We can solve this problem using a system of linear equations. Let's define:

• $x$ as the number of liters of the 20% acid solution.
• $y$ as the number of liters of the 50% acid solution.

From the problem, we know two things:

1. The total volume of the solution must be 60 liters: $x + y = 60$

2. The total amount of acid in the final solution must be 40% of 60 liters, which is 24 liters of acid. The acid in the 20% solution is $0.20x$, and the acid in the 50% solution is $0.50y$. So, the equation for the acid content is: $0.20x + 0.50y = 24$

Now we have a system of two equations:

$\text{1. } x + y = 60$ $\text{2. } 0.20x + 0.50y = 24$

Solving the System:

1. From the first equation:
   Solve for $y$: $y = 60 - x$

2. Substitute $y = 60 - x$ into the second equation: $0.20x + 0.50(60 - x) = 24$ Simplify the equation: $0.20x + 30 - 0.50x = 24$ Combine like terms: $-0.30x + 30 = 24$ Subtract 30 from both sides: $-0.30x = -6$ Divide by -0.30: $x = \frac{-6}{-0.30} = 20$

3. Substitute $x = 20$ back into the first equation to find $y$: $20 + y = 60$ $y = 60 - 20 = 40$

Final Answer:

• Gabe needs 20 liters of the 20% acid solution and 40 liters of the 50% acid solution.

Would you like more details or have any questions?

Here are some follow-up questions you might find helpful:

1. What would happen if the final solution needed to have a higher acid concentration (e.g., 60%)?
2. How can we solve this kind of mixture problem using a matrix method?
3. What would happen if there was a third solution available with a different concentration?
4. How would you modify the equations if the total amount of solution was different (e.g., 50 liters)?
5. How can these methods be applied in real-world problems, like chemical manufacturing?

Tip: In mixture problems, always set up two equations: one for the total volume and one for the amount of the component (acid, in this case).