Algebra 1 topics - Earlham College
Algebra 1 topics
I. Basic Operations
A. Order of Operations ()
1. Do things in Parentheses First. Example:
|[pic] | |6 × (5 + 3) |= |6 × 8 |= |48 | |
|[pic] | |6 × (5 + 3) |= |30 + 3 |= |33 |(wrong) |
2. Exponents (Powers, Roots) before Multiply, Divide, Add or Subtract. Example:
|[pic] | |5 × 22 |= |5 × 4 |= |20 | |
|[pic] | |5 × 22 |= |102 |= |100 |(wrong) |
3. Multiply or Divide before you Add or Subtract. Example:
|[pic] | |2 + 5 × 3 |= |2 + 15 |= |17 | |
|[pic] | |2 + 5 × 3 |= |7 × 3 |= |21 |(wrong) |
4. Otherwise just go left to right. Example:
|[pic] | |30 ÷ 5 × 3 |= |6 × 3 |= |18 | |
|[pic] | |30 ÷ 5 × 3 |= |30 ÷ 15 |= |2 |(wrong) |
5. How Do I Remember? PEMDAS ! “Please Excuse My Dear Aunt Sally”
|P |Parentheses first |
|E |Exponents (i.e. Powers and Square Roots, etc.) |
|MD |Multiplication and Division (left-to-right) |
|AS |Addition and Subtraction (left-to-right) |
6. Examples
a. Example: How do you work out 3 + 6 × 2 ?
Multiplication before Addition:
First 6 × 2 = 12, then 3 + 12 = 15
b . Example: How do you work out (3 + 6) × 2 ?
Parentheses first:
First (3 + 6) = 9, then 9 × 2 = 18
c. Example: How do you work out 12 / 6 × 3 ?
Multiplication and Division rank equally, so just go left to right:
First 12 / 6 = 2, then 2 × 3 = 6
d. Oh, yes, and what about 7 + (6 × 52 + 3) ?
|7 + (6 × 52 + 3) | |
|7 + (6 × 25 + 3) |Start inside Parentheses, and then use Exponents First |
|7 + (150 + 3) |Then Multiply |
|7 + (153) |Then Add |
|7 + 153 |Parenthesis completed, last operation is an Add |
|160 |DONE ! |
B. Associative, Commutative and Distributive Laws
mutative Laws
The "Commutative Laws" just mean that you can swap numbers over and still get the same answer when you add or multiply.
a + b = b + a
a × b = b × a
Examples:
|You can swap when you add: |3 + 6 = 6 + 3 |
|You can swap when you multiply: |2 × 4 = 4 × 2 |
2. Associative Laws
The "Associative Laws" mean that it doesn't matter how you group the numbers (i.e. which you calculate first) when you add or multiply.
(a + b) + c = a + (b + c)
(a × b) × c = a × (b × c)
Examples:
|This: |(2 + 4) + 5 = 6 + 5 = 11 |
|Has the same answer as this: |2 + (4 + 5) = 2 + 9 = 11 |
|This: |(3 × 4) × 5 = 12 × 5 = 60 |
|Has the same answer as this: |3 × (4 × 5) = 3 × 20 = 60 |
Uses:
Sometimes it is easier to add or multiply in a different order:
|What is 19 + 36 + 4? |
|19 + 36 + 4 = 19 + (36 + 4) = 19 + 40 = 59 |
Or even rearrange a little:
|What is 2 × 16 × 5? |
|2 × 16 × 5 = (2 × 5) × 16 = 10 × 16 = 160 |
3. Distributive Law
The "Distributive Law" is the BEST one of all, but needs careful attention.
