Math Matters: Why Do I Need To Know This?
Bruce Kessler, Department of Mathematics
Western Kentucky University
Episode Eleven
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Amortized loans – Monthly mortgage payments
Objective: To illustrate the role of mathematics in home mortgage financing.
We specifically consider the problem of calculating the regular equal monthly
payment of an ordinary amortized loan. In addition to showing how to do this,
we examine the amount of interest paid over the life of the loan, and show how
small additions to the monthly payments can dramatically lower the time needed
to pay back the loan.
Hello, and welcome to “Math Matters: Why do I need to know this?” This is a program
where we look at the entry-level mathematics that we teach here at Western, and try to apply
that to everyday life. I’ve got some neat examples I want to show you today that deal with
everything from consumer math to just general kind of allocation-of-resources kind of things,
so let’s get right after it.
The first thing I would like to talk to you about today is, I’d like to say a little bit about
amortized loans. Now, that’s probably a, it could be a word you are not familiar with, but
basically what I’m talking about is a home loan, okay? Most folks – now you’re maybe too
young to worry about it now – but most folks are going to at some point in their life own a
home, and chances are you are not going to be able to walk up and pay cash for it. If you had
to do that, nobody would have a house. Donald Trump would have a house and a few other
people. So you get a loan and you end up using the house and the property as collateral;
that is, that is called a mortgage. And then the types of mortgages that are available, most
of those are what are called amortized loans. And that just simply means that you’ve got
regular scheduled payments that include a payment to both the principal and the interest on
the principal. And even along those lines, you have some choices as to what you are doing.
The most common type, I believe, would be the equal payment type of amortized loan. The
payments, as the name says, are equal, but what happens is the percentage of that payment
that goes to principal and interest changes throughout the life of the loan. You start off
paying quite a bit of interest, because you owe a lot of money, and then as you start to pay
off the debt, you have more and more of the money going to pay off the principal and less
and less of it going to pay off. You just pay interest on what you owe, but the payment itself
stays the same. It’s just throughout the life of it, it kind of, the amount that goes to each
changes. (Figure 1)
Another approach to doing this would be what is called an equal principal type of amortized
loan. Okay, the idea here is you simply take whatever amount you borrowed, divided by
however many months you are going to have this loan, and you are going to pay that amount
of principal every month. Then, on top of that, you are going to pay interest on what you
owe. So as throughout the life of the loan, the amount you pay to principal stays the same,
but then the amount you pay on interest goes down. So then, the payment actually goes
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Figure 1, Segment 1
down throughout the life of the loan. And that’s an, that’s an okay way to do it. Most folks
still, I think, kind of prefer this equal payment thing where they just have the same payment
throughout. They lock in their interest rate and payment. (Figure 1)
There is another scheme to doing this, and I say scheme because I’m not real big on
this idea, but there is this idea of reverse amortization. And what that means is, you start
with low payments – this is something that might be done for someone is having a hard
time affording a house right now – but you start with a very low payment, and initially that
payment is actually less than the interest you owe on the amount of money you borrowed.
So what happens there is, because your interest is less than what you actually owe, the part
you aren’t paying, it’s tacked onto the principal. So the payments over a period of time go up
and you end up having to pay interest on more than you actually borrowed. You are kind of
betting on the future that you do something good and you’re making lots of money. It’s okay,
it’s legal. Just understand that you could be getting yourself into a bit of fix there. (Figure 1)
The one I want to look at today because it’s, this is the one that is actually difficult to
calculate the payment for, is this equal payment type of amortized loan. We actually did
something similar to this last week, if you go back and check episode ten. We talked about
a person who had a balance on a credit card. They quit using the card and they wanted to
pay within, you know, however many regular monthly payments. So we did a bunch of work
actually last week – I’m not going to go back and kind of show you where this formula comes
from – but we said alright, well, the amount you would have to pay per month then is x
and it’s based on these amounts: the A0 is the initial amount, the first balance, this m is
the monthly interest rate, so you take the annual rate r and divide it by 12 to get m, and
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then n is the number of payments you want to make, okay? That is exactly the situation we
talked about. You’ve got money borrowed and you’re having to pay interest on the money
you’ve borrowed, but you’re reducing that amount every month, so it’s the same formula,
same situation. (Figure 2)
Figure 2, Segment 1
Let’s say, here’s an example, you’re borrowing $100,000 at 6% annual interest over thirty
years. So the monthly rate would be 21 %. And if you go through and plug all these crazy things
into the formula – 1 plus this would give you that, the number of payments would be 12 per
year for 30 years so that’s 360, I think – this works out to be %599.55. Now that works out
so that over the life of the loan that last payment is almost the almost exactly the pay-off.
