1. ## Simplifying Algebraic expressions & Adding fractions with unlike denominators

I have two questions, both of which I am stuck on:

Problem #1:

Instructions state, 'Simplify the expression:'

y - x / (x^2 + xy - 2y^2)

Problem #2:

Instructions state, 'Perform the indicated operation and simplify if possible:'

3 / (x^2 - 9) + 2 / (x - 3)

I've determined that x^2 - 9 can be factored to (x + 3) (x - 3) but I don't know where to take it from here. Do I have to find an LCD to solve this?

2. ## Re: Simplifying Algebraic expressions & Adding fractions with unlike denominators

Hello, Odonsky!

$\text{1. Simplify: }\:\frac{y - x}{x^2 + xy - 2y^2}$

$\text{Factor: }\:\frac{-(x-y)}{(x-y)(x+2y)}$

$\text{Reduce: }\:\frac{-({\color{red}\rlap{/////}}x-y)}{({\color{red}\rlap{/////}}x-y)(x+2y)} \;=\;\frac{\text{-}1}{x+2y}$

$\text{2. Simplify: }\:\frac{3}{x^2 - 9} + \frac{2}{x - 3}$

$\text{Factor: }\:\frac{3}{(x-3)(x+3)} + \frac{2}{x-3} \;=\;\frac{3}{(x-3)(x+3)} + \frac{2}{x-3}\cdot\frac{x+3}{x+3}$

. . . . . $=\;\frac{3}{(x-3)(x+3)} + \frac{2(x+3)}{(x-3)(x+3)} \;=\;\frac{3 + 2(x+3)}{(x-3)(x+3)}$

. . . . . $=\;\frac{3+2x+6}{(x-3)(x+3)} \;=\;\frac{2x+9}{(x-3)(x+3)}$

3. ## Re: Simplifying Algebraic expressions & Adding fractions with unlike denominators

You are right find LCD of denominator which is ( x+3)(x-3) now we combine the fractions and get [ 3 + 2 (x+3)]/(x+3)(x-3) = (2x +9)/(x^2-9)

4. ## Re: Simplifying Algebraic expressions & Adding fractions with unlike denominators

Originally Posted by Soroban
Hello, Odonsky!

$\text{Factor: }\:\frac{-(x-y)}{(x-y)(x+2y)}$

$\text{Reduce: }\:\frac{-({\color{red}\rlap{/////}}x-y)}{({\color{red}\rlap{/////}}x-y)(x+2y)} \;=\;\frac{\text{-}1}{x+2y}$
Thank you for the detailed response!

However, I do not quite understand why you factored a negative 1 into the numerator. Would you please care to explain?

Thank you.

5. ## Re: Simplifying Algebraic expressions & Adding fractions with unlike denominators

Odonsky has correctly done what he has. he has just taken out (-1) common so that he gets the same factor in numerator i.e., ( x-y) which is also there in the denominator for cancellation.