It means that you get the same answer when you multiply a group of numbers by something as when you do each multiplication separately, like this:
(a + b) × c = a × c + b × c
Examples:
|This: |(2 + 4) × 5 = 6 × 5 = 30 |
|Has the same answer as this: |2×5 + 4×5 = 10 + 20 = 30 |
|This: |(6 - 4) × 3 = 2 × 3 = 6 |
|Has the same answer as this: |6×3 - 4×3 = 18 - 12 = 6 |
Uses:
Sometimes it is easier to break up a difficult multiplication:
|What is 204 × 6? |
|204 × 6 = 200×6 + 4×6 = 1,200 + 24 = 1,224 |
Or to combine:
|What is 6 × 16 + 4 × 16? |
|6 × 16 + 4 × 16 = (6+4) × 16 = 10 × 16 = 160 |
C. Log functions, powers, exponents
1. Laws of Exponents
a. Here are the Laws (explanations follow):
|Law |Example |
|x1 = x |61 = 6 |
|x0 = 1 |70 = 1 |
|x-1 = 1/x |4-1 = 1/4 |
| | |
|xmxn = xm+n |x2x3 = x2+3 = x5 |
|xm/xn = xm-n |x4/x2 = x4-2 = x2 |
|(xm)n = xmn |(x2)3 = x2×3 = x6 |
|(xy)n = xnyn |(xy)3 = x3y3 |
|(x/y)n = xn/yn |(x/y)2 = x2 / y2 |
|x-n = 1/xn |x-3 = 1/x3 |
| | |
|[pic] |[pic] |
b. Laws Explained
The first three laws above (x1 = x, x0 = 1 and x-1 = 1/x) are just part of the natural sequence of exponents. Have a look at this example:
|Example: Powers of 5 |
| |.. etc.. | |[pic] |
|52 |1 × 5 × 5 |25 | |
|51 |1 × 5 |5 | |
|50 |1 |1 | |
|5-1 |1 ÷ 5 |0.2 | |
|5-2 |1 ÷ 5 ÷ 5 |0.04 | |
| |.. etc.. | | |
You will see that positive, zero or negative exponents are really part of the same pattern, i.e. 5 times larger (or smaller) depending on whether the exponent gets larger (or smaller).
The law that xmxn = xm+n
With xmxn, how many times will you end up multiplying "x"? Answer: first "m" times, then by another "n" times, for a total of "m+n" times.
Example: x2x3 = (xx) × (xxx) = xxxxx = x5
So, x2x3 = x(2+3) = x5
The law that xm/xn = xm-n
Like the previous example, how many times will you end up multiplying "x"? Answer: "m" times, then reduce that by "n" times (because you are dividing), for a total of "m-n" times.
Example: x4-2 = x4/x2 = (xxxx) / (xx) = xx = x2
(Remember that x/x = 1, so every time you see an x "above the line" and one "below the line" you can cancel them out.)
This law can also show you why x0=1 :
Example: x2/x2 = x2-2 = x0 =1
The law that (xm)n = xmn
First you multiply x "m" times. Then you have to do that "n" times, for a total of m×n times.
Example: (x3)4 = (xxx)4 = (xxx)(xxx)(xxx)(xxx) = xxxxxxxxxxxx = x12
So (x3)4 = x3×4 = x12
The law that (xy)n = xnyn
To show how this one works, just think of re-arranging all the "x"s and "y" as in this example:
Example: (xy)3 = (xy)(xy)(xy) = xyxyxy = xxxyyy = (xxx)(yyy) = x3y3
The law that (x/y)n = xn/yn
Similar to the previous example, just re-arrange the "x"s and "y"s
Exemple: (x/y)3 = (x/y)(x/y)(x/y) = (xxx)/(yyy) = x3/y3
The law that [pic]
To understand this, just remember from fractions that n/m = n × (1/m):
Example: [pic]
And That Is It
If you find it hard to remember all these rules, then remember this:
you can always work them out if you understand the three ideas at the top of this page.
Oh, One More Thing ... What if x= 0?
|Positive Exponent (n>0) |0n = 0 |
|Negative Exponent (n 0 and n > 0
Basically, what we are saying here is that another way to write the log of a product is to take the log of the first base and add it to the log of the second base.
Hmmmm, why don’t I just take the product of their logs??????
Wait a minute, I remember my teacher saying above that logs are another way to write exponents - WHENEVER I WAS MULTIPLYING LIKE BASES, I ADDED MY EXPONENTS - SO I’M GOING TO HAVE TO ADD MY LOGS - EUREKA!!!!
Note that even though m and n are not the bases of the log itself, they can each be written as base b to an exponent, because of the definition of logarithms.
Here is a quick illustration of how this property works:
[pic]
Property 2 - Quotient Rule m > 0 and n > 0
Basically, what we are saying here is that another way to write the log of a quotient is to take the log of the numerator and subtract the log of the denominator.
So here, we have to remember that when we were dividing like bases, we subtracted our exponents - so we do the same type of thing with our logs.
Here is a quick illustration of how this property works: [pic]
Property 3 - Power Rule m > 0
Basically, what we are saying here is that whenever you have a 2nd exponent inside the log - remember the log itself is an exponent - then you can pull it out front and multiply it times the log.
Wow, that looks a little different, but again it comes from the fact that logs are another way to write exponents.
Remember that when we had a base raised to 2 powers that we would multiply those 2 exponents together. That is what we are doing here. Again, even though m is not the base of the log, it can be written as b to an exponent (based on the log definition) and the log itself is an exponent so we have a double exponent - so we multiply our exponents together.
