfrac{de - bf}{ad - bc} \right) + b \left( \dfrac{af - ce}{ad - bc} \right) = e$

$\implies \dfrac{ade - abf}{ad - bc} + \dfrac{baf - bce}{ad-bc} = e$

$\implies \dfrac{ade - abf + baf - bce}{ad - bc} = e$

$\implies ade - bce = ead - bce$

$\implies 0 = 0$

Since $0 = 0$ is always true (infinite number of solutions), doesn't that mean that my calculations are correct?

Answer 1

A better way would be to substitute and ONLY manipulate the left hand side of the equation (leave off the right side), until it simplifies to exactly e. There are a lot of ways to falsely derive 0=0, but if you can rearrange the equation to be exactly e and eliminated the possibility that you're dividing by 0 or square rooting negative numbers (or other incorrect steps) then it will be much more correct.

What you've done is started with an unknown thing and shown something known to be true follows from it. That isn't a valid proof strategy. (In fact, a false statement can prove any statement. If I started with 0=1 then I could prove that 1 = 1, which is true, but 0=1 is false).

You need to start with your expression and manipulate it, without using equations, to show that it equals e.