Answer
First solution: $ 25\%$ acid
Second solution: $ 10\%$ acid
Work Step by Step
Let x be the percentage of acid in the first solution,
and x be the percentage of acid in the second solution.
Then
300x+600y=$0.15$(300+600)$\qquad$ (ml acid)
$300x+600y=135$
and
$ 100x+500y=0.125(100+500)\qquad$ (ml acid)
$100x+500y=75$
We have the system
$\left\{\begin{array}{ll}
300x+600y=135 & /\\
100x+500y=75 & /\times(-3)
\end{array}\right.$
(eliminates x)
$\left\{\begin{array}{ll}
300x+600y=135 & /\\
-300x-1500y=-225 & /add
\end{array}\right.$
$-900y=-90\qquad/\div(-900)$
$ y=0.1=10\%$
Back-substitute
$100x+500y=75$
$100x+500(0.1)=75$
$100x+50=75\qquad/-50$
$100x=25\quad/\div 25$
$ x=0.25=25\%$
First solution: $ 25\%$ acid
Second solution: $ 10\%$ acid
