# Addition and substraction of algebraic fractions -urgent questions

• Sep 19th 2010, 11:18 PM
PythagorasNeophyte
Addition and substraction of algebraic fractions -urgent questions
A very good day to everyone here on Math Help Forum!
I have a few questions related to algebraic fractions. They are as the following:

1. $\displaystyle \dfrac{x^2}{y}$ $\displaystyle + \dfrac{y}{x^2}$

2. $\displaystyle \dfrac{x}{x+16} $$\displaystyle +$$\displaystyle \dfrac{5}{x^2 +x-12}$

3. $\displaystyle \dfrac{5}{2x^2 +5x +3}$$\displaystyle - \dfrac{1}{2x^2 +3x -2} My workings are shown below: 1. \displaystyle \dfrac{x^2}{y} \displaystyle + \dfrac{y}{x^2} \displaystyle = \dfrac{x^2y+x^2y}{x^2y} \displaystyle =\dfrac{2x^2y}{x^2y} \displaystyle =2 ? 2. \displaystyle \dfrac{x}{x+16}$$\displaystyle +$$\displaystyle \dfrac{5}{x^2 +x -12} \displaystyle =\dfrac{x}{x^2 -4^2} \displaystyle + \dfrac{5}{(x+4)(x-3)} \displaystyle =\dfrac{x}{(x-4)(x+4)}$$\displaystyle + \dfrac{5}{(x+4)(x-3)}$

$\displaystyle =\dfrac{x(x-3)+5(x-4)}{(x-4)(x+4)(x-3)}$

$\displaystyle =\dfrac{x^2+2x-20}{(x-4)(x+4)(x-3)}$

$\displaystyle =\dfrac{(x+6)(x-4)}{(x-4)(x+4)(x-3)}$

$\displaystyle = \dfrac{x+6}{(x+4)(x-3)}$ ?

3.$\displaystyle \dfrac{5}{2x^2 +5x +3}$$\displaystyle - \dfrac{1}{2x^2 +3x -2}$

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

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

This is where I am stuck.

If you happened to notice any mistakes and blunders, please pinpoint them and explain how it should be done.
Thank you so much for taking your time!
• Sep 20th 2010, 12:14 AM
Prove It
Quote:

Originally Posted by PythagorasNeophyte
A very good day to everyone here on Math Help Forum!
I have a few questions related to algebraic fractions. They are as the following:

1. $\displaystyle \dfrac{x^2}{y}$ $\displaystyle + \dfrac{y}{x^2}$

Incorrect. To get a common denominator, multiply the top and bottom of each fraction by the same thing (i.e. a cleverly disguised 1)...

$\displaystyle \frac{x^2}{y} + \frac{y}{x^2} = \frac{x^2}{y}\left(\frac{x^2}{x^2}\right) + \frac{y}{x^2}\left(\frac{y}{y}\right)$

$\displaystyle = \frac{x^4}{x^2y} + \frac{y^2}{x^2y}$

$\displaystyle = \frac{x^4 + y^2}{x^2y}$.
• Sep 20th 2010, 02:01 AM
Educated
2. $\displaystyle \dfrac{x}{x+16} + \dfrac{5}{x^2 +x -12} = \dfrac{x}{x+16} + \dfrac{5}{(x+4)(x-3)}$

You will then need to find a cross multiply to get a common denominator.

Note: $\displaystyle \dfrac{x}{x+16} \ne \dfrac{x}{x^2 - 4^2}$

3. Expand the numerator, gather up like terms and then factorise again.
• Sep 20th 2010, 02:19 AM
PythagorasNeophyte
Hello Educated,

You said that $\displaystyle \dfrac{x}{x+16} + \dfrac{5}{x^2 +x -12} = \dfrac{x}{x+16} + \dfrac{5}{(x+4)(x-3)}$ needs to be cross multiplied. But it isn't an equation. It's about the addition of algebriac fractions.

I wonder if $\displaystyle x+16$ can be factorised into $\displaystyle x^2+4^2$ then into $\displaystyle (x+4)(x+4)$. If it's possible, then the LCM would be $\displaystyle (x+4)(x+4)(x-3)$.

Otherwise, the LCM would be $\displaystyle (x+16)(x+4)(x-3)$.

Which of these answers could be correct?
• Sep 20th 2010, 02:32 AM
Educated
LCM is $\displaystyle (x+16)(x+4)(x-3)$

No it is not an equation, but you can still cross multiply the equation to get a common denominator to simplify it:

$\displaystyle \dfrac{x(x+4)(x-3)}{(x+16)(x+4)(x-3)} + \dfrac{5(x+16)}{(x+16)(x+4)(x-3)}$

I was just trying to state that x + 16 doesn't equal x^2 + 4^2. You cannot factorise x + 16.