• June 30th 2008, 11:43 AM
dhju
i have to show how: 1+((4x)/(2x-5))-((15)/((2x^2)-7x+5))
is equal to: (3x+2)/(x-1)
• June 30th 2008, 12:05 PM
dhju
im stupid
dont worry iv solved it (Headbang)
• June 30th 2008, 01:37 PM
masters
Quote:

Originally Posted by Jonboy
I think you mean:

${\color{blue}{1}} + \frac{4x}{2x-5}-\frac{15}{2x^2} - 7x + {\color{blue}{5}}$

Jonboy, you misread the original post. Should've been:

$1 + \frac{4x}{2x-5}-\frac{15}{2x^2 - 7x + 5}$

Then it's easy after that. OP said he already figured it out, so why bother.
• June 30th 2008, 01:43 PM
dhju
im a she, but yeh apart from that ur right. thanks for helpin anyway. Also, how do you write fractions without having to use all those brackets??
• June 30th 2008, 02:24 PM
masters
Quote:

Originally Posted by dhju
im a she, but yeh apart from that ur right. thanks for helpin anyway. Also, how do you write fractions without having to use all those brackets??

Sorry, Ms dhju. Hard to tell gender with these screen names.

For assistance using LaTex, go here http://www.mathhelpforum.com/math-help/latex-help/ and ask any question you want after reading the first thread (Latex Tutorial). You can also click on some else's code in any post and see how it was done.
• June 30th 2008, 04:25 PM
Jonboy
Quote:

...why bother?
for someone in the future who saw this post and wondered how it's done.
• July 1st 2008, 08:03 AM
masters
Quote:

Originally Posted by Jonboy
for someone in the future who saw this post and wondered how it's done.

You're right. I retract that. It's just that you had done so much work on it already.....so for being too abrupt, I will finish it:

$1 + \frac{4x}{2x-5}-\frac{15}{2x^2 - 7x + 5}$

$\frac{(2x-5)(x-1) + 4x(x-1)-15}{(2x-5)(x-1)}$

$\frac{2x^2-7x+5+4x^2-4x-15}{(2x-5)(x-1)}$

$\frac{6x^2-11x-10}{(2x-5)(x-1)}$

$\frac{(3x+2)(2x-5)}{(x-1)(2x-5)}$

$\boxed{\frac{3x+2}{x-1}}$