# Thread: Addition and substraction of algebraic fractions -urgent questions

1. ## 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!

2. 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)...

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

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

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

3. 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.

4. 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?

5. 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.