Class 2 -- Algebra basics

I. Algebraic Expressions
   A. What is an expression?
   B. Plugging in
   C. Combining like terms
   D. Distributing
II. Constructing Algebraic Expressions
    A. Price and "coin" problems
    B. Interest and rebates
    C. Distance and speed
    D. Geometry problems
    E. Number problems

I. Algebraic Expressions

A. What is an expression?

Arithmetic is a way of dealing with numbers of known values. Algebra
gives us a way of writing down general rules that can be applied to
any particular number.

The basic idea of algebra is the unknown, a symbol that stands for a
number. We use can do everything to the unknown we could do to
numbers: add to it, subtract from it, multiply it, divide it, square
it, etc.

This ability to maniuplate unknown or arbitrary numbers makes algebra
a powerful method of solving problems, and of representing
relationships in the real world.

The basic tool of algebra is the algebraic expression. An algebraic
expression just represents a rule for manipulating a number.

Consider a basic example: suppose I run a store, and I want to take $5
off the price of each item. How do I do compute the new price? It's
easy: I just take the old price and subtract $5. Algebraically, I
write this as a recipe: new price = p - 5, where p is the old price.

Another example: how do I compute the tax on an item? Suppose there is
a 5% tax. The way I compute the tax is a take the price of the item
and multiply by 5%: tax = 5% x p = 0.05p.

Note two things here: first, it turned the percent into a decimal,
i.e. 5% = 0.05. Second, when I wrote the second line, I omitted the
times sign. This is an algebra convention: if you have two symbols
next to one another, without anything in between, that means to
multiply them.

Algebraic formulas can also involve more than one quantity. For
example, suppose I have a rectangular room and I want to calculate its
area, i.e. how many square feet the room it. The way I do this is I
take the length of the room and multiply by its width. An algebraic
formula for this operation is area = lw, where l is the length and w
is the width.


B. Plugging in

Given an algebraic expression, we can read it as a recipe for how to
do a calculation given some numbers. This process is called "plugging
in", because we plug a number into our algebraic formula and see what
comes out. All we do is replace the variables by the given numbers and
evaluate.

Here's an example: the formula for the surface area of a sphere is
surface area = 4 pi r^2, where r is the radius and pi ~= 3.14. In
words, this means to compute the square of the radius, then multiply
by 4, then multiply by pi. (Recall that to square something is to
multiply it by itself.)

Using this rule, I can calculate the total surface area of the
Earth. The radius of the earth is about 4000 mi. I square this and get
16 million square miles. Then I multiply by 4 and 64 million. Then I
multiply by pi and get about 200 million square miles.

An important concept in plugging in is what order to do operations
in. For example in the formula for the Earth, I would not have gotten
the same answer if I had multiplied by 4 and pi, then squared.

Here's a simpler example. Consider two expressions 2x^2 and
(2x)^2. The first one means square the number, then multiply by 2, and
the second means multiply by 2 and then square. To see those aren't
the same, suppose we choose our number to be 5.

2x^2 = 2 (5^2) = 2 (25) = 50
(2x)^2 = (2 . 5)^2 = 10^2 = 100.

Obviously not the same!

The rule here is parentheses, then exponents, then
multiplication/division, then addition/subtraction. A useful mnemonic
device for this is "Please excuse my dear Aunt Sally": PEMDAS.

Let's practice this. Plug into the following:

-- ab + a for a = 3, b = 2
-- a(b+a) for a = 3, b = 2
-- x - y^2 for x = 4, y = 2
-- x + (-y)^2 for x = 4, y = 2


C. Combining Like Terms

Evaluating algebraic expressions for given numbers is fine, but it's
not necessarily easier than just working with the numbers to begin
with.

However, one thing that makes algebraic expressions useful is that we
can manipulate and simplify them. One way of simplifying algebraic
expressions is combining like terms.

By "like terms", we mean terms that have the same unknown, the same
variable. Consider for example the expression:

3x + 4x

What does this mean in English? 

Notice that 3x just means x plus iteself 3 times, and 4x just means x
plus itself four times. Well, if we write that out, it's just

x + x + x + x + x + x + x

Clearly, a simpler way to write this is just 7x.

We can do similar things with any expressions where the variable is
the same. However, we can't combine dissimilar variables.

Consider, for example:

5x + 3y - 2x - 8y

How could we combine like terms here?

Note that we can't combine the x and the y terms. They represent
different unknowns, and there's no way to write them together as a
single term.


D. Distribution

The opposite of combining like terms is distributing
terms. Distribution means breaking up an algebraic expression into
simpler pieces. For example, suppose we had the expression

4 (x + y)

The 4 gets multiplied by x and y, so we can simplify by writing out
the multplication explicitly:

4 (x + y) = 4x + 4y

This is called distributing the 4 over the x and y.

