say we're going to add "x" gallons of water, now how many gallons of alcohol is in pure water? well 0%, and 0% of "x" is (0/100) * x = 0.
the 3 gallons we know are 100% alcohol, how many gallons of alcohol is in it? well (100/100) * 3 = 3.
the resulting mix will have x + 3 gallons total and it's going to be 15% alcohol, how many gallons of alcohol is that? (15/100) * (x+3).
now, the amount of alcohol never increased, so the original 3 gallons of only alcohol will remain even after the water has been added, so.
[tex]\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{gallons of }}{amount}\\ \cline{2-4}&\\ \textit{pure water}&x&0.00&0\\ \textit{pure alcohol}&3&1.00&3\\ \cline{2-4}&\\ mixture&x+3&0.15&0.15(x+3) \end{array} \begin{array}{llll} \\[3em] \leftarrow \textit{this amount}\\\\ \leftarrow \textit{must be equal to this one} \end{array} \\\\\\ 3 = 0.15(x+3)\implies 3=0.15x+0.45\implies 2.55=0.15x \\\\\\ \cfrac{2.55}{0.15}=x\implies 17=x[/tex]