There is a little bit of rounding involved. The actual payment would probably be more than
that. You have to realize that most folks when they do this they put stuff into escrow. They
store things there like real estate taxes, homeowners insurance, sometimes even mortgage
insurance depending on what your down payment was. So the thing is, if you take that
amount multiplied by 360 payments, you’ve actually paid back twice what you have borrowed:
$215,000 plus. That’s scary, but that is just kind of the system – that’s the way things work.
You’re actually not, you’re not paying anything you didn’t owe. You have a lot of money
borrowed, and so that’s in the process of paying it back. That’s what you are paying back.
(Figure 3)
How can you lower this amount? Well, you can do this: you can borrow over less time.
Set this up for 15 years rather than 30. What changes in this formula would be the 15, and
that gets you up to a monthly payment of $843.86. Now this is the prohibitive part of this:
that may be too high. You may not be able to afford a house at that price, so you do, you
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Figure 3, Segment 1
will end up spending less money over time, but you have to spend so much per month and
it may be prohibitive. I think this works out to...yeah, $150,000 you pay back. So you are
saving 64, right at $64,000, but again, you have to make that payment. (Figure 4)
The other thing you can do is you can add a little bit to each payment, and this is kind of
like going in-between. Your monthly payment is this and if you graph this crazy thing, you,
the blue curve here is this thing in terms of n, and then right at 360 that’s where we hit this
level of $599.55. (Figure 5) If we add $10 each month and bump that up, well, the curve
moves. It hits the curve a little bit to the left and it’s more like 344 payments to do that. It
shortens the loan by fifteen months. I say fifteen instead of sixteen because there is a bit of
stuff leftover you would actually have to go another month to pay it off, but you’re going to
send in an amount that is about six thousand dollars less than if you do the standard amount.
(Figure 6) If you go twenty over per month, it will take it down about 30 months and you
will have saved about $11,386. The good side of that is that you don’t have to make the big
payments unless you want to. You know, you are kind of tacking on when you want to do
things, so it’s not a bad system actually to get the longer loan and pay it off in a shorter
amount of time. (Figure 7)
I did not get into how I solved for the new n values in the two cases where we are adding
on to the regular payment because of the time constraints and because of the messiness of
the calculations. There are several approaches. We could certainly hard-code the situation
into a spreadsheet or other computer application and simply see how many months it would
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Figure 4, Segment 1
Figure 5, Segment 1
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Figure 6, Segment 1
Figure 7, Segment 1
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take. We can also simply solve the equation
100000(0.005)(1 + 0.005)n
= 599.55 + 10
(1 + 0.005)n − 1
for the new value of n in the first example. Even this answer can be found mechanically by
using a graphing calculator to graph both the left and right hand sides (replacing n with
x on TI calculators,) and finding the first coordinate of the intersection (as illustrated in
Figure 6.) To find it algebraically,
500(1.005n )
= 609.55
1.005n − 1
500(1.005n ) = 609.55(1.005n ) − 609.55
609.55 = 109.55(1.005n )
609.55
12191
=
109.55
2191
n ln 1.005 = ln 12191 − ln 2191
ln 12191 − ln 2191
≈ 344.125 monthly payments.
n=
ln 1.005
The work for the second example is analogous.
1.005n =
I flashed up a bunch of stuff very quickly there, so let me put up some summary pages
for what we talked about.
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Summary page 1, Segment 1
Summary page 2, Segment 1
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Summary page 3, Segment 1
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Least common multiple and greatest common divisor
– Using fractions
Objective: To illustrate the relevance of the concepts of least common multiple and greatest common divisor. We carefully define both concepts (and the
concept of relatively prime) and demonstrate various techniques for finding each.
I want to come back to something that is not algebraic. I’ll give your algebra a break
there just for a second, and I want to talk about greatest common divisors and least common
multiples. And this is kind of a number theory idea, but it’s a useful idea in that it helps
us handle fractions. People are kind of odd about fractions. You know, even people who like
math, and I have to say I’m kind of in this group, I avoid fractions like a plague because they
are tedious to deal with. However, I prefer them in a lot of cases to decimals things, okay?
