A pharmacist has 40% and 80% of iodine solutions on hand. How many liters of each iodine solution will be required to produce 4 liters of a 50% iodine mixture?
.
Let x = liters of 40% iodine
then
4-x = liters of 80% iodine
Using algebra:
.40x + .80(4-x) = .50(4)
.40x + 3.20-.80x = 2
3.20-.40x = 2
x = 4 liters (40% iodine)
80% iodine:
4-x = 4-4 = 0 liters needed (80% iodine)
D because you can just take the 0. off of 0.18!
Answer:
im pretty sure it is a. (2,-3)