Often we use distribution in conjunction with combining like
terms. For example:

4 (x + y) + 2 (x - y) = 4x + 4y + 2x - 2y = 6x + 2y

Here's another example:

3a - 2(a + b) = 3a - 2a - 2b = a - 2b

Note a bit of trickiness here: the - sign applies to both the a and
the b in the parentheses. You can check that this is necessary by
plugging in:

for a = 2 and b = 1, 3a - 2(a+b) = 3(2) - 2(2+1) = 6 - 6 = 0
equivalently: a - 2b = 2 - 2(1) = 2 - 2 = 0

On the other hand, if you didn't keep the - sign, you would have said

3a - 2(a+b) = 3a - 2a + 2b = a + 2b
so a + 2b = 2 + 2(1) = 2 + 2 = 4

Obviously that's not correct!

Here are some practice ones for you:

-- simplify 3(x+y) - 2(x + 2y)
-- simplify (3a)^2 + a(2a + 1)
-- simplify 3 (x - 2(x + 2y))


II. Constructing Algebraic Expressions

The trickiest part of algebra is usually constructing algebraic
expressions to express a given operation. We can think of this as a
problem of translation. Given a certain operation in English, what is
the translation into algebra.

Like translating languages this is something of an art, and it
requires finesse. There are no absolute rules. The only thing to do
is practice a lot, in a lot of situations.


A. Price and "coin" problems

One common situation where one has to come up with algebraic
expressions is in situtations involving prices. Here's an example:

Sodas cost $1.50 each, and slices of pizza are $2.50 each. Write an
algebraic expression for the total cost of pizza and sodas. The
solution is cost = 1.50s + 2.50p, where s = number of sodas bought and
p = number of slices of pizza.

Here's another example: I have $20 and I buy some coffee for me and my
friends. Each cup of coffee is $1.25. How much money do I have left?
The answer is 20 - 1.25c, where c is how many cups of coffee I buy.

A closely related type of problem is a "coin" problem: a problem
involving items of different value. For example: what is the total
value of d dimes and n nickels?

Example: A case of soda contains 12 cans. Write an algebraic
expression for the number of cans.


B. Interest and rebates

Closely related to price problem are interest and rebate problems,
which involve computing percentages of amounts of money. An example:
A bank account earns 3% interest per year. Write an expression for the
amount of interest an account containing d dollars earns.

The solution is 0.03d. Note that we have converted the percent into a
decimal, which we always do in writing algebraic expressions.

Here's a tricky, closely related problem: a bank account earns 3%
interest per year. How much money in total will it contain at the end
of the year?

The solution is 1.03d. One way to see this is to note that the amount
it contains will be equal to the amount originally deposited (called
the principle) plus the interest. Thus we have d + 0.03d, which is
1.03d following the rules we just learned.

Practice example: I have a bank account containing b dollars that
earns 2% interest and a savings bond for s dollars that earns 5%
interest. How much money do I have a year later?

A rebate problem is quite similar. For example, what is the cost of an
item whose original price is p if it is sold for 10% off?

The solution is 0.9p. To see why, think of this as the reverse of the
interest problem. The discount is 10% of the original price or
0.1p. The price you have to pay is the original price minus the
discount: p - 0.1p = 0.9p.

Practice example: a store is selling shirts for 5% off and pants for
15% off. What is the total price of a purchase of shirts with an
original price of s dollars and pants with an original price of p
dollars?

Tricky example: a store is selling shirts for p percent off. What is
the price of shirts with an orignal total price of $100? $200?


C. Distance and speed

Another example where algebraic expressions come up is with distances
and times. Here's a simple example: a car is going 60 mph. How far
does it travel in h hours?

This can be phrased in different ways, depending on what is given. A
related problem, for example, is: a car is going 60 mph. How long does
it take to travel m miles?

These can get more complicated, of course, but the general rule is
that distance = rate x time, or time = distance / rate. Some examples:

-- A plane is traveling at 500 mph. How long does it take to travel m
   miles?

-- A plane is traveling at 500 mph and is making a 5000 mile trip. How
   far has it gone after h hours? How much distance does it have left
   to go?


D. Geometry problems

Yet another example is problems involving geometry, and geometric
formulas for area, perimeter, and things like that. An example is: the
area of a triangle is half the base time the height. Write an
algebraic expression for the area.

Practice examples:

-- A rectangular fence is twice as long as it is wide. If its width is
   w, what is its total length?

-- A farmer has 500 feet of fencing, and is using it to make a
   square enclosure. If the enclosure is s feet on a side, how many
   feet of fencing does the farmer have left?

-- A farmer is building a square enclosure s feet on a side, and
   fencing costs $2.50 per foot. How much does it cost to make an
   enclosure of a given side length?


E. Number problems

Finally, algebraic expressions can be given purely as operations on
numbers. Here are some examples:

-- Add 2, double the result, then subtract 4

-- Divide by 5, subtract 2 from the result, then multiply by 3

-- Subtract 1, square the result, then add 1