Here is a quick illustration of how this property works: [pic]
|Example 1: Expand [pic] as much as possible. Evaluate without a calculator where possible. |
When they say to "expand", they mean that they've given you one log expression with lots of stuff inside it, and they want you to use the log rules to take the log apart into lots of separate logs, each with only one thing inside. That is, they've given you one log with a complicated argument, and they want you to convert this to many logs, each with a simple argument.
Note how there is no base written. Does that mean there is no base? Not in the least.
What would the base be in this problem? If you said 10 you are correct. This is known as the common log.
|[pic] | |
| |*Use the product rule |
| |*Use the definition of logs to simplify |
| |*2 is the exponent needed on 10 to get 100 |
|[pic]Example 2: Expand [pic] as much as possible. Evaluate without a calculator where possible. |
Note how there is a ln and no base written.
What would the base be in this problem? If you said e you are correct. This is known as the natural log.
|[pic] | |
| |*Use the quotient rule |
| |*Use the definition of logs to simplify |
| |*4 is the exponent needed on to get e^4 |
When they tell you to "simplify" or “condense” a log expression, this means they will have given you lots of log terms, each containing a simple argument, and they want you to combine everything into one log with a complicated argument. "Simplifying" or “condensing” is the opposite of "expanding".
|Example 3: Condense [pic] into one logarithmic expression. Evaluate without a calculator where possible. |
This time we are going in reverse of what we did in examples 1 - 4. However, you can use the same properties we used on them. You can use those properties in either direction.
What is the base in this problem? This time the base is e. Make sure that you keep that same base throughout the problem.
|[pic] |*Use the product rule |
|Example 4: Condense [pic] into one logarithmic expression. Evaluate without a calculator where possible. |
Again we are going in reverse of what we did in examples 1 - 4.
What is the base in this problem? This time the base is e. Make sure that you keep that same base throughout the problem.
|[pic] | |
| |*Use the power rule |
| | |
| |*Use the product rule |
| | |
II. Simplifying Algebraic expressions
A. Expanding: “Expanding” means removing the ( ) ... but you have to do it the right way! ( ) are called "parentheses" or "brackets" Whatever is inside the ( ) needs to be treated as a "package". So when you multiply, you have to multiply by everything inside the "package". Note that in Algebra putting two things next to each other usually means to multiply. So 3(a+b) means to multiply 3 by (a+b)
• Example: a(b+c) = ab + ac
• Example: 3(x+6) = 3x + 3(6) = 3x + 18
You will often need to multiply negatives (a negative times a positive gives a negative, but multiplying two negatives gives a positive):
• Example: -a(b-c) = -ab + ac
• Example: -3(x-6) = -3x + 3(6) = -3x + 18
• Example: 2x(x-6) = 2x2 -2(6) = 2x2 -12
If there are three terms inside, they’re treated exactly the same way!
• Example: a(2-b+6) = 2a – ab +6a
Fractions in Algebra: You can add, subtract, multiply and divide fractions in algebra in the same way that you do in simple arithmetic.
1. Adding Fractions:
To add fractions there is a simple rule:
[pic]
Example:
|[pic] |
2. Subtracting Fractions
Subtracting fractions is very similar to adding, except that the + is now -
[pic]
Example:
|[pic] |
3. Multiplying Fractions
Multiplying fractions is the easiest one of all, just multiply the tops together, and the bottoms together:
[pic]
Example:
|[pic] |
4. Dividing Fractions
To divide fractions, first "flip" the fraction you want to divide by, then use the same method as for multiplying:
[pic]
Example:
[pic]
Adding and Subtracting Polynomials
A polynomial looks like this:
|[pic] |
|example of a polynomial |
|this one has 3 terms |
To add polynomials you simply add any like terms together .. so what is a like term?
Like Terms: “Like terms” are terms whose variables (and their exponents such as the 2 in x2) are the same. In other words, terms that are “like” each other.
