# Thread: You have to take a look

1. ## You have to take a look

Well...here is the original problem:

$\displaystyle 2+\frac{3}{2x+1}$

...and my aproach:

$\displaystyle 2+\frac{3}{2x+1}$

=$\displaystyle \frac{2}{1}+\frac{3}{2x+1}$

Now the lowest common factor would be $\displaystyle 2x+1$ and we rewrite the expression as:

=$\displaystyle \frac{2(2x+1)}{1(2x+1)}+\frac{3}{2x+1}$

Which is the same as:

=$\displaystyle \frac{4x+2+3}{2x+1}$

And finally I have:

=$\displaystyle \frac{4x+5}{2x+1}$

Now my questions is Can this expression be further simplified? As I see we could still make 4x and 2x compensate to obtain

=$\displaystyle \frac{2x+5}{1}$

=$\displaystyle 2x+5$

but the book I'm working with states the solution is just

$\displaystyle \frac{4x+5}{2x+1}$

Could it be that simplification works when numerator and denominator imply multiplication only? Mmm...

2. Originally Posted by Alienis Back
Well...here is the original problem:

$\displaystyle 2+\frac{3}{2x+1}$

...and my aproach:

$\displaystyle 2+\frac{3}{2x+1}$

=$\displaystyle \frac{2}{1}+\frac{3}{2x+1}$

Now the lowest common factor would be $\displaystyle 2x+1$ and we rewrite the expression as:

=$\displaystyle \frac{2(2x+1)}{1(2x+1)}+\frac{3}{2x+1}$

Which is the same as:

=$\displaystyle \frac{4x+2+3}{2x+1}$

And finally I have:

=$\displaystyle \frac{4x+5}{2x+1}$

Now my questions is Can this expression be further simplified? As I see we could still make 4x and 2x compensate to obtain

=$\displaystyle \frac{2x+5}{1}$

=$\displaystyle 2x+5$

but the book I'm working with states the solution is just

$\displaystyle \frac{4x+5}{2x+1}$

Could it be that simplification works when numerator and denominator imply multiplication only? Mmm...
bingo!

3. It cannot be simplified after you reach $\displaystyle \frac {4x+5}{2x+1}$ because they're followed by an addition and not by a multiplication.