Basic Math & Pre-Algebra For Dummies, 2nd Edition (2014)

Part V. The X-Files: Introduction to Algebra

Chapter 23. Putting Mr. X to Work: Algebra Word Problems

IN THIS CHAPTER

Solving algebra word problems in simple steps

Choosing variables

Using charts

Word problems that require algebra are among the toughest problems that students face — and the most common. Teachers just love algebra word problems because they bring together a lot of what you know, such as solving algebra equations (Chapters 21 and 22) and turning words into numbers (see Chapters 6, 13, and 18). And standardized tests virtually always include these types of problems.

In this chapter, I show you a five-step method for using algebra to solve word problems. Then I give you a bunch of examples that take you through all five steps.

Along the way, I give you some important tips that can make solving word problems easier. First, I show you how to choose a variable that makes your equation as simple as possible. Next, I give you practice organizing information from the problem into a chart. By the end of this chapter, you’ll have a solid understanding of how to solve a wide variety of algebra word problems.

Solving Algebra Word Problems in Five Steps

Everything from Chapters 21 and 22 comes into play when you use algebra to solve word problems, so if you feel a little shaky on solving algebraic equations, flip back to those chapters for some review.

Throughout this section, I use the following word problem as an example:

In three days, Alexandra sold a total of 31 tickets to her school play. On Tuesday, she sold twice as many tickets as on Wednesday. And on Thursday, she sold exactly 7 tickets. How many tickets did Alexandra sell on each day, Tuesday through Thursday?

Organizing the information in an algebra word problem by using a chart or picture is usually helpful. Here’s what I came up with:

At this point, all the information is in the chart, but the answer still may not be jumping out at you. In this section, I outline a step-by-step method that enables you to solve this problem — and much harder ones as well.

Here are the five steps for solving most algebra word problems:

1. Declare a variable.

2. Set up the equation.

3. Solve the equation.

4. Answer the question that the problem asks.

5. Check your answer.

Declaring a variable

As you know from Chapter 21, a variable is a letter that stands for a number. Most of the time, you don’t find the variable x (or any other variable, for that matter) in a word problem. That omission doesn’t mean you don’t need algebra to solve the problem. It just means that you’re going to have to put x into the problem yourself and decide what it stands for.

When you declare a variable, you say what that variable means in the problem you’re solving.

Here are some examples of variable declarations:

· Let m = the number of dead mice that the cat dragged into the house.

· Let p = the number of times Marianne’s husband promised to take out the garbage.

· Let c = the number of complaints Arnold received after he painted his garage door purple.

In each case, you take a variable (m, p, or c) and give it a meaning by attaching it to a number.

Notice that the earlier chart for the sample problem has a big question mark next to Wednesday. This question mark stands for some number, so you may want to declare a variable that stands for this number. Here’s how you do it:

Let w = the number of tickets that Alexandra sold on Wednesday.

Whenever possible, choose a variable with the same initial as what the variable stands for. This practice makes remembering what the variable means a lot easier, which will help you later in the problem.

For the rest of the problem, every time you see the variable w, keep in mind that it stands for the number of tickets that Alexandra sold on Wednesday.

Setting up the equation

After you have a variable to work with, you can go through the problem again and find other ways to use this variable. For example, Alexandra sold twice as many tickets on Tuesday as on Wednesday, so she sold 2w tickets on Tuesday. Now you have a lot more information to fill in on the chart:

You know that the total number of tickets, or the sum of the tickets she sold on Tuesday, Wednesday, and Thursday, is 31. With the chart filled in like that, you’re ready to set up an equation to solve the problem:

Solving the equation

After you set up an equation, you can use the tricks from Chapter 22 to solve the equation for w. Here’s the equation one more time:

For starters, remember that 2w really means w + w. So on the left, you know you really have w + w + w, or 3w; you can simplify the equation a little bit, as follows:

The goal at this point is to try to get all the terms with w on one side of the equation and all the terms without w on the other side. So on the left side of the equation, you want to get rid of the 7. The inverse of addition is subtraction, so subtract 7 from both sides:

You now want to isolate w on the left side of the equation. To do this, you have to undo the multiplication by 3, so divide both sides by 3:

Answering the question

You may be tempted to think that, after you’ve solved the equation, you’re done. But you still have a bit more work to do. Look back at the problem, and you see that it asks you this question:

How many tickets did Alexandra sell on each day, Tuesday through Thursday?

At this point, you have some information that can help you solve the problem. The problem tells you that Alexandra sold 7 tickets on Thursday. And because w = 8, you now know that she sold 8 tickets on Wednesday. And on Tuesday, she sold twice as many on Wednesday, so she sold 16. So Alexandra sold 16 tickets on Tuesday, 8 on Wednesday, and 7 on Thursday.

Checking your work

To check your work, compare your answer to the problem, line by line, to make sure every statement in the problem is true:

In three days, Alexandra sold a total of 31 tickets to her school play.

That part is correct because 16 + 8 + 7 = 31.

On Tuesday, she sold twice as many tickets as on Wednesday.

Correct, because she sold 16 tickets on Tuesday and 8 on Wednesday.

And on Thursday, she sold exactly 7 tickets.

Yep, that’s right, too, so you’re good to go.

Choosing Your Variable Wisely

Declaring a variable is simple, as I show you earlier in this chapter, but you can make the rest of your work a lot easier when you know how to choose your variable wisely. Whenever possible, choose a variable so that the equation you have to solve has no fractions, which are much more difficult to work with than whole numbers.

For example, suppose you’re trying to solve this problem:

Irina has three times as many clients as Toby. If they have 52 clients altogether, how many clients does each person have?

