Many word problems will give rise to systems of equations --- that is, a pair of equations like this:

You can solve a system of equations in various ways. In many of the
examples below, I'll use the * whole equation
approach*. To review how this works, in the system above, I could
multiply the first equation by 2 to get the y-numbers to match, then
add the resulting equations:

If I plug into , I can solve for y:

In some cases, the whole equation method isn't necessary, because you can just do a substitution. You'll see this happen in a few of the examples.

The first few problems will involve items (coins, stamps, tickets) with different prices. If I have 6 tickets which cost $15 each, the total cost is

If I have 8 dimes, the total value is

This is common sense, and is probably familiar to you from your experience with coins and buying things. But notice that these examples tell me what the general equation should be: The number of items times the cost (or value) per item gives the total cost (or value). This is where I get the headings on the tables below.

You'll see that the same idea is used to set up the tables for all of these examples: Figure out what you'd do in a particular case, and the equation will say how to do this in general.

* Example.* Calvin has $8.80 in pennies and
nickels. If there are twice as many nickels as pennies, how many
pennies does Calvin have? How many nickels?

In this kind of problem, it's good to do everything in cents to avoid having to work with decimals. So Calvin has 880 cents total.

Let p be the number of pennies. There are twice as many nickels as pennies, so there are nickels. I'll arrange the information in a table.

Be sure you understand *why* the equations in the pennies and
nickels rows are the way they are: The number of coins times the
value per coin is the total value. If the words seem too abstract to
grasp, try some examples:

If you have 3 nickels, they're worth cents.

If you have 4 nickels, they're worth cents.

If you have 5 nickels, they're worth cents.

So if you have nickels, they're worth cents.

The total value of the coins (880) is the value of the pennies * plus* the value of the nickels. So I add the first
two numbers in the last column, then solve the resulting equation for
p:

Calvin has 80 pennies.

Therefore, he has nickels.

* Tables for problems.* I'll often arrange the
equations for word problems in a * table*, as I
did above. Roughly:

1. The *number* of things will go in the first column. This
might be the number of tickets, the time it takes to make a trip, the
amount of money invested in an account, and so on.

2. The *value per item* or *rate* will go in the second
column. This might be the price per ticket, the speed of a plane, the
interest rate (in percent) earned by an investment, and so on.

3. The *total value* or *total amount* will go in the
third column. This might be the total cost of a number of tickets,
the distance travelled by a car or a plane, the total interest earned
by an investment, and so on.

With this arrangement:

There are many correct ways of doing math problems, and you don't have to use tables to do these problems. But they are convenient for organizing information --- and they give you a pattern to get started with problems of a given kind (e.g. interest problems, or time-speed-distance problems).

In some cases, you *add* the numbers in some of the columns in
a table. In other cases, you set two of the numbers in a column
equal, or subtract one number from another. There is no general rule
for telling which of these things to do: You have to think about what
the problem is telling you.

* Example.* A total of 78 seats for a concert
are sold, producing a total revenue of $483. If seats cost either
$2.50 or $10.50, how many $2.50 seats and how many $10.50 seats were
sold?

Suppose x of the $2.50 seats and y of the $10.50 seats were sold.

The first and third columns give the equations

Multiply the second equation by 10 to clear decimals:

Solve the equations by multiplying the first equation by 25 and subtracting it from the second:

Then , so . Thus, 42 of the $2.50 seats and 36 of the $10.50 seats were sold.

* Example.* Tickets to a concert cost either
$12 or $15. A total of 300 tickets are sold, and the total receipts
were $4140. How many of each kind of ticket were sold?

The first and third columns give the equations

Multiply the first equation by 15 and subtract equations:

Then

There were 120 tickets sold for $12 each and 180 tickets sold for $15 each.

* Example.* An investor buys a total of 360
shares of two stocks. The price of one stock is $35 per share, while
the price of the other stock is $45 per share. The investor spends a
total of $15000. How many shares of each stock did the investor buy?

Let x be the number of shares of the $35 stock and let y be the number of shares of the $45 stock.

The first and third columns give

Multiply the first equation by 45, then subtract the second equation:

Since , I have . The investor bought 120 shares of the $35 stock and 240 shares of the $45 stock.

The next problem is more complicated than the others, since it
involves solving a system of *three* equations with
*three* variables. You'll see that I do it by substitution. If
you take more advanced courses (such as * linear
algebra*), you'll learn methods for solving systems like these
which are like the whole equation method. They involve representing
the equations using * matrices*.

* Example.* Phoebe has some 32-cent stamps,
some 29-cent stamps, and some 3-cent stamps. The number of 29-cent
stamps is 10 less than the number of 32-cent stamps, while the number
of 3-cent stamps is 5 less than the number of 29-cent stamps. The
total value of the stamps is $9.45. How many of each stamp does she
have?

I will do everything in cents.

I'll let x be the number of 32-cent stamps, let y be the number of 29-cent stamps, and let z be the number of 3-cent stamps. Here's the table.

The last column says

The number of 29-cent stamps is 10 less than the number of 32-cent stamps, so

The number of 3-cent stamps is 5 less than the number of 29-cent stamps, so

I want to get everything in terms of one variable, so I have to pick a variable to use. Since the last two equations both involve y, I'll do everything in terms of y.

z is already solved for in terms of y, since . I'll solve for x in terms of y:

Plug and into and solve for y:

Then

Phoebe has 20 32-cent stamps, 10 29-cent stamps, and 5 3-cent stamps.

The next problem is about numbers. The setup will give two equations, but I don't need to solve them using the whole equation approach as I did in other problems. Since one variable is already solved for in the second equation, I can just substitute for it in the first equation.

* Example.* The sum of two numbers is 90. The
larger number is 14 more than 3 times the smaller number. Find the
numbers.

