I guess this is some basic question, but I cannot see what I'm doing wrong. This is the equation I have:
$$\frac{1}{ax+1}+\frac{1}{bx+1}-c=0$$
and I want to convert it to quadratic equation like so:
$$\frac{(bx+1)+(ax+1)}{(ax+1)(bx+1)}-c=0$$ $$\frac{bx+ax+2}{abx^2+ax+bx+1}-c=0$$ then we multiply both sides with $abx^2+ax+bx+1$ and retain restriction that $ax\neq-1, bx\neq-1$ $$bx+ax+2-c(abx^2+ax+bx+1)=0; ax\neq-1, bx\neq-1$$ $$bx+ax+2-cabx^2-cax-cbx-c=0; ax\neq-1, bx\neq-1$$ $$x^2(-cab)+x(b+a-ca-cb)-c+2=0; ax\neq-1, bx\neq-1$$
Now, if I draw 2 graphs for these 2 functions by $x$ it seems it is different. How is this the case?
Notice that you get a quadratic equation, so there are two roots. One of them is precisely the point you're avoiding (eg, ax = -1)