now, let's say we'll use "x" oz of the 14% solution, and "y" oz of the 7% solution.
how much hydrochloric acid is there in the "x" oz of 14%? well, is just 14% of x, or (14/100) * x, or 0.14x.
how much hydrochloric acid is there in the "y" oz of 7%? well is just 7% of y, or (7/100) * y, or 0.07y, the rest may be water or some other substance.
now, the mixture needed is 28 oz, and is only 12% of hydrochloric acid, since is 12% hydrochloric acid and the rest water or such, how much hydrochloric acid is there in it? well, is just (12/100) * 28, or 3.36 oz.
[tex]\bf \begin{array}{lccclll}
&\stackrel{ounces}{amount}&\stackrel{\%~acid}{quantity}&\stackrel{ounces~acid}{quantity}\\
&------&------&------\\
\textit{14\% sol'n}&x&0.14&0.14x\\
\textit{7\% sol'n}&y&0.07&0.07y\\
------&------&------&------\\
mixture&28&0.12&3.36
\end{array}
\\\\\\
\begin{cases}
x+y=28\implies \boxed{y}=28-x\\
0.14x+0.07y=3.36\\
----------\\
0.14x+0.07\left( \boxed{28-x} \right)=3.36
\end{cases}
\\\\\\
0.14x-0.07x+1.96=3.36\implies 0.07x=1.4
\\\\\\
x=\cfrac{1.4}{0.07}\implies x=20[/tex]
how many ounces of the 7% solution? well y = 28 - x.