Let L be the larger number and let S be the smaller number. The sum is 90:

The larger number is 14 more than 3 times the smaller number:

Plug into the first equation and solve:

Then . The numbers are 19 and 71.

The next set of examples involve * simple
interest*. Here's how it works. Suppose you invest $600 (the * principal*) in an account which pays simple interest. At the end of one interest period,
the interest you earn is

You now have dollars in your account.

Notice that you multiply the amount invested (the principal) by the interest rate (in percent) to get the amount of interest earned.

By the way --- How does "percent" fit the pattern of the
earlier problems, where I had things like "dollars per
ticket" or "cents per nickel"? In fact,
"percent" is short for "per centum", and
*centum* is the Latin word for a hundred. So "4
percent" means "4 per 100". Since "per"
translates to division, I get , as you probably know from earlier math courses.

* Example.* $6000 is divided between two
accounts, one paying interest and the other
paying interest. At the end of one interest
period, the interest earned by the account exceeds the interest earned by the account by $65. How much was invested in each
account?

I have

Rewrite the equations:

Clear the decimals by multiplying the second equation by 100:

Multiply the first equation by 3 and subtract equations to solve for y:

Then

$2500 was invested at and $3500 was invested at .

* Example.* Bonzo invests some money at interest. He also invests $1700 more than twice that
amount at interest. At the end of one interest
period, the total interest earned was $278. How much was invested at
each rate?

The last column gives an equation which can be solved for x:

Then , so $2100 was invested at and $5900 was invested at .

There are various kinds of * mixture problems*.
The first few involve mixtures of different things which cost
different amounts per pound. For instance, if you have 4 pounds of
candy which costs $2 per pound, the total cost of the candy is

In other words, the number of pounds times the price per pound is the total cost.

* Example.* Calvin mixes candy that sells for
$2.00 per pound with candy that costs $3.60 per pound to make 50
pounds of candy selling for $2.16 per pound. How many pounds of each
kind of candy did he use in the mix?

The first and third columns give two equations:

Multiply the first equation by 2 and subtract equations:

Then

He used 45 pounds of the $2 candy and 5 pounds of the $3.60 candy.

* Example.* Phoebe wants to mix raisins worth
$1.60 per pounds with nuts worth $2.45 per pound to make 17 pounds of
a mixture worth $2 per pound. How many pounds of raisins and how many
pounds of nuts should she use?

Suppose she uses x pounds of raisins and y pounds of dried fruit.

The first and third columns give the equations

Multiply the second equation by 100 to clear the decimals. This gives

Solve the equations by multiplying the first equation by 160 and subtracting it from the second:

Hence, and . She needs 8 pounds of raisins and 9 pounds of nuts.

Mixture problems do not always wind up with two equations to solve. Here's an example where the setup gives a single equation.

* Example.* How many pounds of coffee worth $4
per pound must be mixed with 20 pounds of Elmer's Glue worth $2 per pound to obtain a mixture
worth $3.60 per pound?

Let x be the number of pounds of coffee. Set up a table:

The last line says . Solve for x:

You will need 80 pounds of coffee.

An * alloy* is a mixture of different kinds of
metals. Suppose you have 50 pounds of an alloy which is silver. Then the number of pounds of (pure) silver in
the 50 pounds is

That is, the 50 pounds of alloy consists of 10 pounds of pure silver and pounds of other metals.

Notice that you multiply the number of pounds of alloy by the percentage of silver to get the number of pounds of (pure) silver.

* Example.* Phoebe mixes an alloy containing
silver with an alloy containing silver to make 100 pounds of an alloy with silver. How many pounds of each kind of alloy did she
use?

The first and third columns give two equations:

Multiply the second equation by 100 to clear decimals:

Multiply the first equation by 14 and subtract equations:

Then

She used 60 pounds of the alloy and 40 pounds of the alloy.

Other mixture problems involve * solutions*. For
instance, a solution may be acid, or alcohol. What does this mean? Suppose you have 80
gallons of a solution which is acid. Then the number of gallons of (pure) acid in
the solution is

So you can think of the 80 gallons of solution as being made of 16 gallons of pure acid and gallons of pure water.

Notice that you multiply the gallons of solution by the percentage of acid to get the number of gallons of (pure) acid.

* Example.* How many gallons of each of a acid solution and an acid solution must be mixed to produce 50 gallons of
a acid solution?

The first and third columns give the equations

Multiply the second equation by 10 to clear the decimals:

Multiply the first equation by 6 and subtract equations:

Then

Use 15 gallons of the solution and 35 gallons of the solution.

* Example.* Amounts of a alcohol solution and a alcohol solution are to be mixed to produce 24
gallons of a alcohol solution. How many gallons
of the alcohol solution and how many gallons of
the alcohol solution should be used?

Suppose x gallons of the alcohol solution and y gallons of the alcohol solution are used.

The first and third columns give the equations

Multiply the second equation by 100 to clear decimals:

Solve the equations by multiplying the first equation by 65 and subtracting the second:

Then , so . Thus, 16 gallons of the solution and 8 gallons of the solution must be used.

Two angles are * complementary* if their sum is
--- that is, if they add up to a right
angle.

For example, and are complementary, because

* Example.* Two angles are complementary, and
the larger one is more than 3 time
the smaller one. Find the angles.

Let L be the larger angle, and let S be the smaller angle. The angles are complementary:

The larger one is more than 3 time the smaller one:

Plug into and solve for S:

Then . The smaller angle is and the larger angle is .

* Example.* Two angles are complementary. One
angle is less than twice the other. Find
the two angles.

The sum is :

One angle is less than twice the other:

Plug into the first equation and solve:

Then . The angles are and .

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Copyright 2012 by Bruce Ikenaga