I’ll give you an example of how fractions are nicer. You know, which is easier? Is it
easier to deal with the fraction 31 , which is okay, or would you prefer to do something like
this? You know, a repeating decimal that goes on forever, okay? Well, this is much more
compact right here. A much more instinctive way of doing this, plus, if you cut this off at any
point, you are rounding. It’s an approximation. So in that situation, this is better. There
is a way to do this with a bar over it that says this repeating part keeps going, but even that
can be messy. If I look at a fraction like 17 and then I look at its decimal equivalent, okay, I
think it starts repeating here after the one. So this part repeats again. So my point is that
it’s very tedious and in some instances to deal with decimals, fractions it’s much nicer to
deal with the fraction itself. If that denominator down there doesn’t have all prime factors
of twos and fives, then you cannot write that as a power of ten in the denominator and you
are going to have a repeating decimal. (Figure 1)
So we need to know how to deal with fractions. Of course, they pop up in all things
that we do. They just require some skills that we just don’t need with the decimals. For
example, to simplify fractions we need to be able to calculate that greatest common factor
of the numerator and denominator. And then, if we are adding and subtracting, we get a
common denominator. We get pieces of the same side and that means we really need to find
the least common multiple of the two denominators. (Figure 2)
So what I want to talk to you about is I’m assuming that we are going to be dealing with
fractions, and that you need to be able to deal with fractions and whatever you do. So let’s
talk about some methods for actually finding these things. The greatest common divisor, and
I’m using some notation that is pretty obvious, GCD, that’s greatest common divisor of two
numbers. Okay what I am talking about is the largest counting number that I can divide into
each of of the two numbers a and b with no remainder. So, clean division if you will. And
then, if the biggest number that will divide into both is one, then we say they are relatively
prime. (Figure 3)
So for example if I am dealing with a fraction, and I want to simplify this fraction, I look
at those two numbers and say “What is the greatest common divisor of those two numbers?”
and probably in your head, “Well, 12 is the biggest number I can think of that would divide
both. Okay, so then what you do is say alright 36 is 3 times 12 and 48 is 4 times 12 and
then we can get rid of the 12s and we’re left with three-fourths. We know we are in simplest
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Figure 1, Segment 2
Figure 2, Segment 2
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Figure 3, Segment 2
form at that stage because 3 and 4 are relatively prime. The only thing I can divide cleanly
into each of them is one. (Figure 3)
That is a neat little trick, but what if the numbers were worse? This is what I want to
look at – methods for finding that greatest common divisor when you can’t necessary do it
in your head. Now I’m going to stick with the two numbers 36 and 48. I don’t want to get
messy stuff here – I want you to understand the process. So let’s talk about some methods
for finding these things.
The most obvious way of doing this is to simply list out all the different divisors of 36 and
48, which I have done here. And then you look for the biggest one. The biggest one I see here
is 12. So that’s the greatest common divisor. Okay now, that’s obvious, but it’s also tedious.
If these are large enough, they will have a great big number of divisors. Another method that
is sometimes more effective is to use the prime factorization of each one. Write each number
as a product of primes. 36 for example is 4 times 9 which is 22 times 32 . And 48 is, you can
go through and divide by all the twos and you get 4 factors of two and are left with a factor
of 3, and that gives you 48. And then what you will do is look at the lowest occurrence of
each prime and what I mean by that is the lowest exponent. Two to the two and three to the
one. You take those two things and multiple them, and that gives you the greatest common
divisor. Okay, that’s typically an effective method for doing things. (Figure 4)
There’s a sneaky trick that we attribute to Euclid called the Euclidean algorithm, where
we take the two numbers and we do this: we take the big number and divide by the smaller of
the two and you get the remainder. The greatest common divisor of these two will be equal
to the greatest common divisor of the smaller of the two and the remainder. And then, just
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Figure 4, Segment 2
keep doing that until you get clean division. And when that happens, the number that you
divided by, that’s your greatest common divisor. Okay, so that’s the sneaky trick. (Figure 5)
There’s even a graphical approach to doing this, and this is really slick. You take a
rectangle or you can use graphing paper and set up a 36 by 48 rectangle. And then starting
at (0, 0), just send out a diagonal line and just bounce it around until it hits a corner and
then what you do is look at that little spot where those two things meet, where the lines meet,
that’s the greatest common divisor 12. This is actually patented. I find this hard to believe,
but this was patented by a guy back in 1961. He has a patent number and everything for this.
