# Thread: I need help beautifying this big ugly expression

1. ## I need help beautifying this big ugly expression

I'm stuck on how this question can go on:

"The expression:
$\displaystyle \frac{\frac{1}{2x-1}+\frac{1}{y}}{\frac{x}{2xy-y}}$
is equivalent to..."
I can flip the denominator of the whole fraction to multiply and factorize the y out but I can't seem to figure out a way to get past this:

$\displaystyle \frac{2x^2y-xy+4x^3-2x^3-2x-2x^2+1}{(2x^2-1)(x)}$

The answer to the question comes out as $\displaystyle \frac{y+2x-1}{x}$

So I'm not sure how the first equation comes out as that ^

2. Originally Posted by Kataangel I'm stuck on how this question can go on:

"The expression:
$\displaystyle \frac{\frac{1}{2x-1}+\frac{1}{y}}{\frac{x}{2xy-y}}$
is equivalent to..."
I can flip the denominator of the whole fraction to multiply and factorize the y out but I can't seem to figure out a way to get past this:

$\displaystyle \frac{2x^2y-xy+4x^3-2x^3-2x-2x^2+1}{(2x^2-1)(x)}$

The answer to the question comes out as $\displaystyle \frac{y+2x-1}{x}$

So I'm not sure how the first equation comes out as that ^
Instead of flipping, multiplying, and doing all that other rather tedious stuff, we can do it much more easily:

$\displaystyle \frac{\frac{1}{2x-1}+\frac{1}{y}}{\frac{x}{2xy-y}}$

Let's get a common denominator in our numerator (god that sounds confusing):

$\displaystyle \frac{\frac{y}{2xy-y}+\frac{2x-1}{2xy-y}}{\frac{x}{2xy-y}}$

$\displaystyle \frac{\frac{y+2x-1}{2xy-y}}{\frac{x}{2xy-y}}$

$\displaystyle {\frac{y+2x-1}{2xy-y}} \cdot {\frac{2xy-y}{x}}$

$\displaystyle {\frac{y+2x-1}{x}}$

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