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Hi I have a simple algebra question which i can work out like the following $$\begin{align} x - 8 & = x/3 + 1/6 \\ 6(x - 8) & = 6(x/3 + 1/6) \\ 6x - 48 & = 6x/3 + 6/6 \\ 6x - 48 & = 2x + 1 \\ 6x & = 2x + 49 \\ 4x & = 49 \\ x & = 49/4 \end{align}$$

However, if I were to first remove the $8$ from the right hand side: $$\begin{align} x - 8 & = x/3 + 1/6 \\ x & = x/3 + 49/6 \\ 3x & = x + 49/6 \\ 2x & = 49/6 \\ \end{align}$$ How do I go from $2x = 49/6$ to $x = 49/4$?

I'm pretty confused so any guidance would be appreciated.

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up vote 5 down vote accepted

A correction is needed in the third step of your second argument: $3x=x +\color{red}3\cdot 49/6$.

That is from the second step: $$ x={x\over3}+{49\over 6} $$ you multiply both sides by 3 (and on the right, the "entire side"): $$ 3\cdot x=3\cdot \bigl( {x\over3}+{49\over 6}\bigr) $$to obtain $$ 3x= 3\cdot{x\over3} +3\cdot{49\over 6} , $$ or $$ 3x= x + {49\over 2} ; $$ which will give you the same solution as before, $x=49/4$.

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Why is that? Could you please show me the correct steps involved. –  user346443 Apr 24 '12 at 2:28
    
Awesome, thanks alot. –  user346443 Apr 24 '12 at 3:07
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