# Basic algebraic equation question

Ok thats very basic but dont get it , I have the following equation $R-A X-B \dfrac{A X}{2 B}=0$ the right result of it is $R- \dfrac{3 A X}{2}=0$

well, now when I attempt to solve it I do the following: $A X+\dfrac{A X}{B}=R$, so what I have done here is cancel the $B$ that is multiplying with one of the $B$'s which are dividing, I don't see why that is wrong, but I guess that's where I make the mistake.. Can anyone tell me why please? Cheeers!

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"one of the B's"? There's more than one B dividing? –  User58220 Apr 3 '13 at 0:17
yeah well is divided by 2*B , so B*B, 2 B's. cheers –  Maximilian1988 Apr 3 '13 at 0:18
2*B is not the same as B*B. –  Code-Guru Apr 3 '13 at 0:29
goooosh...embarrassed... :$! thaanks – Maximilian1988 Apr 3 '13 at 0:19 I have done things equally embarrassing more times than I can count. – bob.sacamento Apr 3 '13 at 0:21 add comment You have: $$R - A X - B \cdot \frac{A X}{2 B} = R - A X - \frac{A B X}{2 B} = R - A X - \frac{A X}{2} = R - A X \left(1 + \frac{1}{2} \right) = R - \frac{3 A X}{2}$$ - how did you edit it so it looks that good? – Maximilian1988 Apr 3 '13 at 0:22 Look at the help for formatting, in particular here. You should be able to ask to edit your own post and look at how I wrote it. – vonbrand Apr 3 '13 at 0:24 ok sweet! Ill have a look at it – Maximilian1988 Apr 3 '13 at 0:27 Stupid braino. Thanks, fixed. – vonbrand Apr 3 '13 at 0:31 BTW, your final$X = \dfrac{2 R}{3 A}\$. –  vonbrand Apr 3 '13 at 0:32