# Expression simplification

• September 26th 2013, 07:21 AM
bobred
Expression simplification
Hi

I have these two expressions

$A+B=D$ and $D=\frac{\text{i}2kA}{\text{i}k-\alpha}$

I need to find $B$, so I substitute $D$ into the first expression. The problem is I don't know how to get it into the form below

$B=A\frac{(\text{i}k+\alpha)}{(\text{i}k-\alpha)}$

Thanks
• September 26th 2013, 08:05 AM
votan
Re: Expression simplification
Quote:

Originally Posted by bobred
Hi

I have these two expressions

$A+B=D$ and $D=\frac{\text{i}2kA}{\text{i}k-\alpha}$

I need to find $B$, so I substitute $D$ into the first expression. The problem is I don't know how to get it into the form below

$B=A\frac{(\text{i}k+\alpha)}{(\text{i}k-\alpha)}$

Thanks

You can't. Think of it this way: A + B = D is B = D - A
• September 26th 2013, 08:57 AM
Shakarri
Re: Expression simplification
$B=A\frac{(\text{i}k+\alpha)}{(\text{i}k-\alpha)}$

From this you get $B(ik-\alpha)=A(ik+\alpha)$

The real numbers on the left have to equal the real numbers on the right and the same goes for imaginary numbers
• September 26th 2013, 09:03 AM
votan
Re: Expression simplification
Quote:

Originally Posted by Shakarri
$B=A\frac{(\text{i}k+\alpha)}{(\text{i}k-\alpha)}$

From this you get $B(ik-\alpha)=A(ik+\alpha)$

The real numbers on the left have to equal the real numbers on the right and the same goes for imaginary numbers

The poster does not say imaginary numbers. When he specifies what his question is about we will be able to handle this properly.
• September 26th 2013, 09:27 AM
bobred
Re: Expression simplification
Hi

I find B by hand easy enough, but Maple gives the expression below and I don't know how

$B=A\frac{(\text{i}k+\alpha)}{(\text{i}k-\alpha)}$

The expressions come from boundary values.
• September 26th 2013, 11:49 AM
Shakarri
Re: Expression simplification

With $A+B=D$ and $D=\frac{\text{i}2kA}{\text{i}k-\alpha}$

You get
$A+B=\frac{\text{i}2kA}{\text{i}k-\alpha}$

$B=\frac{\text{i}2kA}{\text{i}k-\alpha}-A$

Factorize out A

$B=A\times(\frac{\text{i}2k}{\text{i}k-\alpha}-1)$

Get a common denominator and simplify.
• September 26th 2013, 01:01 PM
bobred
Re: Expression simplification
Hi, thanks I have got to the point above its just the next step i'm having trouble with.