Applications

The systems of linear equations can be used for a variety of problems such as coin, geometry, age, work, mixture, investment, motion, and digits. That is why we should exert effort in understanding them because they can be helpful to real life situations like other math concepts.

**Age Word Problem**

*In January of the year 2000, my husband John was eleven times as old as my son William. In January of 2012, he will be three times as old as my son. How old was my son in January of 2000?*

Let "J " stand for my husband John's age, and let "W " stand for William's age. Then J = 11W in the year 2000. In the year 2012, John and William will be twelve years older, so their ages will be J + 12 and W + 12. Also, John will be three times as old as William, so J + 12 = 3(W + 12). So now you've got two equations, each with two variables:

J = 11W J + 12 = 3W + 36

If you know how to solve systems of equations, you can proceed with those techniques. Otherwise, use the first equation to simplify the second; since J = 11W, plug in "11W " for "J " in the second equation:

J + 12 = 3W + 36 (11W) + 12 = 3W + 36 11W – 3W = 36 – 12 8W = 24 W = 3

Remember that the problem did not ask for the value of the variable W; it asked for the age of a person. So the answer is: William was three years old at the beginning of the year 2000.

**Coin Word Problem**

**Your uncle walks in, jingling the coins in his pocket. He grins at you and tells you that you can have the coins if you can figure out how many of each kind of coin he is carrying. You're not too interested until he tells you that he's been collecting those gold-tone dollar coins. The twenty-six coins in his pocket are dollars and quarters, and they add up to seventeen dollars. How many of each coin does he have? Here's how you figure it out:**

There are twenty-six coins in total. Some of them are dollar coins; let "d" stand for the number of dollar coins. The rest of the coins are quarters; let "q" stand for the number of quarters. Then d + q = 26.

If your uncle has only one quarter, then 25×1 = 25 cents comes from quarters. If he has two quarters, then 25×2 = 50 cents comes from quarters. Since he has q quarters, then 25×q = 25q cents comes from quarters.

For the dollar part to work, you'll have to convert to cents. That is, one dollar is one hundred cents. Since he has d dollars, then he has 100d cents from the dollar coins.

He has seventeen dollars in total, or 1700 cents, part of which is from quarters and part of which is from dollars. To help keep things straight, I'll set up a table:

If you know how to solve systems of linear equations, you can see that you have the following system:

d + q = 26 100d + 25q = 1700

If you don't know about systems yet, don't worry: we're nearly done. Since we have two equations related to the same situation, I'll solve one of them and plug the result into the other one. For instance, starting with the two equations::

d + q = 26 100d + 25q = 1700

...I'll solve the first equation above to get:

q = 26 – d

(I could have solved the first equation for d instead of for q; the end result would have been the same, though the computations would have looked different in the middle. Choosing to solve for q in this case was purely a matter of taste.)

Then I'll plug this into the second equation in place of q to get:

100d + 25(26 – d) = 1700

Then I'll solve:

100d + 25(26 – d) = 1700 100d + 650 – 25d = 1700 75d + 650 = 1700 75d = 1050 d = 14

In other words, fourteen of the coins are dollar coins. Since the remainder of the twenty-six coins are quarters, there are 12 quarters.

*The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended? *

Now, though, you can use systems of equations, which means you can use two variables.

a + c = 2200 4a + 1.5c = 5050

Then solve the system for the number of adults and the number of children:

a = 2200 – c

4(2200 – c) + 1.5c = 5050

8800 – 4c + 1.5c = 5050

8800 – 2.5c = 5050

–2.5c = –3750

c = 1500

a = 2200 – (1500) = 700

There were 1500 children and 700 adults.

**Distance Problem**

*A 555-mile, 5-hour plane trip was flown at two speeds. For the first part of thtrip, the average speed was 105 mph. Then the tailwind picked up, and the remainder of the trip was flown at an average speed of 115 mph. For how long did the plane fly at each speed? *

Let's set up a grid:

Using "d = rt", the first row gives us d = 105t and the second row gives us:

555 – d = 115(5 – t)

Since the two distances add up to 555, add the two distances:

555 = 105t + 115(5 – t)

Then we get:

555 = 105t + 575 – 115t 555 = 575 – 10t –20 = –10t 2 = t

According to our grid, "t" stands for the time spent on the first part of the trip, so the answer is "The plane flew for two hours at 105 mph and three hours at 115 mph."

You can add distances and you can add times, but you cannot add rates. Think about it: If you drive 20 mph on one street, and 40 mph on another street, does that mean you averaged 60 mph?

**Digit Problem**

*The sum of the digits of a two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the number.*

I'll use "t" for the "tens" digit of the original number and "u" for the "units" (or "ones") digit. Keeping in mind that the tens digit stands for "ten times this value", I then have:

t + u = 7 10u + t = 10t + u + 27

(The "10t + u" on the right-hand side of the second equation represents the original number.) First I'll simplify the second equation:

10u + t = 10t + u + 27 9u – 9t = 27 u – t = 3

This gives me:

u + t = 7 u – t = 3

Adding down, I get 2u = 10, so u = 5. Then t = 2. Checking, this means that the original number was 25 and the new number (gotten by switching the digits) is 52. Since 52 – 25 = 27, this solution checks out.

The number is 25.

**Business Problem**

*A landscaping company placed two orders with a nursery. The first order was for 13 bushes and 4 trees, and totalled $487. The second order was for 6 bushes and 2 trees, and totalled $232. The bill does not list the per-item price. What is the cost of one bush and of one tree?*

I could try to add the bushes and trees, to get 19 bushes and 6 trees, but this wouldn't get me anywhere, because I don't have subtotals for the bushes and trees. So I'll pick variables ("b" for the number of bushes and "t" for the number of trees) and set up a system of equations:

13b + 4t = 487 6b + 2t = 232

Multiplying the second row, I get:

13b + 4t = 487 –12b – 4t = –464

This says that b = 23. Back-solving, I get that t = 47.

Bushes cost $23 each; trees cost $47 each.