# Thread: Expression simplification

1. ## 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

2. ## Re: Expression simplification

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

3. ## 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

4. ## Re: Expression simplification

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.

5. ## 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.

6. ## 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.

7. ## Re: Expression simplification

Hi, thanks I have got to the point above its just the next step i'm having trouble with.