I don’t know maybe I owe him a nickel since I’ve showed this to you, but I don’t know. I
hope not. But, that does work. (Figure 5)
The other thing was the least common multiple. We have similar tricks for that. Now
what I’m talking about here is I want smallest counting number that is divisible by both a and
b, so I have kind of turned things around. And we use that when we start adding fractions.
Now what I could do to get a common denomiator is to multiply these two thing together.
That is a big number, which I don’t want to have to deal with, so I’m going to find the least
common multiple. And you can kind of do this in your head. I get both of these denominators
to 144 and whatever I multiply by to do that, I multiply by in the top and in the bottom. And
then I add things up, equalizes things up and that’s 55 over 144. That is relatively prime so
you are done. Okay? (Figure 6)
When finding the least common multiple, you can list things out until you get a common
kind of thing here. That’s one way of doing it. You can also look at the prime factorizations
and take the highest occurrence of each prime. That would be two to the fourth and three
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Figure 5, Segment 2
Figure 6, Segment 2
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squared. Multiplying those up gives you the least common multiple, or here is another trick.
If you notice, as long as there are two numbers there, the other numbers that I have circled in
purple, these are, that would give me the greatest common divisor. (Figure 7) So the product
of all those things gives me the product of a and b. So to find the least common multiple, I
can just divide. You know just take this to the other side and divide by it. So in this case
I already and this is good if I already know the greatest common divisor. We already know
that is 12 so then do your little division. 3 times 48 equals 144. (Figure 8)
Figure 7, Segment 2
Okay, those are some neat little tricks that would help you deal with fractions. I ran
through those quickly, so let me flash up so summary pages that will recap what we talked
about.
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Figure 8, Segment 2
Summary page 1, Segment 2
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Summary page 2, Segment 2
Summary page 3, Segment 2
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Summary page 4, Segment 2
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Systems of two linear equations – Two equations,
two unknowns
Objective: To introduce the concept of a solution to a system of two linear equations, both graphically and algebraically, give methods for finding such solutions,
and show the usefulness of the techniques by examining a real-world situation.
The last thing I would like to talk to you about today is an application of a system of
linear equations. Now this is something we cover in our 116 class. We actually cover it
in our 096 class near the end. The idea here is that you have a couple unknowns that you
need to solve for and you at least have a couple of conditions that give you some information,
enough information to where you can solve for the two unknowns. There are a ton of different
applications where these kind of, where this situation pops up. The only thing that would
make it rare would be if you said well, most times you have more than two things. Most of
the time it’s three things you are worried about or it’s four things that you are worried about.
Today I am going to restrict myself to talking about two things at a time.
You know I am going to show you one example, but you can apply that to anything.
You’ve got kid one and kid two and you want to give them so much stuff. How much each
do you give each kid? Which by the way, I have four kids, so numbering your kids instead
of naming them, it’s much easier to remember. You just remember, you know, the trauma,
kind of inducing which one came first – much easier to remember. Or name them all George
like Foreman did. By the way, this is off topic but the last time we had kids we had a name
picked out for a girl and then we found out we were having two girls. So we liked the name
Sarah, and then we found out we were having twins and I said “Oh, that’s easy. We’ll just
do Sarah1 and Sarah2 ,” but for some reason my wife didn’t go with it, you know, she didn’t
like that system. So, anyways, it would have worked for me.
We are going to set up the two conditions with the two variables. That will give us a
system of linear things and I’m assuming here we aren’t squaring things, we are not raising
them to powers, we are just dealing with linear things. It’s a lot more complicated if you
start to square things. Here’s an example. Adapt this to your everyday life but I’m trying
to do something real general, very everyday. A coffee shop owner wants to mix an expensive
brand of coffee, an expensive type of coffee at $9.00 per pound with a less expensive coffee to
create a blend that will sell better. You know, the expensive cost too much, the other one is
considered sub par, you know, not as good as the rest, so we are trying to mix a batch here,
a 50 pound batch that is somewhere in the middle, okay? So there are two things that I need
to figure out: how much of the expensive coffee, and I’ll call that x; how much of the cheaper
coffee, and I will call that y. And there is enough information here for me to solve for the
two conditions. You know I worry about the weight. The sum of the two weights needs to be
50 pounds, so x + y equals 50. Well, the price, if I take nine pounds, nine dollars per pound
on the one and 5 dollars per pound on the other, I would like to make that same amount of
money by just simply charging $6.50 or 6.5 for each of the fifty pounds of the mixed variety.
