dansmath > lessons > beginning algebra

Arithmetic (The basic operations and what order to do them)
Prealgebra (Introduction to symbols and expressions)
Beginning Algebra (Simplifying, solving, and graphing)

[ top of page ] Click & choose a topic or just scroll & learn!
• Simplifying Expressions
• Basic Factoring Techniques
• Solving Linear Equations
• Solving Word Problems
• Coordinates and Graphing

• Simplifying Expressions [ top of page ] . . . 9/97 - revised 10/07

We saw in the prealgebra section that we can combine like terms, like 2n + 3n = 5n.

We also used the distributive law to expand numbers; here we distribute the 7:

7*13 = 7(10 + 3) = 7*10 + 7*3 = 70 + 21 = 91.

Also this works with variables; we state the law usually as a(b + c) = ab + ac.

In a typical algebra problem you're asked to expand something like 5(x + 7).

5(x + 7) = 5(x) + 5(7) = 5x + 35. This problem was. . . no problem!

When faced with an expression like 4x + 5(3x - 12), what do we do first?
Let's see: PEMDAS says work in parentheses first, but 3x and 12 are unlike.
Hmm, let's try the distributive law again, but just with the multiplied 5:
4x + 5(3x - 12)
= 4x + 5(3x) - 5(12)
= 4x + 15x - 60
= 19x - 60 . That's our answer, simplified.

What about (4x + 5)(3x - 12) ? Is this the same as 4x + 5(3x - 12) ?

No, the parentheses change it. Here we can use the distributive law twice:

(4x + 5)(3x - 12)

= (4x + 5)(3x) - (4x + 5)(12)

= 12x^2 + 15x - 48x - 60 (remember to change the sign on that last term)

= 12x^2 - 33x - 60 . That worked, but it was long.

Is this the only way? No. The best way? No. Use the "FOIL system":

First, Outside, Inside, Last.

(4x + 5)(3x - 12)
. . . . F . . . . . . O . . . . . . I . . . . . . L . . . .
= (4x)(3x) - (4x)(12) + (5)(3x) - (5)(12)
= 12x^2 - 48x + 15x - 60 = 12x^2 - 33x - 60. Better!

Here's another example: . . .
(n + 3)(n - 3) = n^2 - 3n + 3n - 9 = n^2 - 9.
Notice the "middle terms" cancel, and we're left with what's called the difference of two squares.

In general, (a + b)(a - b) = a^2 - b^2. Also see the basic factoring and more factoring sections.

[ beginning algebra | top of page ]

Basic Factoring [ more factoring | top of page ] . . . 10/07

We saw the distributive law work to expand things out:
6(2x + 7) = 6 * 2x + 6 * 7 = 12x + 42.

The steps can be reversed to give a factorization:
12x + 42 = 6 * 2x + 6 * 7 = 6(2x + 7).

This is called factoring out a common factor. In this case the common factor was 6.

Example: Factor the expression 8 x^3 + 20 x^2.

Look for the common factor:
8 x^3 = 2*2*2*x*x*x and 10x^2 = 2*2*5*x*x.

The common part is 2*2*x*x = 4 x^2. Write it as
8 x^3 + 20 x^2 = (4 x^2)(2x) + (4 x^2)(5) = (4 x^2)(2x + 5).

That's factored!

Another method that works if we're lucky is called the grouping method. This works by
grouping up equal size blocks of terms and factoring out a common factor from each, then
hoping the "left-over factors" are equal. That's where the luck, or sometimes skill, comes in.

Example: Factor by grouping: 12x^2 - 9xy + 8x - 6y .

There's no common factor for all terms, but we can split the pairs:
12x^2 - 9xy + 8x - 6y = 3x(4x - 3y) + 2(4x - 3y) = (3x + 2)(4x - 3y) .

[ beginning algebra | top of page ]

Solving Linear Equations [ top of page ] . . . 9/97 . . . revised 11/07

An equation has to have an equals sign, as in 3x + 5 = 11 .

A solution to an equation is a number that can be plugged in for the variable to make
a true number statement.

For example, putting 2 in for x above in 3x + 5 = 11 gives
3(2) + 5 = 11 , which says 6 + 5 = 11 ; that's true! So 2 is a solution.

But how to start with the equation, and get (not guess) the solution?
We use some principles of equality, such as:
* Adding the same thing (number or variable term) to both sides of an equation
* Subtracting the same thing (number or variable term) from both sides
* Multiplying or dividing both sides by a non-zero quantity.
These all keep the equation "balanced" like a scale.

3x + 5 = 11 . . . our given equation
. . . - 5 . . . - 5 . . . .subtract 5 from each side to get constants on the right
3x . . . = 6 . . . . . . . . . . the intermediate result

3x / 3 = 6 / 3 . . divide both sides by 3 to isolate the x
. . . . x = 2 . . . . . . . the solution (same as before!) . . . . . . We've solved the equation.

The thing that makes this equation linear is that the highest power of x is x^1 ; no
x^2 or other powers, variables in the denominator, square roots, or other funny stuff.
(For "quadratic equations" go to intermediate algebra).

[ beginning algebra | top of page ]

Solving Word Problems [ top of page ] . . . 11/07

A word problem is a mathematical equation phrased in regular language.
Students groan when we tackle word problems, but they also ask,
"What are we gonna use this stuff for?" The answer is, word problems!

There's no one rule for solving all word problems, but you'll increase your
chance of success if you follow a good strategy. Try this series of steps:

 STRATEGY TO TRANSLATE AND SOLVE WORD PROBLEMS Step 1. Read the problem. Figure out what quantities are known, and what are unknown.   Step 2. Decide "What am I being asked to find?" Write this unknown down and then give it a letter name. Put other quantities in terms of this variable.   Step 3. Translate the sentences into mathematical symbols. Look for key words, and form the equation that relates what you want to what you know.   Step 4. Now use algebra techniques to solve the equation for the unknown quantity.   Step 5. Restate your answer in English, using Step 2, to give a clear answer to the main question that was asked.   Step 6. Check your value for the unknown in the original words of the problem. Now you're done!

 Example: (Step 1) If promotional buttons (such as the dansmathcast button at the right) cost me 40 cents each, plus a one-time setup charge of \$25, how many buttons can I get for \$73? (2) We let n = number of buttons; the cost C will be given by (remember 40 cents = 0.40 dollars) (3) Equation is C = 0.40 n + 25.00. Replace C by our cost: 73 = 0.4 n + 25 (4) Subtract 25 from each side: 0.4 n = 73 - 25 = 48 ; now divide by 0.4: n = 48 / 0.4 = 120. (5) I can buy 120 buttons and spread them around like candy to promote my show!

Coordinates and Graphing [ top of page ] . . . 9/97, revised 8/01

A point on the screen you're looking at (like this red one: .) has a "location"
which is measured by how many pixels across and down it is from the upper left corner.
These are its "screen coordinates."

In math, the coordinates of a point in the plane are measured in relation to a "central"
point, the origin, first to the right, then up. The coords are listed as (x, y) for (over, up).
In the picture, the origin is at the + and the red dot has coords (x, y) = (5, 2).

``` . ^y
..4|......
..3|......
..2|......
..1|......
--0+------>x
.-10123456
.-2|......```

Coordinates are also used in writing equations for graphs; we can have a relation between
x and y, and translate that into the language of pictures.
In the first two examples, the functions are "linear" so the graphs are straight lines.
 The x and y coords add up to 2. The y is always twice the x. General function, at most one y per x.

More graphs and their equations are available in the functions and graphs section.

Well, that's it for now. Check back often for new stuff!
Click below for other topics, or visit the ask dan page!
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