The general information in the right-hand frame mentions coins and packages. Bills ($5, $10, etc.), stamps, and tickets could be added to the list. In each case, we count the items by type to get the number of each type. We also have a number associated with each item, depending on the type. For example, all of the dimes in a collection of coins have the number 10 (cents) associated with them, and all the packages of one weight type have a weight associated, such as 20 lbs. We are interested in both the number of items of each type and the number of the whole collection. We also are interested in the value of the items in each type and in the whole collection.
Example 1: Coins. A collection of 30 coins (quarters, dimes, and nickels) is made up of 12 quarters, 5 dimes, and the rest in nickels. Use the variables q, d, and n for the numbers of each type of coin. How are q, d, and n related to the total of 30 coins?
The numbers of each type should add to 30.
Express this algebraically.
30 = q + d + n
Each coin has a different value, a fact we use when we determine the value of 12 quarters, for example.
What is the value of 12 quarters (in cents)?
The value of 12 quarters is 12 * 25 = 300 cents.
More generally, what is the value of "q" quarters?
Since each has the value of 25 cents, we can multiply the number by 25. We don't know the number, but have labeled it as q. Consequently, q quarters has the value q * 25.
Express the value of d dimes in a similar way.
Value of d dimes = d * 10.
Express the value of n nickels.
Value of n nickels = n * 5.
We were given the number of quarters and dimes in this collection of 30 coins. How do we find the number of nickels?
The numbers of coins have to add up to 30.
30 = q + d + n = 12 + 5 + n
Solve for n.
Subtract 17 from both sides to get
30 - 17 = 17 + n - 17, or
n = 13.
Now determine the value of the coins. How is the total value related to the values of the quarters, dimes, and nickel?
(Total value of the collection, T) =
(sum of values of the various types)
T = q * 25 + d * 10 + n * 5
Substitute our q = 12, d = 5, and n = 13 to find the total value.
T = 12 * 25 + 5 * 10 + 13 * 5, or
T = 300 + 50 + 65 = 415 cents.
Summarize these results.
The collection of 30 coins consists of 12 quarters, 5 dimes, and 13 nickels. The value of the collection = $4.15
Example 2: Packages. Now let's consider a truckload of packages of two types: light-weight packages weighing 20 lbs and heavy packages weighing 30 lbs. Suppose that the truck must carry 100 of the 30-lb packages in this load, and the rest of the load capacity of 7000 lb is to be filled with 20-lb packages. Let L be the number of 20-lb and H be the number of 30-lb packages. We need to determine how many of the 20-lb packages can be carried.
Express the weight of the 100 heavy packages.
W30 = 100 * 30 = 3000
Express the weight of the L packages weighing 20 lbs each.
W20 = L * 20.
Can W20 be as large as we please?
No.
How is it limited?
The sum of the two weights should be 7000 lbs.
Express this using our symbols W30 and W20.
W20 + W30 = 7000.
Substitute the above results for W20 and W30.
L * 20 + 3000 = 7000
Solve for L.
Subtract 3000 from both sides to get
20L = 4000. Divide both sides by 20 to get
L = 200
Summarize these results.
The 7000-lb load can consist of the required 100 of the 30-lb packages and 200 of the 20-lb packages.
Example 3: Tickets: Ticket problems are similar to the coin problem. We are interested in the number of tickets in each price group, the value or price of each type (reserved, general admission, etc.) and the total value of a collection of tickets. Consider the following problem.
Orchestra seats to a show are $80 and balcony seats are $60. If a theater club spends $3760 on the purchase of 56 seats, how many orchestra seats were purchased?
What is the unknown in this problem?
The number of orchestra seats purchased is the unknown.
Define a variable for this problem.
Let x = the number of orchestra seats.
Express the number of balcony seats purchased in terms of x.
Since 56 seats were purchased in total,
the number of balcony seats is 56 - x.
Express the dollars spent on orchestra seats.
Since each seat costs $80, x seats must cost x times as much, or 80x dollars.
Express the dollars spent on balcony seats.
Each seat costs $60. The number of seats is (56 - x). Multiply these together to get the cost of the seats:
60 * (56 - x) dollars.
How are these two amounts of money related to the $3760?
When added together, they are equal to the $3760.
Express this in terms of x.
80x + 60*(56 - x) = 3760
We need to solve for x. First distribute the factor of 60.
80x + 3360 - 60x = 3760
Collect like terms on the left side.
20x + 3360 = 3760
Subtract 3360 from both sides.
20x = 3760 - 3360, or
20x = 400
Solve for x.
Summarize these results.
20 orchestra seats and
56 - 20 = 36 balcony seats were purchased.
Check in the original words.
20 seats @ $80 = $1600
36 seats @ $60 = $2160
Total: 56 seats @ $3760. It checks!
Example 4: Sale of two types of items. An appliance store sells two types of units at prices of $500 and $750. In one week, the total sales of the two types was 28 units, and the receipts were $16000.
We want to express the receipts in terms of the number of $500 units sold.
Express the receipts, R, in terms of the numbers and prices.
R = (# of cheaper units * price of cheaper unit)
+ (# of expensive units * price of expensive unit).
Let x = # of cheaper units. Express the # of expensive units in terms of x.
Since 28 units were sold in total,
28 = x + (# of expensive units), or
# of expensive units = 28 - x
Now express the receipts in terms of x.
R = x * 500 + (28 - x) * 750
Distribute to remove parentheses.
R = 500 * x + 21000 - 750 * x
Collect the x terms.
R = -250 * x + 21000
If $16000 were received for the sale of these units, include that data in the equation.
16000 = -250 * x + 21000
Subtract 21000 from both sides.
16000 - 21000 = -250 * x +21000 - 21000
Or
-5000 = -250 * x
Divide both sides by -250.
, or
20 = x
Interpret this result.
20 units were sold for $500 and 28 - 20 = 8 units were sold for $750.
Check this result.
We'll use our receipts equation:
R = 20 * $500 + 8 * $750
R = $10,000 + $6000 = $16000