So we have
[tex] \dfrac{1}{7}-3( \dfrac{3}{7n}- \dfrac{2}{7})[/tex]
The first step is to distribute the –3
[tex]\dfrac{1}{7}+( \dfrac{-3*3}{7n}- \dfrac{-3*2}{7})[/tex]
Next, we simplify within the parentheses. In order to do this, we must have a common denominator.
[tex]\dfrac{1}{7}+( \dfrac{-9}{7n}- \dfrac{-6}{7}* \dfrac{n}{n} ) \\ \\ \\ \\ \dfrac{1}{7}+( \dfrac{-9}{7n}- \dfrac{-6n}{7n} ) \\ \\ \\ \\ \dfrac{1}{7}+ \dfrac{-9+6n}{7n}[/tex]
Then, we find a common denominator again and simplify.
[tex]\dfrac{1}{7}* \dfrac{n}{n} + \dfrac{-9+6n}{7n} \\ \\ \\ \\ \dfrac{n-9+6n}{7n} \\ \\ \\ \\ \dfrac{7n-9}{7n} [/tex]
And we're done. While it may be tempting to try to cancel 7n, this is not actually allowed. Try substituting in a non-zero number for n and you'll see why.
