by Donald Proctor

Most people in our vocation would opt for a dose of cod liver oil ahead of a dose of algebra, and few of them have never tasted cod liver oil — they just know it can’t be as bad as algebra, with all those silly rules to be memorized. In the version of algebra we’ll be exploring, there is only one main rule, and you’ve already seen it earlier: If two things that are equal are treated equally, they’re still equal. If not treated equally, they won’t be equal any more.

The technique for this logical version of algebra requires, first of all, writing an equation or statement in which we know both sides of the equation are equal to each other even if there are values we don’t know in the equation.

Let’s demonstrate this concept in an example: You have been given a 525-ft length of three-eighths-inch rot-resistant nylon rope and told to fashion a new 150-ft halyard for the flag pole and to use all the remainder to make five lifelines for flotation rings for the aeration basins. How long should each lifeline be?

Step No. 1: Write a statement (equation) that involves what is known and not known:

Total rope length = halyard length + five lifeline lengths.

Step No. 2: Substitute values for as much as you can:

525 ft = 150 ft + 5 lifelines.

Step No. 3: Treat both sides of the equation equally — subtract 150 ft from both sides:

525 ft (–150 ft) = 150 ft (–150 ft) + 5 lifelines.
Or 375 ft = 5 lifelines.

Step No. 4: Treat both sides of the equation equally — divide both sides by 5:

375 ft ÷ 5 = 5 lifelines ÷ 5. Or 75 ft = 1 lifeline.

You may be muttering that you could have found that answer by using logic and skipped algebra altogether. Not true, my friend. Algebra is just logic written down in an understandable format. Unfortunately, algebra usually ends up with a mess of x’s and y’s instead of descriptive variable such as “lifelines,” but that is a matter of choice, not necessity.

Let’s try another example: A chlorine rapid-mix basin is to be built as a square basin with a liquid depth equal to two-thirds of its width. The basin volume must be 890 ft³. What should the dimensions be?

Step No. 1: Write the starting equation from what is known but involving what is not known. If, length × width × depth = volume, then using W to represent the length of a side, the equation becomes:

W × W × 2/3W = 890 ft³.
That equation can be consolidated. 2/3W³ = 890 ft³.

Now we treat both sides of the equation equally and divide by 2/3 (or multiply by 3/2).

2/3W³ × 3/2 = 890 ft³ × 3/2. Or W³ = 1335 ft³.

Find the cube root of each side.

W = 11.01 ft. So, 2/3W = 7.3 ft.

So the chlorine rapid-mix basin should be about 11 ft wide, 11 ft long, and 7.3 ft deep.

There are many possible things that we might do to both sides of an equation that would fit the “treat equal sides equally” rule, but a bit of logical thinking will usually indicate what needs to be done to get progressively simpler equations until, finally, the remaining equation is the desired answer.

©2006 Water Environment Federation. All rights reserved.