The key sentence in the problem is “Irina has three times as many clients as Toby.” It’s significant because it indicates a relationship between Irina and Toby that’s based on either multiplication or division. And to avoid fractions, you want to avoid division wherever possible.

Whenever you see a sentence that indicates you need to use either multiplication or division, choose your variable to represent the smaller number. In this case, Toby has fewer clients than Irina, so choosing t as your variable is the smart move.

Suppose you begin by declaring your variable as follows:

Let t = the number of clients that Toby has.

Then, using that variable, you can make this chart:

No fraction! To solve this problem, set up this equation:

Irina + Toby = 52

Plug in the values from the chart:

3t + t = 52

Now you can solve the problem easily, using what I show you in Chapter 22:

Toby has 13 clients, so Irina has 39. To check this result — which I recommend highly earlier in this chapter! — note that 13 + 39 = 52.

Now suppose that, instead, you take the opposite route and decide to declare a variable as follows:

Let i = the number of clients that Irina has.

Given that variable, you have to represent Toby’s clients using the fraction , which leads to the same answer but a lot more work.

Solving More-Complex Algebraic Problems

Algebra word problems become more complex when the number of people or things you need to find out increases. In this section, the complexity increases to four and then five people. When you’re done, you should feel comfortable solving algebra word problems of significant difficulty.

Charting four people

As in the previous section, a chart can help you organize information so you don’t get confused. Here’s a problem that involves four people:

Alison, Jeremy, Liz, and Raymond participated in a canned goods drive at work. Liz donated three times as many cans as Jeremy, Alison donated twice as many as Jeremy, and Raymond donated 7 more than Liz. Together the two women donated two more cans than the two men. How many cans did the four people donate altogether?

The first step, as always, is declaring a variable. Remember that, to avoid fractions, you want to declare a variable based on the person who brought in the fewest cans. Liz donated more cans than Jeremy, and so did Alison. Furthermore, Raymond donated more cans than Liz. So because Jeremy donated the fewest cans, declare your variable as follows:

Let j = the number of cans that Jeremy donated.

Now you can set up your chart as follows:

This setup looks good because, as expected, there are no fractional amounts in the chart. The next sentence tells you that the women donated two more cans than the men, so make a word problem, as I show you in Chapter 6:

Liz + Alison = Jeremy + Raymond + 2

You can now substitute into this equation as follows:

3j + 2j = j + 3j + 7 + 2

With your equation set up, you’re ready to solve. First, isolate the algebraic terms:

3j + 2j – j – 3j = 7 + 2

Combine like terms:

j = 9

Almost without effort, you’ve solved the equation, so you know that Jeremy donated 9 cans. With this information, you can go back to the chart, plug in 9 for j, and find out how many cans the other people donated: Liz donated 27, Alison donated 18, and Raymond donated 34. Finally, you can add up these numbers to conclude that the four people donated 88 cans altogether.

To check the numbers, read through the problem and make sure they work at every point in the story. For example, together Liz and Alison donated 45 cans, and Jeremy and Raymond donated 43, so the women really did donate 2 more cans than the men.

Crossing the finish line with five people

Here’s one final example, the most difficult in this chapter, in which you have five people to work with.

Five friends are keeping track of how many miles they run. So far this month, Mina has run 12 miles, Suzanne has run 3 more miles than Jake, and Kyle has run twice as far as Victor. But tomorrow, after they all complete a 5-mile run, Jake will have run as far as Mina and Victor combined, and the whole group will have run 174 miles. How far has each person run so far?

The most important point to notice in this problem is that there are two sets of numbers: the miles that all five people have run up to today and their mileage including tomorrow. And each person’s mileage tomorrow will be 5 miles greater than his or her mileage today. Here’s how to set up a chart:

With this chart, you’re off to a good start to solve this problem. Next, look for that statement early in the problem that connects two people by either multiplication or division. Here it is:

Kyle has run twice as far as Victor.

Because Victor has run fewer miles than Kyle, declare your variable as follows:

Let v = the number of miles that Victor has run up to today.

Notice that I added the word today to the declaration to be very clear that I’m talking about Victor’s miles before the 5-mile run tomorrow.

At this point, you can begin filling in the chart:

As you can see, I left out the information about Jake and Suzanne because I can’t represent it using the variable v. I’ve also begun to fill in the Tomorrow column by adding 5 to my numbers in the Today column.

Now I can move on to the next statement in the problem:

· But tomorrow … Jake will have run as far as Mina and Victor combined… .

I can use this to fill in Jake’s information:

In this case, I first filled in Jake’s tomorrow distance (17 + v + 5) and then subtracted 5 to find out his today distance. Now I can use the information that today Suzanne has run 3 more miles than Jake:

With the chart filled in like this, you can begin to set up your equation. First, set up a word equation, as follows:

Jake tomorrow + Kyle tomorrow + Mina tomorrow + Suzanne tomorrow + Victor tomorrow = 174

Now just substitute information from the chart into this word equation to set up your equation:

17 + v + 5 + 2v + 5 + 17 + 17 + v + 8 + v + 5 = 174

As always, begin solving by isolating the algebraic terms:

v + 2v + v + v = 174 – 17 – 5 – 5 – 17 – 17 – 8 – 5

Next, combine like terms:

5v = 100

Finally, to get rid of the coefficient in the term 5v, divide both sides by 5:

You now know that Victor’s total distance up to today is 20 miles. With this information, you substitute 20 for v and fill in the chart, as follows:

The Today column contains the answers to the question the problem asks. To check this solution, make sure that every statement in the problem is true. For example, tomorrow the five people will have run a total of 174 miles because

42 + 45 + 17 + 45 + 25 = 174

Copy down this problem, close the book, and work through it for practice.