I'm presented with the equation $\frac{a+b}{a} = \frac{b}{a+b}$

Performing cross multiplication yields $a^2+2ab+b^2 = ab$

Subtracting $ab$ from both sides, we get $a^2+ab+b^2 = 0$

Multiplying both sides by (a-b) and simplifying:

$(a-b)(a^2+ab+b^2) = 0 * (a-b)$

$a^3 - b^3 = 0$

$a^3 = b^3$

$a = b$

Substituting into the original equation, we finally arrive at $2 = \frac{1}{2}$ ... something has obviously gone terribly wrong. Where did I mess up?

Also it is worth noting that the original problem explicitly states that $a ≠ b$


You multiplied with $0=a-b$. That is wrong. You are just recovering this solution. Just to give you a similar example, say you have $$x-1=0$$ If I multiply both sides by $x+1$, I get $$x^2-1=0$$ or $$x=\pm 1$$Obviously $x=-1$ is not a solution. You introduced it when you multiplied with $x+1$.

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  • $\begingroup$ Right. But that introduces a new solution. $a^3=b^3$ has one real solution that you introduced when you multiplied by $(a-b)$ and two complex solutions. $\endgroup$ – Andrei Apr 25 at 19:04
  • $\begingroup$ Just use the formula for quadratics to find $a$ and $b$ $\endgroup$ – Andrei Apr 25 at 19:05
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    $\begingroup$ So it's an extraneous solution, just discard it? $\endgroup$ – Saif Taher Apr 25 at 19:09
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    $\begingroup$ Yes, since you already checked that's not a solution of the original equation $\endgroup$ – Andrei Apr 25 at 19:11

Here $$(a-b)(a^2+ab+b^2) = 0 (a-b)$$ When $ a=b$. both side becomes $0$

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  • $\begingroup$ It is explicitly stated in the original problem that a is not equal to b. That's why I'm so confused $\endgroup$ – Saif Taher Apr 25 at 19:00
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    $\begingroup$ As you are saying that $a\ne b \implies (a-b)\ne 0$ then $a^3-b^3=0\implies (a-b)(a^2+ab+b^2)=0\implies (a^2+ab+b^2)= 0$ and then your first equation also doesn't give any absurd results. $\endgroup$ – Nimu Basak Apr 25 at 19:10
  • $\begingroup$ Sorry I get it now thx. $\endgroup$ – Saif Taher Apr 25 at 19:48

You made a logical error.

You started with an equation that you had no idea if it had any solution. Then you made a series of transformations and found out that the only possible solution (in the real numbers) could be $a=b$. Then you put that into the original equation and found that $a=b$ leads to something that is untrue. Alternatively, the problem statement removed the possibility of $a=b$ outright.

So the only logical conclusion is that this equation has no real solution!

You can easily see that this is actually true, because at some time in between you arrived at


Multiplying that by $2$ leads to


This can only be true if $a=b=0$, which is impossible because the orignal problem statement had $a$ and $a+b$ as denominators!

The core of Andrei's answer is that the transformations you used to modify the equation step by step do not always produce equations with the exact same set of solutions. That's already true for the first step, which yieled


It certainly has the solution $a=b=0$, which isn't a solution of the equation you started with.

The reason is that some kinds of transformations, like multiplying both sides of an equation with a number, or squaring both sides of an equation, are not equivalent transformations. That means the transformed equation has all the solutions of the untransformed, but it may have more.

So in order to check if what you found as solution of your many times transformed equation is actually a solution of the original equation, you have to substitute the found solution into the original equation.

In this case, it lead to the result that non of the found solutions is actually a solution of the orignal equations.

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