For the following problem, complete the table to the right. Then solve the problem using a system of equations.
A $40 \%$ dye solution is to be mixed with a $60 \%$ dye solution to get $120 \mathrm{~L}$ of a $50 \%$ solution. How many liters of the $40 \%$ and $60 \%$ solutions will be needed?
Complete the table below.
\begin{tabular}{|c|c|c|}
\hline Liters of solution & Percent (as a decimal) & Liters of pure dye \\
\hline $\mathrm{x}$ & 0.40 & \\
\hline $\mathrm{y}$ & 0.60 & \\
\hline 120 & 0.50 & \\
\hline
\end{tabular}
(Simplify your answers. Use integers or decimals for any numbers in the expression.)

Step 1 ：Let's denote the amount of the 40% solution as x and the amount of the 60% solution as y.

Step 2 ：We can set up two equations based on the information given. The first equation is based on the total volume of the solution, which is 120L. This gives us: \(x + y = 120\)

Step 3 ：The second equation is based on the total amount of dye in the solution. The amount of dye in the 40% solution is 0.4x, and the amount of dye in the 60% solution is 0.6y. The total amount of dye in the 120L solution is 50% of 120L, or 60L. This gives us: \(0.4x + 0.6y = 60\)

Step 4 ：We can solve this system of equations to find the values of x and y. The solution to the system of equations is x = 60 and y = 60.

Step 5 ：This means that we need 60L of the 40% solution and 60L of the 60% solution to get 120L of a 50% solution. This makes sense because the final solution is exactly in the middle of the two initial solutions in terms of dye concentration, so we need equal amounts of each.

Step 6 ：Final Answer: We need \(\boxed{60}\) liters of the 40% solution and \(\boxed{60}\) liters of the 60% solution.