MINI-LESSON on Systems of Equations Applications Type 2
Remember that a system of equations is two equations that give you two different facts about the same two variables. In order to solve these equations, you can use addition, substitution, or graphing.
The type 2 problems are called COMBINATION problems. You will be given data about two values combined and asked to find each value. In these type of problems you will often see the words total, sum, together, combined—or some other word that means to add. In addition, you will usually see the word each. With combination problems, both equations in the system will be addition equations (with two terms on the left, and one term on the right). Let’s see why….
Often times, you can calculate these problems by brute force and trial & error. Obviously, it is easier to find the answer using algebra. ALEKS teaches you to make a list of known data for these problems. This is an excellent strategy because the labels on that list will help you know how to write the equations. Here is an example:
Tickets for senior citizens cost $4, and tickets for adults cost $7. The total number of tickets sold was 25 for total box office receipts of $151. How many of each type of ticket were sold?
Here is a list of what we know and what we want to find out:
How many senior citizen tickets were sold? s tickets
How many adult tickets were sold? a tickets
Total number of tickets sold 25 tickets
Cost of senior ticket 4 dollars per ticket
Cost of adult ticket 7 dollars per ticket
Total box office receipts 151 dollars
Interesting – we have three of the same label, two of a second label, and one with a final label. The solution to this problem is in tickets. So our first equation will use the three data points measured in tickets:
a + s = 25 [tickets + tickets = tickets]
Notice that the labels are the same? Now look at the remaining labels—per or each is a keyword meaning division. We have two dollars per ticket and the total is in dollars. We write this in such a way that the ticket labels cancel out so we have all labels the same. Let me show you:
Notice how the labels are the same? The two remaining labels that were the same multiply by the tickets (and the ticket labels cancel) to form the remaining single label.
4s + 7a = 151
Now I have my two equations.
Let’s do another problem:
Two rainstorms occurred in one week in a certain area. The first storm lasted 40 hours, and the second storm lasted 15 hours, for a total of 1325 mL of rain. What was the rate of rainfall for each of the two storms if the sum of the two rates was 55 mL per hour?
Let’s use x and y for the two types of rainstorms. Let’s make a list of what we know:
What is the rate of the first storm? x mL per hour
What is the rate of the second storm? y mL per hour
Total rate of both storms 55 mL per hour
First storm lasted 40 hours
Second storm lasted 15 hours
Total rainfall 1325 mL
Because the label we want to find is mL per hour, I’ll use the three data items with that same label (notice that there will always be three data items with the same label):
x + y = 55 [mL per hour + mL per hour = mL per hour]
Notice that the labels are the same? Now what else do I know about x and y—let’s look only at the labels:
x is mL/hour y is mL/hour 40 is hours 15 is hours 1325 is mL
The second equation is:
40x + 15y = 1325 [mL/hour · hours + mL/hour · hours = mL]
One more example:
There is a gasoline truck and a diesel truck owned by a towing company. Last week, the sum of the gas mileage for the two trucks is 60 miles per gallon. The gasoline truck consumed 30 gallons of gasoline and the other truck used 15 gallons of diesel fuel. The two trucks have logged a total of 1500 miles for the week. What was the average gas mileage (in miles per gallon) for each truck during last week?
Here is our list (starting, as always, with what we want to find):
x miles per gallon
y miles per gallon
Sum of both trucks 60 miles per gallon
amount of gas used by x 30 gallons
amount of diesel used by y 15 gallons
Total miles for both trucks 1500 miles
The first equation is always easy to find: x + y = whatever is the same label (60)
x + y = 60
The second equation uses the two remaining data items with the same label as coefficients of the variables (so the labels can cancel) and their sum gives the remaining label:
30x + 15y = 1500
So I guess it really doesn’t matter what other words are in the problem, when you determine the data in a list form, look for three data items, two data items, and one data item:
x + y = (whatever is same data item as x and y)
ax + by = (whatever is the lone data item label)