# Thread: How to rearrange to make D the subject?

1. ## How to rearrange to make D the subject?

Hi,

For some reason i'm struggling to rearrange a lot of equations where i'm attempting to make a variable that appears twice in the equation the subject. At the moment i am attempting to rearrange this equation to make D the subject. May someone please show me how to do this in steps please?

$A=B*D*(1/(C+D))$

And if possible, i would like some tips on how to make variables that appear twice the subject please?

Any help is much appreciated
Rob

2. ## Re: How to rearrange to make D the subject?

You have $A=\frac{BD}{C+D}$

A good first step would be to multiply by the denominator to get

$A(C+D) = BD$

Expand the brackets

$AC+AD=BD$

Rearrange a little

$AC=BD-AD$

factorise, and I'll leave the rest to you.

3. ## Re: How to rearrange to make D the subject?

Hello, Rob!

$\text{Solve for }D\!:\;\;A \:=\:\frac{BD}{C+D}$

Since there is a $D$ in the denominator, multiply by $(C+D)$

. . $A(C+D) \:=\:BD \quad\Rightarrow\quad AC + AD \:=\:BD$

Get all $D$-terms on one side, the rest on the other side.

. . $AD - BD \:=\:-AC$

Factor out the $D.$

, . $(A-B)D\:=\:-AC$

Solve for $D$ . . . divide by $(A-B)$

. . $D \;=\;\frac{-AC}{A-B}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

. . but you may see it in a different form.

Multiply the fraction by $\frac{\text{-}1}{\text{-}1}$

. . $D \;=\;\frac{\text{-}1}{\text{-}1}\cdot\frac{-AC}{A-B} \;=\;\frac{AC}{B-A}$

4. ## Re: How to rearrange to make D the subject?

Originally Posted by a tutor
$AC=BD-AD$

factorise, and I'll leave the rest to you.
I managed to get it up to this point but for some reason i just couldn't think about how to do the next step. It was just so simple. I kind of kick myself for not being able to figure it out. lol

Much appreciated to you both.