A pharmacist has an 18% alcohol solution. How much of this solution and how much water must be mixed together to make 10 liters of a 12% alcohol solution? Answer:

We need a table to do this. 3 rows and 3 columns. Across the top the columns are number of liters, % alcohol, and total. We will label the first row stuff with alcohol, the second row stuff with water, and the third row will be the mix of alcohol and water. In the first column we will put an x for alcohol since we don't know how much alcohol we have in the mix. In the second column we will put the decimal equivalent of the percent alcohol, which is .18. In the total column we will put the product of those 2, which is .18x. Now for the second row, water. We will put a y in the first column since we don't know how much water we have in the mix. In the second column we will put a 0, since water doesn't have any alcohol in it. In the third column we will put the product of the 2 which will be 0. In the third row, the mix row, we put a 10 since we want 10 L of this mix. We will put the decimal equivalent of how much alcohol we want in this mix which is .12. In the total column we put the product of those 2 which is 1.2. Since we are adding the alcohol and the water, x and y, to get a total of 10, this is our first equation. x+y=10. In the total column we will add the percent alcohol in the alcohol and the water to get 1.2. .18x + 0 = 1.2, or just .18x = 1.2. Let's divide both sides by .18 to find the amount of alcohol is in this mix. We get 6 2/3 liters of alcohol. If there is a total of 10 L of this mix, 10 - 6 2/3 = water, and the amount of water then is 3 1/3 L.