A dehydrated patient needs a 6.4% saline IV. Unfortunately, the hospital only has bags of 6% and 8% saline solutions. How many liters of each of these solutions should be mixed together to yield 1 liter of the desired concentration? Solution You have a system of two equations, with two unknowns. Let x = the liters you need of the 6% solution Let y = the liters you need of the 8% solution x + y = 1 .06x + .08y = .064 Solve for x in the first equation to get x = 1 - y Plug this value into your second equation to get: .06(1-y) + .08y = .064 Solve for y to get y = .2 liters Since x + y has to equal 1 liter, then x must equal .8 liters. So you will need .8 liters of the 6% solution and .2 liters of the 8% solution. Hope that helps, let me know if you have any questions, good luck! .