Question

I can't figure this out and yet it's just a simply question. Not the best at math so please help me​

Answer 1

`6=\textit{oz of solution at x\%}\\\\ ~~~~~~ x\%~of~6\implies \cfrac{x}{100}(6)\implies \cfrac{6x}{100} \\\\\\ 9=\textit{oz of solution at 0\%}\\\\ ~~~~~~ 0\%~of~9\implies \cfrac{0}{100}(9)\implies 0 \\\\\\ \textit{15 oz of solution at 16\%}\\\\ ~~~~~~ 16\%~of~15\implies \cfrac{16}{100}(15)\implies 2.4 \\\\[-0.35em] ~\dotfill`

`\begin{array}{lcccl} &\stackrel{oz}{quantity}&\stackrel{\textit{\% of oz that is}}{\textit{acid only}}&\stackrel{\textit{oz of}}{\textit{acid only}}\\ \cline{2-4}&\\ \textit{1st Sol'n}&6&(x)/(100)&(6x)/(100)\\ \textit{2nd Sol'n}&9&0&0\\ \cline{2-4}&\\ mixture&15&0.16&2.4 \end{array}~\hfill \begin{cases} 6 + 9 = 15\\\\ (6x)/(100)+0=2.4 \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 2nd equation}}{\cfrac{6x}{100}=2.4}\implies \cfrac{3x}{50}=\cfrac{24}{10}\implies 30x=1200 \\\\\\ x=\cfrac{1200}{30}\implies \boxed{x=\stackrel{\%}{40}}`