A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive mixture. The premium antifreeze solution contains 70% pure antifreeze. The company wants to obtain 280 gallon of a mixture that contains 30% pure antifreeze. How many gallons of water and how many gallons of premium antifreeze solution must be mixed?

In order to figure out how much of both solutions are needed, we need to know just how much antifreeze is needed the desired mixture. We can do this by multiplying the amount of the mixture by its percentage of pure antifreeze, in this case 30%. By multiplying 280 by 0.3 (the decimal value of 30%), we get 84 gallons, meaning we need to add enough 70% pure antifreeze to end up with 84 gallons of antifreeze in our final solution.

Next, we need to figure out how much of our 70% pure antifreeze solution is needed to have the appropriate amount of pure antifreeze in the final solution. In order to solve for this, we can set up the equation: 0.7x = 84 (essentially, what value x is necessary to have 70% of it equal 84?). From here, we can divide both sides by 0.7 to isolate x and find that it equals 120 gallons, meaning we need 120 gallons of 70% pure antifreeze to equal the amount of antifreeze that would be in 280 gallons of 30% antifreeze.

Finally, we know that the entire solution is 280 gallons, so the rest of the solution must be filled with the only other liquid we are given: water. By subtracting 280 and 120 (our gallons of 70% antifreeze needed for the required antifreeze content), we get that 160 gallons of water is needed to create the 30% antifeeze mixture.