I got this question from a friend of mine, and my solution that I gave to my friend was wrong. He did not explain why it was wrong, he only gave the answer.
The question is:
How many solutions does following problem have: $$\left(a-x\right)\left(b-x\right)=\left(1-ax\right)\left(1-bx\right)$$
My solution (that was wrong) is this;
Step 1 - write the equation
$\left(a-x\right)\left(b-x\right)=\left(1-ax\right)\left(1-bx\right)$
Step 2 - simplify the left side of the equation
$\left(a-x\right)\left(b-x\right)=ab-ax-bx+x^2=ab-x\left(a+b\right)+x^2$
Step 3 - simplify the right side of the equation
$\left(1-ax\right)\left(1-bx\right)=1-bx-ax+abx^2=1-x\left(a+b\right)+abx^2$
Step 4 - simplify the equation even more by putting the right side to the left side
$\left(ab-1\right)+\left(1-ab\right)x^2=0$
Step 5 - divide the equation by (1-ab)
$\frac{\left(ab-1\right)}{\left(1-ab\right)}+x^2=0$
Step 6 - solve for x^2
$x^2=\frac{\left(ab-1\right)}{\left(ab-1\right)}=1\:\Longrightarrow x=\pm 1$
So my answer based on this calculation would be that there are to answers for x. But this is still wrong? Why is this wrong?
Here is the real answer:
\begin{align} Number\:of\:solutions\:can't\:be\:decided. \end{align}
Division $n/d$ as an operation is only defined when $d \neq 0$. So you have two possibilities when you reach step 4. Either $1 - ab = 0$ or $1 - ab \neq 0$. If $1-ab \neq 0$, then the rest of your solution works, so you have 2 solutions. If $1-ab = 0$, then the equation from step 4 becomes $0 = 0$, which is true regardless of the value of $x$; in that case, you get one solution for each real number (so infinitely many)!