Adding fractions in algebra

**Grich** · December 17th 2008, 03:01 AM

Okay, I have the expression:
$ \frac{6}{2(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2} $
now I need help to understand how these fractions are added together.
I have the solution to it, but upon analysing it I am a bit confused:
$ \frac{6}{2(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2} $
$ = \frac{6}{2(x - 1)(x + 1)} \times \frac{3(x - 1)}{3(x - 1)} + \frac{1}{6(x - 1)^2} \times \frac{x + 1}{x + 1} \\ $
$ = \frac{18(x - 1)}{6(x - 1)^2(x + 1)} + \frac{x + 1}{6(x - 1)^2(x + 1)} $
$ = \frac{18(x - 1) + x + 1}{6(x - 1)^2(x + 1)} $

Why are we multiplying $\frac{6}{2(x - 1)(x + 1)}$ by $\frac{3(x - 1)}{3(x - 1)}$ and how did we get the conclusion to use $3(x - 1)$ in the first place.
Same with $\frac{1}{6(x - 1)^2}$ and $x + 1$ .

I get the rest of it but those two parts of the solution don't make sense to me!?

Are they using an Least Common Denominator (LCD) or something?

Thanks in advance.

**masters** · December 17th 2008, 10:09 AM

Originally Posted by Grich

Okay, I have the expression:
$ \frac{6}{2(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2} $
now I need help to understand how these fractions are added together.
I have the solution to it, but upon analysing it I am a bit confused:
$ \frac{6}{2(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2} $
$ = \frac{6}{2(x - 1)(x + 1)} \times \frac{3(x - 1)}{3(x - 1)} + \frac{1}{6(x - 1)^2} \times \frac{x + 1}{x + 1} \\ $
$ = \frac{18(x - 1)}{6(x - 1)^2(x + 1)} + \frac{x + 1}{6(x - 1)^2(x + 1)} $
$ = \frac{18(x - 1) + x + 1}{6(x - 1)^2(x + 1)} $

Why are we multiplying $\frac{6}{2(x - 1)(x + 1)}$ by $\frac{3(x - 1)}{3(x - 1)}$ and how did we get the conclusion to use $3(x - 1)$ in the first place.
Same with $\frac{1}{6(x - 1)^2}$ and $x + 1$ .

I get the rest of it but those two parts of the solution don't make sense to me!?

Are they using an Least Common Denominator (LCD) or something?

Thanks in advance.

Hello Grich,

I wouldn't do it that way. Let me show you my approach and see if you can follow it.

$\frac{6}{2(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2}$

First, simplify the first term of your sum by dividing 2 into 6.

$\frac{3}{(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2}$

Find the LCD: $6(x-1)^2(x+1)$

Now multiply each term of your sum by the LCD and make appropriate cancellations (divisions)

$6(x-1)^2(x+1)\left(\frac{3}{(x-1)(x+1)}\right)+6(x-1)^2(x+1)\left(\frac{1}{6(x-1)^2}\right)$

Now, put the simplified numerator over your LCD and finish up

$\frac{18(x-1)+x+1}{6(x-1)^2(x+1)}$

**Grich** · December 17th 2008, 03:36 PM

Thanks masters, that was great help. I knew it was something to do with the LCD!
Here is how I solved the solution:
Find the LCD, which is:
$6(x - 1)^2(x + 1)$
The working out:
$= \left(\frac{6}{2(x - 1)(x + 1)} \times \frac{3(x - 1)}{3(x - 1)}\right) + \left(\frac{1}{6(x - 1)^2} \times \frac{(x + 1)}{(x + 1)}\right)$
$ = \frac{18(x - 1)}{6(x - 1)^2(x + 1)} + \frac{(x + 1)}{6(x - 1)^2(x + 1)} $
Therefore it will equal:
$ \frac{18(x - 1) + (x + 1)}{6(x - 1)^2(x + 1)} $
I have to remember to simplify the prackets though!
$ \frac{18(x - 1) + x + 1}{6(x - 1)^2(x + 1)} $

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