I'm having a problem with my solution to a textbook exercise:
Find the Cartesian equation of the curve given by this parametric equation: $$x = \frac{t}{2t-1}, y = \frac{t}{t+1}$$
The textbook's answer is $y = \frac{x}{3x-1}$
My solution is different. I understand how the textbook got to its solution, but I can't find where I made my mistake. Can anyone spot my error below?
$$x = \frac{t}{2t-1} = \frac{t}{2t} - \frac{t}{1}$$ $$\implies x = \frac{1}{2} - t$$ $$\implies x - \frac{1}{2} = -t$$ $$\implies t = -x + \frac{1}{2}$$
Sub this into $y = \frac{t}{t+1} \implies y = \frac{-x + \frac{1}{2}}{-x + \frac{1}{2} + 1}$ $$= \frac{-x + \frac{1}{2}}{-x + \frac{1}{2}} + \frac{-x + \frac{1}{2}}{1}$$ $$= 1 - x + \frac{1}{2}$$ $$= -x + \frac{3}{2}$$
So $y = -x + \frac{3}{2}$
I suspect my error is when I split my fractions up, but if so, why can't I do it like that?
Many thanks!
As you said Danny,
$$\frac{a}{b+c} \neq \frac{a}{b} + \frac{a}{c}$$
and Theo pointed out with a simple example why:
$$0.5 = \frac{1}{2} = \frac{1}{1+1} \neq \frac{1}{1} + \frac{1}{1} = 2 \; .$$
It is a common mistake however, so tempting that few people have resisted making it.
P.S: Note you made the mistake twice, once in the formula with $x$ and once in the one with $y$.