Simplifying fractions

mcroldan08 · #1 $Old$ November 9th, 2009, 04:41 PM

This is the problem

Simplify the following equation:
(5x/(2x^2+7x+3))-((x-3)/(2x^2-3x-2))+((2x+1)/(x^2+x-6))

The way i did it was factoring the denominators then adding them, therefore getting common terms where the denominator would be (2x+1)(x+3)(x-2). After that, I don't know what to do next, I'm stumped XD

Would appreciate the help..

Soroban · #2 $Old$ November 9th, 2009, 05:07 PM

Hello, mcroldan08!

Quote:

Simplify the following expression:

. . $\frac{5x}{2x^2+7x+3} - \frac{x-3}{2x^2-3x-2} + \frac{2x+1}{x^2+x-6}$

It sounds like your intentions were correct.
But since you didn't show us what you did, we can't tell if you made any errors.

We have: . $\frac{5x}{(x+3)(2x+1)} - \frac{x-3}{(x-2)(2x+1)} + \frac{2x+1}{(x-2)(x+3)}$

You are correct . . . the LCD is: . $(x+3)(2x+1)(x-2)$

And we must convert the fractions to the LCD.

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

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

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

. . $= \;\frac{5x^2 - 10x - x^2 + 9 + 4x^2 + 4x + 1}{(x+3)(2x+1)(x-2)}$

. . $= \;\frac{8x^2 - 6x + 10}{(x+3)(2x+1)(x-2)}$

. . $= \;\frac{2(4x^2 - 3x + 5)}{(x+3)(2x+1)(x-2)}$

mcroldan08 · #3 $Old$ November 9th, 2009, 06:34 PM

That is actually what I got... however, I was wondering if this can be simplified further. This is actually a multiple choice question and the choices are:

a.) 4/(x+3)
b.) 4/(x-3)
c.) 2/(x-3)
d.) 2/(x+3)

Could it be that the none of the choices are correct?