A mixture of 9% disinfectant solution is to be made from 11% and 8% disinfectant solutions.

How much of each solution should be used if 33 gallons of 9% solution are needed?

Method 1 (shortcut): Note that 9% is 1/3 of the way from 8% to 11%, so the amount of 11% solution needed is 1/3 of 33 gallons, or 11 gallons.
The amount of 8% solution needed is 33 - 11 = 22 gallons.

Method 2 (traditional method, using algebra):
Let x = amount of 11% solution needed, in gallons.
Then 33 - x = amount of 8% solution needed, in gallons.

The amount of disinfectant in the 11% solution is 0.11x.
The amount of disinfectant in the 8% solution is 0.08(33 - x).
The total amount of disinfectant needed is 0.09(33) gallons.

Therefore, we have
0.11x + 0.08(33 - x) = 0.09(33)
11x + 8(33 - x) = 9(33)
11x + 8(33) - 8x = 9(33)
3x = 33
x = 11.

So 11 gallons of 11% solution is needed; 33 - 11 = 22 gallons of 8% solution is needed.

Whatever method is used to find the solution, it is a good idea to check that the solution works.

Check:
11 gallons of 11% solution has 0.11(11) = 1.21 gallons of disinfectant.
22 gallons of 8% solution has 0.08(22) = 1.76 gallons of disinfectant.
33 gallons of 9% solution has a total of 0.09(33) = 2.97 gallons of disinfectant.
The solution works, because 1.21 + 1.76 = 2.97 and 11 + 22 = 33.

Lord bless you today!