Expression simplification

Hi

I have these two expressions

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

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

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

Thanks

Re: Expression simplification

Quote:

Originally Posted by

**bobred** Hi

I have these two expressions

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

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

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

Thanks

You can't. Think of it this way: A + B = D is B = D - A

Re: Expression simplification

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

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

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

Re: Expression simplification

Quote:

Originally Posted by

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

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

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.

Re: Expression simplification

Hi

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

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

The expressions come from boundary values.

Re: Expression simplification

Oh sorry I missread your post.

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

You get

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

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

Factorize out A

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

Get a common denominator and simplify.

Re: Expression simplification

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