$\displaystyle A(B-1)=C$ solve for B The book states the answer is $\displaystyle \frac{C}{A}+1$ but does not show the work involved to solve it.
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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
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.
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 ?
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 $\displaystyle A\not=0$, $\displaystyle A(B-1)=C \Leftrightarrow AB-A=C \Leftrightarrow AB = C + A \Leftrightarrow B=\frac{C}{A}+\frac{A}{A}=\frac{C}{A}+1$
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