let's say we'll mix "x" amount of the 80% solution, how much antifreeze is in it? well is just 80% of antifreeze, and the rest is something else.... what's 80% of "x"? so is (80/100)*x, or 0.8x.
likewise, the mixture will be a 70% solution, and let's say it adds up to "y" amount, so how much antifreeze is in it? well (70/100) * y or 0.7y.
[tex]\bf \begin{array}{lccclll}
&\stackrel{gallons}{amount}&\stackrel{\%}{quantity}&\stackrel{antifreeze}{quantity}\\
&------&------&------\\
\textit{80\% sol'n}&x&0.80&0.8x\\
\textit{10\% sol'n}&100&0.10&10\\
------&------&------&------\\
mixture&y&0.70&0.7y
\end{array}[/tex]
so whatever "x" is, we know that x + 100 = y.
and we know that 0.8x + 10 = 0.7y.
[tex]\bf \begin{cases}
x+100=\boxed{y}\\
0.8x+10=0.7y\\
----------\\
0.8x+10=0.7\left( \boxed{x+100} \right)
\end{cases}
\\\\\\
0.8x+10=0.7x+70\implies 0.8x-0.7x=70-10\implies 0.1x=60
\\\\\\
x=\cfrac{60}{0.1}\implies x=600[/tex]
