A pharmacist has an 18% alcohol solution and a 40% alcohol solution. How much of each should he mix together to make 10 liters of a 20% alcohol solution?

To determine how much of each solution the pharmacist should mix, we can set up a system of equations based on the given information.

Let's assume the pharmacist needs to mix x liters of the 18% alcohol solution and y liters of the 40% alcohol solution to obtain a total of 10 liters.

The first equation can be written to represent the total amount of solution:
x + y = 10 (equation 1)

The second equation can be written to represent the overall alcohol percentage of the mixture:
0.18x + 0.40y = 0.20(10) (equation 2)

Now we can solve this system of equations using the "substitution method" or the "elimination method" to find the values of x and y.

Using the substitution method, we can solve equation 1 for x:
x = 10 - y

Substituting this into equation 2, we get:
0.18(10 - y) + 0.40y = 2
1.8 - 0.18y + 0.40y = 2
0.22y = 0.2
y = 0.2 / 0.22
y ≈ 0.9091

Substituting the value of y back into equation 1, we can solve for x:
x + 0.9091 = 10
x = 10 - 0.9091
x ≈ 9.0909

Therefore, the pharmacist should mix approximately 9.0909 liters of the 18% alcohol solution and 0.9091 liters of the 40% alcohol solution to make 10 liters of a 20% alcohol solution.

.18 x + .40(10 - x) = .20 * 10