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.