so say, we add "x" of the 60% saline solution, now, how much salt is in that solution? now 60% is salt but the rest is something else, but only salt is 60% of "x", or (60/100) * x, or 0.6x.
likewise, for the 10 Liters, if only 50% is salt, then is just (50/100) * 10 or 5.
now... let's check the water whilst we add "y" amount, the water has no salt at all, freshwater for that matter, so, salinity is 0%. or (0/100) * y or 0.
[tex]\bf \begin{array}{lccclll}
&\stackrel{liters}{amount}&\stackrel{\textit{\% of salt}}{quantity}&\stackrel{\textit{total salt}}{quantity}\\
&------&------&------\\
\textit{60\% sol'n}&x&0.60&0.6x\\
\textit{pure water}&y&0.00&0y\\
------&------&------&------\\
mixture&10&0.50&5
\end{array}[/tex]
[tex]\bf \begin{cases}
x+y=10\\
0.6x+0y=5
\end{cases}
\\\\\\
0.6x=5\implies x=\cfrac{5}{0.6}\implies x=\cfrac{25}{3}\implies x=8\frac{1}{3}[/tex]
