Choosing a Strategy

(Reposted from my blog for my students)

I will often tell my students to select strategies that work best for them to solve a problem, rather than focusing on a single specific strategy. What do I mean by this?

I don’t mean that students should use inefficient strategies. One of the things that successful students of mathematics should learn is how to solve problems in efficient ways. In elementary school, students are encouraged to learn the multiplication table because it’s more efficient to remember what \(4 \times 3\) is than to have to calculate it each time.

However, as mathematics problems become more complex in high school, we get more ways of approaching a problem. So saying that there’s a single correct “best” way to approach a problem is misleading. Different students work in different ways, and so what works best for one student won’t necessarily work best for another one.

Here’s an example from algebra. Let’s say the problem is \(4(x + 6) = 2x – 8\). You’re asked to solve for \(x\). What’s your first step?

Many students want to get rid of the parentheses first, using the Distributive Property. That creates \(4x + 24 = 2x – 8\). That’s fine, but as we discussed this week in the Algebra I classes, we need to be careful with the Distrubitive Property. Using it first with, for instance, \(4 + 3(4 – 3)\div 3\) might lead you to write \(4 + 12 – 9\div 3 = 4 + 12 – 3 = 13\), which is the incorrect answer (it’s really \(5\)). So always remember, when using the Distributive Property, to keep the addends (the values that are added together) as a group.

Some students might notice that everything in the problem is divisible by \(2\), so you can factor that out and get rid of it. That creates \(2(x + 6) = x – 4\). The danger here is remembering to factor \(2\) out of each term exactly once. Don’t factor a \(2\) out of the \(6\) inside the parentheses, and remember to factor a \(2\) out of both terms on the right. So both \(2(x + 3) = x – 4\) and \(2(x + 6) = x – 8\) might be things you’d be tempted to do, and they’d both lead to an incorrect solution.

Other students might decide to “move” the \(2x\) to the other side of the equal sign by subtracting it from both sides; others might decide to “move” the \(8\) instead. In general, we math teachers encourage collecting the expressions with the variable you’re solving for first, so we’d generally say to subtract \(2x\) first, and then deal with the constants in a later step, but if a particular student works better with moving the constants first, so be it.

The least efficient algebraic method that I see for solving this is to divide both sides by \(4\) first. This yields \(x + 6 = \frac{2x – 8}{4} = \frac{x}{2} – 2\), which will still lead to the same solution but which introduces fractional forms, which students often struggle with.

However, all of these methods are better than the guess-and-check method that many students learn before high school: Just try a number, then adjust as needed. As we get into more complicated problems, guess-and-check becomes very inefficient. In this case, for instance, the solution is \(x=-16\). How many guesses would it take to get there? Some students might not even think to check negative values. And not all of our solutions will be integers. How do you guess-and-check to solve \(4.5x Р8.3 = 2.1x + 13\)?

So when students hear me say that I’ll accept different strategies so long as they produce the correct solution, students should also be mindful that strategies that take less time, take fewer steps, or work for a wider array of problems are better than those that take more time, take more steps, or only work for a small group of problems.

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