Algebra Mini-Series #4: Use Multiplication in Equations Wisely

This is part four of our weeklong series on improving your algebra skills. Our goal for this series is to provide a useful resource for both students and teachers, so that this article can be used in the classroom, for test prep, or to help yourself practice and master skills you never learned. With these tips, we hope you can make your algebra cleaner, faster, and more intuitive. Check back tomorrow for another informative article on some aspect of elementary algebra!

Back to: Part 3: Using Substitution to Solve Equations

Forward to: Part 5: How to Use Fractions with Algebra Correctly

Five Tricks to Improve Your Algebra Skills

Part 4: Know When (and how) To Multiply Large Numbers Together

Oftentimes big numbers can make a simple algebra problem nasty. This, of course, should never be something to hold you back, but students aren’t usually taught how to deal with numbers in equations. Rather it’s assumed that this is the easy part and that the algebra involved is the meat of the problem. If you’re ever faced with a situation where it’s the numbers that are tripping you up, here are two easy ways which can help you to deal with large numbers more quickly and effectively.

A Quick Trick to Multiply Large Numbers

Quick! What’s 19*32? Did you guess 608? Or did you grab a sheet of paper and start working it out by hand? Typically guessing isn’t going to get you anywhere, and you don’t want to start writing out long multiplication in the middle of a math test. Here’s an easier way to do this: rewrite the numbers as (20-1)(30+2) and distribute (or FOIL) them out. 19*32 = 20*30 + 2*20 – 1*30 – 1*2 = 600 + 40 – 30 – 2 = 608. I bet you never thought you’d use the distributive property this way! (This is actually a fault of our educational system—most students are taught so much abstraction that they forget all the X and Ys they’re dealing with are actually numbers.)

After breaking the number up, we didn’t have to do any long multiplication at all, but mostly addition and subtraction. The reason that I chose 20 – 1 for 19 is that we would like our digits to all be as small as possible—you could also try 10 + 9 and this would work too. In fact, this way prevents you from having to deal with negatives, so it’s really your choice. One advantage of the first method is that you’ll never have to deal with multiplying digits larger than 5.

Let’s try this on the following problem:

Solve the following for y:
22x = (y + x)/17 – 1

First, move the 1 to the left side:

22x + 1 = (y + x)/17

Next, clear the denominator by multiplying through:

22x*17 + 1*17 = y + x

At this point, we unfortunately have to multiply out 22*17 since we’re going to have to subtract x from the left side. So let’s do it the “easy” way: (20 + 2)(20 – 3) = 400 – 60 + 40 – 6 = 374.

374x + 17 = y + x

373x + 17 = y

Unfortunately, this was the easiest way to do this problem, as we can see from the fact that 373 is prime. But sometimes it’s possible to avoid having to do this multiplication altogether. Take the following for example:

If y = 17/21, then (y*60 + 17/7)/3 =?

Plugging in y, we have (17*60/21 + 17/7)/3. But let’s not multiply 17*60 just yet..rather, distribute the /3 to get (17*60/21)/3 + (17/7)/3 = 17*20/21 + 17/21. Again, don’t multiply 17*20 and add 17—this is just 17*20 + 17 = 17*21. (Don’t forget what multiplication means!) So our answer is 17*21/21 = 17.

While the cases where this is useful may at first seem limited, waiting to multiply numbers until the end can greatly reduce the amount of mental grinding you have to do to solve a problem and will often result in cancelled factors. Using this in conjunction with the above technique should make dealing with numbers in your equations a lot simpler.

Challenge problem: 211*39=?

Solution: 211 = 200 + 10 + 1, 39 = 40 – 1, so 211*39 = (200 + 10 + 1)*(40 – 1) = (200 + 10 + 1)*40 – (200 + 10 + 1) = 8000 + 400 + 40 – 200 – 10 – 1 = 8000 + 200 + 30 – 1 = 8229.