Simplifying Complex Fractions

[kinseystark]

I have a homework page due tomorrow and I don't know what I'm doing.
The problem is:
((4/x^2-9)+(2/x-3))/((1/x+3)+(1/x+3))

Thank you! I know you have to find a common denominator, but I'm not sure how or i just don't know.

 

[Jason76]

Your bracketing is faulty; should be:
[4 / (x^2 - 9) + 2 / (x - 3)] / [1 / (x + 3) + 1 / (x + 3)]

Do you know how to factor x^2 - 9 ?
If not, you need classroom help...

\(\displaystyle \frac{4}{(x^{2}- 9)} + \frac{2}{(x - 3)} \)

Can't make out Latex so division bar goes here

\(\displaystyle \frac{1}{(x + 3) + 1} + \frac{1}{(x + 3)}\)

Ok, my book says:

1. Add or subtract the fractions in the numerator and denominator seperately (the seperate fractions are known as secondary fractions)

2. Invert and multiply the denominator of the complex fraction by the numerator of the complex fraction.

Another method:

1. Find the LCD of the two secondary fractions

2. Find the LCD of the complex fraction

3. Multiply both secondary fractions by the LCD of the complex fraction.

The LCD of the numerator is \(\displaystyle (x^{2} - 9)(x - 3)\)

The LCD of the denominator is \(\displaystyle (x + 3)(1)\)

Am I on the right track here?

 

[Deleted member 4993]

Numerator:

\(\displaystyle \frac{4}{(x^{2}- 9)} + \frac{2}{(x - 3)} \)

\(\displaystyle \ = \ \frac{4}{(x+3)(x-3)} + \frac{2}{(x - 3)} \)

\(\displaystyle \ = \ \frac{4}{(x+3)(x-3)} + \frac{2}{(x - 3)} \)

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

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

Can't make out Latex so division bar goes here:

Denominator:

\(\displaystyle \frac{1}{(x + 3) + 1} + \frac{1}{(x + 3)}\)

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

and continue...

Ok, my book says:

1. Add or subtract the fractions in the numerator and denominator seperately (the seperate fractions are known as secondary fractions)

2. Invert and multiply the denominator of the complex fraction by the numerator of the complex fraction.

Another method:

1. Find the LCD of the two secondary fractions

2. Find the LCD of the complex fraction

3. Multiply both secondary fractions by the LCD of the complex fraction.

.

 

[Jason76]

I don't understand the 4th red line from the top, in regards to the numerator.

\(\displaystyle \frac{2(x + 5)}{(x + 3)(x - 3)}\)

 

[srmichael]

I don't understand the 4th red line from the top, in regards to the numerator.

\(\displaystyle \frac{2(x + 5)}{(x + 3)(x - 3)}\)

\(\displaystyle 4+2(x+3)=4+2x+6=2x+10=2(x+5)\)

Make sense?

 

[Jason76]

\(\displaystyle 4+2(x+3)=4+2x+6=2x+10=2(x+5)\)

Make sense?

Right, I see the factoring makes the difference. However, obviously, it's not recognizable at first sight.

Also, how do you make Latex numbers so big?

 

[soroban]

Hello, kinseystark!

\(\displaystyle \text{Simplify: }\:\dfrac{\dfrac{4}{x^2-9} + \dfrac{2}{x-3}}{\dfrac{1}{x+3} + \dfrac{1}{x+3}}\)

Your LCD is too large.

Factor the denominators (if possible).

The denominators are: .\(\displaystyle (x-3)(x+3),\;(x-3),\; (x+3),\; (x+3)\)

The LCD is \(\displaystyle (x-3)(x+3).\)


Multiply top and bottom by the LCD:

. . \(\displaystyle \displaystyle \frac{(x-3)(x+3)\left(\dfrac{4}{(x-3)(x+3)} + \dfrac{2}{x-3}\right)} {(x-3)(x+3)\left(\dfrac{1}{x+3} + \dfrac{1}{x+3}\right)} \;=\;\frac{4 + 2(x+3)}{1(x-3) + 1(x-3)} \)


. . . \(\displaystyle =\;\dfrac{4+2x+6}{x-3+x-3} \;=\;\dfrac{2x+10}{2x-6} \;=\;\dfrac{2(x+5)}{2(x-3)} \;=\;\dfrac{x+5}{x-3} \)

 

[Jason76]

Hello, kinseystark!


Your LCD is too large.

Factor the denominators (if possible).

The denominators are: .\(\displaystyle (x-3)(x+3),\;(x-3),\; (x+3),\; (x+3)\)

The LCD is \(\displaystyle (x-3)(x+3).\)


Multiply top and bottom by the LCD:

. . \(\displaystyle \displaystyle \frac{(x-3)(x+3)\left(\dfrac{4}{(x-3)(x+3)} + \dfrac{2}{x-3}\right)} {(x-3)(x+3)\left(\dfrac{1}{x+3} + \dfrac{1}{x+3}\right)} \;=\;\frac{4 + 2(x+3)}{1(x-3) + 1(x-3)} \)


. . . \(\displaystyle =\;\dfrac{4+2x+6}{x-3+x-3} \;=\;\dfrac{2x+10}{2x-6} \;=\;\dfrac{2(x+5)}{2(x-3)} \;=\;\dfrac{x+5}{x-3} \)

Of course, you can also solve this problem by :

1. Adding the secondary fractions in the numerator

2. Adding the secondary fractions in the denominator

3. Multiplying the total numerator by the reciprocal of the total denominator

And the strategies for subtracting complex fractions are the same.