Basic Math and Pre-Algebra For Dummies (103 page)

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.

 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 all together, 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:

• Irina 	3 t
  Toby 	t

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

• Irina + Toby = 52
  

Plug in the values from the chart:

• 3
  t
  +
  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.

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.

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:

• Jeremy 	j
  Liz 	3 j
  Alison 	2 j
  Raymond 	Liz + 7 = 3 j + 7

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:

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

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

• 3
  j
  + 2
  j
  –
  j
  – 3
  j
  = 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.

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:

•  	Today 	Tomorrow (Today + 5)
  Jake 	 
  Kyle 	 
  Mina 	 
  Suzanne 	 
  Victor 	 

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:

•  	Today 	Tomorrow (Today + 5)
  Jake 	 
  Kyle 	2 v  	2 v + 5
  Mina 	12 	17
  Suzanne 	 
  Victor 	v	v + 5

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:

• Today 	Tomorrow (Today + 5)
  Jake 	17 + v  	17 + v + 5
  Kyle 	2 v  	2 v + 5
  Mina 	12 	17
  Suzanne 	 
  Victor 	v	v + 5

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:

• Today 	Tomorrow (Today + 5)
  Jake 	17 + v  	17 + v + 5
  Kyle 	2 v  	2 v + 5
  Mina 	12 	17
  Suzanne 	17 + v + 3 	17 + v + 8
  Victor 	v	v + 5