A little help?

[joey]

Hey everyone,

today my teacher came up with a problem that I do not understand and i really wish to understand why.

the problem is to show that 2/x(x+2) = 1/x - 1/x+2

For some of you, this sound very easy, but for me, I do not know how to tackle this problem... Any help would be great!

 

[lookagain]

Hey everyone,

today my teacher came up with a problem that I do not understand and i really wish to understand why.

the problem is to show that 2/x(x+2) = 1/x - 1/x+2

For some of you, this sound very easy, but for me, I do not know how to tackle this problem... Any help would be great!

You need grouping symbols around the denominators:

2/[x(x+2)] = 1/x - 1/(x+2)

_____________________________________________



Let A and B be constants belonging the real numbers


Then

A/x + B/(x + 2) = 2/[x(x + 2)] ------->


A(x + 2) + Bx = 2 ------>

Ax + 2A + Bx = 2 ------->



Solve this system:


Ax + Bx = 0
2A = 2
--------------



A + B = 0
A = 1
------------


B = -A


A = 1
B = -1



1/x - 1/(x + 2)

 

[stapel]

...show that 2/x(x+2) = 1/x - 1/x+2

...I do not know how to tackle this problem.

The method to use is called "partial fraction decomposition", and was supposed to have been covered back in algebra (and then again in calculus). You can refresh on the topic here.

 

[HallsofIvy]

IF you were given \(\displaystyle \frac{2}{x(x+2)}\) and wanted to separate it into two fractions, you would use "partial fractions".

But if the problem is just to show that the left side is equal to the right side, the simplest thing to do is to get "common denominators" on the right and subtract the fractions:

\(\displaystyle \frac{1}{x}- \frac{1}{x+ 2}= \frac{x+ 2}{x(x+2)}- \frac{x}{x(x+2)}= \frac{x+2- x}{x(x+ 2)}= \frac{2}{x(x+2)}\)