Thread: addition/subtraction of algebraic fractions

i have to show how: 1+((4x)/(2x-5))-((15)/((2x^2)-7x+5))
is equal to: (3x+2)/(x-1)
someone please help!!

dont worry iv solved it

Originally Posted by Jonboy

I think you mean:

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

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

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

Then it's easy after that. OP said he already figured it out, so why bother.

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??

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.

...why bother?

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

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:

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

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

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

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

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

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