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Number of Variables

Number of variables. Some students have questions about how many variables to use in word problems. To some extent, this is not really a “mathematical” issue, but more a matter of preference. In other words, sometimes students get caught up on what is the right way to do it with just one variable or assigning two variable?

And it turns out, there is no absolute right way. You can do it both ways, many times. It’s really a matter of discerning what works best for you. So I’m gonna show both methods. And it’s very important that you, in your own practice, practice both ways. If you solve a word problem with two variables,, go back and solve the same problem again.

Try it with only one variable. See what works best for you. And as you develop a sense of your own preferences, you’ll be more comfortable with word problems. So here’s a practice video. I’m gonna say pause the video and solve this.

And then we’ll talk about this. Okay. This is a very simple word problem. Frank has $13 more than Glenda does, and together they have 81.

How much does Frank have? So first of all, we’re gonna start approaching this just with two variables. The obvious choices are F and G, for Frank’s money and Glenda’s money. F has, Frank has, so F equals 13 more than Glenda. F equals G plus 13. And then of course their sum is 81.

Those are our two equations. Well, the equation on the left is solved for F right now. Technically, we want to solve for G, because we want to eliminate G, so that our final equation has F. Because F is our target value here. So I solve for G, I substitute that in.

Now I have a single equation with F, my target value. Simplify, divide by 2 and we get F equals 47. That’s how much Glen, that’s how much has. So that’s one way to solve the problem. Let’s think about this again using only one variable. Well, for the one variable, the smart variable would be our target value.

Our target value is Frank. So were gonna use F. That’s gonna be our only variable. Frank has 13 more than Glenda. An so it means that Glenda has 13 less than Frank. So Glenda equals F minus 13.

And so I’m gonna use F minus 13 for Glenda. And then the sum of those two, Frank plus Glenda, F plus, F minus 13, that equals 81. Well, this should look familiar. This is the same equation as we had before. So I’m not gonna solve it again.

It has the same answer. But this shows you, whether you start with one variable or two variables, you really wind up in the same place. You may get there slightly faster if you use one variable instead of two variables. But then again, it can be hard sometimes to see how to set up everything in terms of one variable.

So that’s the trade-off. If you see that you can introduce just one variable and express everything easily in terms of that. Then that might save you a couple steps and a little bit of time. Practice with this and experiment. If you find that when you pick only one variable, you get confused or have trouble expressing the other quantities in terms of that one variable.

Then it might just work better for you always to start with two variables. So again, get out of the mindset that one way is right, one way is wrong. This is about finding your own preference, finding what works best for you. And if you find that one choice is more, is more comfortable for you, is easier for you, then by all means, go with that. What if there are more than two variables?

Well, keep in mind that if we have N variables, we also need N equations to solve for those variables. We did have a lesson on solving three equations with three unknowns. So that’s, that’s particularly tricky math. And that’s typically more advanced than anything you’d need to do on a word problem.

And forget about it for a four equations or five equations. You just don’t see that on the test. If a word problem talks about three or four items, chances are very good that we can relate everything to a single item, which will allow us to create a single equation. So here’s a relatively simple word problem.

So this is a little more simple than you’d see on the test, typically. Pause the video and then we’ll talk about this problem. Okay. Mark is 8 years older than Lisa. And Peter’s age is 3 years less than three times Lisa’s age.

If the sum of their ages is 80, what is Peter’s age? So Mark equals, Mark is, Mark equals 8 years older than Lisa. L plus 8. That’s one equation. The other one a little trickier. Peter is, P equals 3 years less, so we’re subtracting 3.

Three times Lisa’s age. So 3L minus three. That’s Peter’s age. So notice we’re expressing both of these in terms of L. Then the sum of the ages is 80. So we plug in those two top equations.

We can just write one equation in terms of L. And so that’s often what happens here. Even though we have three different quantities originally, we want to get it down very quickly to one equation, with one variable. Now this one, we just do a little cleaning up and simplifying. We get 5L plus 5, equals 80.

5L equals 75. So L equals 15. Now, very good that we use sensible letters here. L equals 15, that’s Lisa’s age. That’s not the answer to the question. We need Peter’s age.

So Peter’s age, we have to take that multiply by 3 and then subtract 3. Well that would be 42. And that’s the answer to the question. Now here’s a word problem that is actually worthy of something you might see on the test.

Pause the video and then we’ll talk about this. Okay. Wendy left the house with D dollars in cash. She made the following purchases. Gas for three, $3 less than one-third of D.

A book for one-sixth of D. Stationery for $6 dollars more than one-sixth of D. Groceries for a quarter of D. After these four purchases, she had $4 left. What is the value of D? Well for, very fortunately, they give us a variable.

So we’ll just stick with that variable and everything is expressed in terms of that variable. So we can say minus 3 minus D, that’s the gas. Plus one-sixth D, that’s the book. Plus one-sixth D plus 6, that’s the stationery. Plus one-quarter D, that’s the groceries.

Plus the $4 left. All that together add up to D. So I’ll do a little simplifying. One-sixth plus one-sixth is one-third. And also neaten up the numbers. They just reduce to plus 7.

Now, I’m gonna find a common denominator, add all those fractions. I get eleven-twelfths D plus 7 equals D. Subtract eleven-twelfths D from both sides. I get 7 equals one-twelfth D, multiplied by 12. D equals 7 times 12, or 84. And that’s the answer.

In summary. If there are two quantities in a word problem, it may save your time to use only one variable. And again, remember this is very much a matter of personal choice. You need to experiment for yourself.

Try it with two variables. Try it with one variable. Do that on several word problems. Find out what works best for you and then go with that. If there are three or more quantities in a word problem, you’ll always want to relate all of them to a single variable and construct a single equation.

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