so... let us call the number... say "a"
twice "a" is 2 * a or 2a
the product of that and 3
2a * 3
now, 5 times "a"
5 * a or 5a
the difference of that and 1/5
5a - 1/5
what the dickens is "a" anyway?
well, we're told both of those quantities are the "same"
so [tex]2a\cdot 3=5a-\cfrac{1}{5}\impliedby \textit{solve for "a"}[/tex]
hint: if you multiply both sides by 5, you'd get rid of the denominators
