1. ## Solving Expression

$\displaystyle A(B-1)=C$ solve for B

The book states the answer is $\displaystyle \frac{C}{A}+1$ but does not show the work involved to solve it.

2. ## Re: Solving Expression

If you start by dividing both sides by A, you get (B-1)=C/A

Therefore by adding 1 to each side, B=(C/A)+1

3. ## Re: Solving Expression

It should be noted that division by A needs to be defined for this answer to be correct! This particular point needs to be emphasised, especially for beginners to elementary algebra.

4. ## Re: Solving Expression

Originally Posted by Mrdavid445
If you start by dividing both sides by A, you get (B-1)=C/A

Therefore by adding 1 to each side, B=(C/A)+1
The problem that trips me up on this particular problem is I try to distribute the A over the parenthesis first. Why does that not work ?

5. ## Re: Solving Expression

Originally Posted by cmf0106
The problem that trips me up on this particular problem is I try to distribute the A over the parenthesis first. Why does that not work ?
It does. Assuming $\displaystyle A\not=0$,

$\displaystyle A(B-1)=C \Leftrightarrow AB-A=C \Leftrightarrow AB = C + A \Leftrightarrow B=\frac{C}{A}+\frac{A}{A}=\frac{C}{A}+1$