# Help LCD for variable fractions

• August 15th 2008, 12:41 PM
cmf0106
Help LCD for variable fractions
Im having problems figuring out how to find the LCD for two problems in particular. If someone would kindly show me what the LCD for each problem is and walk me through the process to get that LCD it would be much appreciated.

1. $2+\frac{x-1}{(x+4)^2}$
&
2. $\frac{6}{2(x-1)(x+1)}+\frac{1}{6(x-1)^2}$

Many thanks

Also here's a problem of similar sort to show you how I normally solve these types of problems if it helps.

$\frac{13}{xy^2}-\frac{6}{yz}
$

$
\frac{xy^2z(\frac{13}{xy^2})-xy^2z(\frac{6}{yz})}{xy^2z}
$

$\frac{13z-6xy}{xy^2z}
$

• August 15th 2008, 02:54 PM
randir
The first problem is easy, once you notice that $2=\frac{2}{1}$
So,
$\frac{2}{1}+\frac{x-1}{(x+4)^2}$
Then multiply the top and bottom of the left hand fraction by (x+4)^2 and the denominators will be the same (1*right hand denominator obviously = right hand denominator, and then balance by multiplying on top)
$\frac{2(x+4)^2}{(x+4)^2}+\frac{x-1}{(x+4)^2} = \frac{2(x+4)^2 + x-1}{(x+4)^2}$

::edit::

I should have started with this, really. Your method of finding LCD differs from mine, but i didn't notice straight away. Mine works without fail (i think) and is a little simpler (or perhaps just seems that way because im used to it). Here it is:
$\frac{A}{B} + \frac{C}{D}$
"Cross multiply". That is, multiply the right fraction by the left denominator, and the left fraction by the right denominator:
$\frac{AD}{BD} + \frac{CB}{DB}$

Eg $\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$

Try this for your second problem. The answer is below, i don't know how to change the colour so that you cant see it until its highlighted. Don't look until you give it a go.

$\frac{6}{2(x-1)(x+1)}+\frac{1}{6(x-1)^2}$
$\frac{36(x-1)^2}{12(x-1)^3(x+1)} + \frac{2(x+1)(x-1)}{12(x-1)^3(x+1)}$