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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!

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    Thanks Theo, much appreciated!2011-05-03

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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 $y$.