How many gallons each of $30 \%$ alcohol and $10 \%$ alcohol should be mixed to obtain $20 \mathrm{gal}$ of $23 \%$ alcohol?
\begin{tabular}{|c|c|c|}
\hline \begin{tabular}{c}
Gallons of \\
Solution
\end{tabular} & \begin{tabular}{c}
Percent \\
(as a decimal)
\end{tabular} & \begin{tabular}{c}
Gallons of \\
Pure Alcohol
\end{tabular} \\
\hline$x$ & $30 \%=0.3$ & \\
\hline$y$ & $10 \%=0.1$ & \\
\hline 20 & $23 \%=$ & \\
\hline
\end{tabular}

How many gallons of $30 \%$ alcohol should be in the mixture? $\square$ gal
How many gallons of $10 \%$ alcohol should be in the mixture? $\square$ gal

Step 1 ：The problem is asking for the amount of 30% alcohol and 10% alcohol needed to make 20 gallons of 23% alcohol. This is a mixture problem that can be solved using a system of linear equations.

Step 2 ：The first equation can be formed from the total volume of the mixture, which is 20 gallons. If we let \(x\) be the volume of the 30% alcohol and \(y\) be the volume of the 10% alcohol, then the first equation is \(x + y = 20\).

Step 3 ：The second equation can be formed from the total volume of pure alcohol in the mixture. The volume of pure alcohol in the 30% solution is \(0.3x\) and in the 10% solution is \(0.1y\). The total volume of pure alcohol in the 23% solution is \(0.23 \times 20 = 4.6\) gallons. So, the second equation is \(0.3x + 0.1y = 4.6\).

Step 4 ：We can solve this system of equations to find the values of \(x\) and \(y\).

Step 5 ：The solution to the system of equations is \(x = 13\) and \(y = 7\). This means that to obtain 20 gallons of 23% alcohol, we need to mix 13 gallons of 30% alcohol and 7 gallons of 10% alcohol.

Step 6 ：Final Answer: The amount of $30 \%$ alcohol that should be in the mixture is \(\boxed{13}\) gallons. The amount of $10 \%$ alcohol that should be in the mixture is \(\boxed{7}\) gallons.