# Solving Expression

• Apr 4th 2013, 01:55 PM
cmf0106
Solving Expression
$A(B-1)=C$ solve for B

The book states the answer is $\frac{C}{A}+1$ but does not show the work involved to solve it.
• Apr 4th 2013, 06:23 PM
Mrdavid445
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
• Apr 4th 2013, 07:10 PM
Gusbob
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.
• Apr 4th 2013, 08:50 PM
cmf0106
Re: Solving Expression
Quote:

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 ?
• Apr 4th 2013, 09:07 PM
Gusbob
Re: Solving Expression
Quote:

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 $A\not=0$,

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