And you may want to multiply that out I think it works out to 325 or something so, you’ve
got two equations, you’ve got two unknowns and we need to find x and y to satisfy both at
the same time. (Figure 1)
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Figure 1, Segment 3
This is where I need to talk you through just a little bit here. There are three basic things
that could happen when we look at two equations and two unknowns. These are linear things,
so each one of the equations will give us a line. It’s possible that the two lines will be parallel,
okay, and if you look at that, you can see that basically these are the same except for a sign
but they are very different when it comes to this constant out here. Well, if those lines are
parallel, you know what parallel means, they are not going to touch each other. And that
means there is no point x and y that satisfies both equations at the same time. So if that
happens, you are not going to have a solution. Oh, bad luck – try something else. (Figure 2)
The other thing that could happen that still, in a way, it’s the same case, because you
have parallel lines, but you have two lines that are identical to each other, okay? You graph
them both and they both cover the same set of points, and you can see that here, that basically
I have scaled, that basically this bottom one is this one times a − 2, − 2 times this, − 2
times that gives me the same line, okay? In this case, you do have to solutions – it’s just
you have a bunch of solutions. Every point on that line is a solution to the system. And
that’s not always helpful either. (Figure 3)
A lot of times you just want one solution. And that’s case three where you have two lines
they are not parallel. If they are not parallel, they have to cross somewhere, okay? And the
two I have drawn cross right there, okay? There’s your solution to the system. And, in fact,
that is a good way actually to solve these things numerically as you go into your calculator:
graph the two lines, and then use your calculator functions to say “Well, okay, what’s the
intersect of the two lines?” That’s okay, but that gives you numerical answers, so again with
answers like this, I would have repeating decimals, okay, and that is actually the solution.
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Figure 2, Segment 3
Figure 3, Segment 3
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(Figure 4)
Figure 4, Segment 3
I would like to talk about methods for doing this symbolically. One out of two possible
methods I can think of is called the substitution method. And this is where you go ahead,
you find one variable in one equation you solve for it. And then, you jam that into the other
equation, okay? For example, in our coffee example, I have gone ahead and taken the liberty
of graphing this, just so you see that we are in this nonparallel case. There is the solution
right here – it looks like it is a little less than twenty for x and a little more than 30 for y,
okay. But I would solve for one of the two variables in one of the two equations and because
I don’t want to generate fractions, I’m going to go ahead and solve for y in the top one.
So that would be 50 − x for y and then you take that and jam it into the y in the second
equation. So here we go. You will clean that up a bit: 9x and minus the 5x would give you
4x, and then this is 250 subtracted from both sides. You get 75 so 4x equals 75, so you divide
or if you convert that to a mixed number, you get 18 34 pounds.
through by the junk. x is 75
4
Then you back substitute. You’ve got a nice formula for x, so take this number, jam it back
up in here and you get y is a 125
or if you prefer, 31 14 pounds. The sum of those is 50 and
4
the value would be the same as if I charged nine there and five for whats with the other one.
(Figure 5)
The other method that I can think of to do this, and it’s the common method, is add
multiples of the equations to each other. So in our coffee problem, I would say let’s eliminate
one of the two variables. I multiply by a negative five here. And that means I need to multiply
by a negative five across the board. And then simply add them. The − 5y and the 5y cancel
each other out I get 4x. I get 75. It’s the same result. It’s going to give me the same answer.
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Figure 5, Segment 3
That gives me the x that I found before and you back substitute. You take this answer then
and take it back up here and substitute for y, and you get the poundage for x and y. (Figure 6)
These are nice little methods and I grant you, that’s kind of a rinky-dink problem. There’s
all kinds of things that I could think of. I don’t have time to go into all the different types
of problems where you have two things. Now next week, I want to look at, what if you had
three, what if you had four, you know, what other methods can you use to solve these things.
I’ve got some summary slides that talk about the things that I talked about and kind of recap
what I did.
Closing
Hope you enjoyed the examples I showed you today and I hope they, I don’t want you
thinking that they are too high-brow or too tough or anything like that. I’m really searching
for a common everyday examples of the mathematics we use in our entry-level courses. If
you have good examples you would like to show me, or tell me about or if you just simply
have questions about things from what you have seen on the program or in your class, go
head and email me or go to the website or email me at mathmatters@wku.edu. I would love
to get some ideas from you. That would be great. With that, I am done I hope to see you
next week.
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Figure 6, Segment 3
Summary page 1, Segment 3
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Summary page 2, Segment 3
Summary page 3, Segment 3
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Summary page 4, Segment 3
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