Examples:
|Terms |Why are they "Like Terms" |
|[pic] |7x |x |-2x |because the variables are all x |
|[pic] |(1/3)xy2 |-2xy2 |6xy2 |because the variables are all xy2 |
2 Adding and Subtracting the Polynomials
Two Steps:
• Place like terms together
• Add (or subtract) the like terms
Example: Add 2x2 + 6x + 5 and 3x2 - 2x – 1
Place like terms together: 2x2 + 3x2 + 6x - 2x + 5 - 1
Add the like terms: (2+3)x2 + (6-2)x + (3-1) = 5x2 + 4x + 4
Example: Subtract 2x2 + 6x + 5 and 3x2 - 2x – 1
Distribute the negative sign: (2x2+6x+5) – (3x2-2x-1)=2x2+6x+5-3x2+2x+1
Place like terms together: 2x2 - 3x2 + 6x + 2x + 5 + 1
Add the like terms: (2-3)x2 + (6+2)x + (5+1) = -x2 + 8x + 6
3 Multiplying the Polynomials
A polynomial looks like this:
|[pic] |
|example of a polynomial |
|this one has 3 terms |
To multiply a polynomial:
• multiply each term in one polynomial by each term in the other polynomial
• add those answers together, and simplify if needed
Let us look at the simplest cases first:
1 term × 1 term (monomial times monomial)
To multiply one term by another term, first multiply the constants, then multiply each variable together and combine the result, like this:
• Example: 2xy(4y) = 2·4·xy·y = 8xy2
(Note: I used "·" to mean multiply. In Algebra we don't like to use "×" because it looks too much like the letter "x")
1 term × 2 terms (monomial times binomial)
Multiply the single term by each of the two terms, like this:
• Example: 2x(x+3xy) = 2x·x+2x·3xy = 2x2+6x2y
2 terms × 2 terms (binomial times binomial)
You can multiply them in any order so long as each of the first two terms gets multiplied by each of the second two terms.
But there is a handy way to help you remember to multiply each term called "FOIL" that stands for "Firsts, Outers, Inners, Lasts":
|[pic] |Firsts: ac |
| |Outers: ad |
| |Inners: bc |
| |Lasts: bd |
So you multiply the "Firsts" (the first terms of both polynomials), then the "Outers", etc.
Let’s try this on more complicated examples:
• Example: (2x-3)(x+2)=2x2 + 4x -3x -6 = 2x2 + x – 6
• Example: (2x-3)(2x+3)=4x2 + 6x -6x -9 = 4x2 - 9
• Example: (2x-3)(ax+b)=2ax2 + 2xb – 3ax -3b
III. Solving Algebra Equations: One Equation, one unknown
Algebra is just like a puzzle where you start with something like "x-2 = 4" and you want to end up with something like "x = 6".
But instead of saying "obviously x=6", use this neat step-by-step approach:
• Work out what to remove to get "x = ..."
• Remove it by doing the opposite (adding is the opposite of subtracting)
• Do that to both sides
Here is an example:
|We want to remove the "-2" |To remove it, do the opposite, in this |Do it to both sides: |Which is ... |Solved! |
| |case add 2: | | | |
|[pic] |[pic] |[pic] |[pic] |[pic] |
Just remember this:
|To keep the balance, what you do to one side of the "=" |
|you should also do to the other side! |
Example:
|X |+ |5 |= |12 |
|Start with: |x + 5 = 12 |
|What you are aiming for is an answer like "x = ...", and the plus 5 is in the way of that! | |
|If you subtract 5 you can cancel out the plus 5 (because 5-5=0) | |
|So, let us have a go at subtracting 5 from both sides: |x+5 -5 = 12 -5 |
|A little arithmetic (5-5 = 0 and 12-5 = 7) becomes: |x+0 = 7 |
|Which is just: |x = 7 |
| |Solved! |
|(Quick Check: 7+5=12) | |
We do the same thing when we have a number divided by x, but you have to know that dividing is the opposite of multiplying. Have a look at this example:
|We want to remove the "4"|To remove it, do the |Do it to both sides: |Which is ... |Solved! |
| |opposite, in this case divide| | | |
| |by 4: | | | |
|[pic] |[pic] |[pic] |[pic] |[pic] |
Example:
|X |/ |3 |= |5 |
|Start with: |x/3 = 5 |
|What you are aiming for is an answer like "x = ...", and the divide by 3 is in the way of that! | |
|If you multiply by 3 you can cancel out the divide by 3 (because 3/3=1) | |
|So, let us have a go at multiplying by 3 on both sides: |x/3 ×3 = 5 ×3 |
|A little arithmetic (3/3 = 1 and 5×3 = 15) becomes: |1x = 15 |
|Which is just: |x = 15 |
|(Quick Check: 15/3 = 5) | |
Example:
|x |/|3 |+ |2 |= |5 |
It might look hard, but not if you solve it in stages.
First let us get rid of the "+2":
|Start with: |x/3 + 2 = 5 |
|To remove the plus 2 use minus 2 (because 2-2=0) |x/3 + 2 -2 = 5 -2 |
|A little arithmetic (2-2 = 0 and 5-2 = 3) becomes: |x/3 + 0 = 3 |
|Which is just: |x/3 = 3 |
Now, get rid of the "/3":
|Start with: |x/3 = 3 |
|If you multiply by 3 you can cancel out the divide by 3: |x/3 ×3 = 3 ×3 |
|A little arithmetic (3/3 = 1 and 5×3 = 15) becomes: |1x = 9 |
|Which is just: |x = 9 |
|(Quick Check: 9/3 + 2 = 3+2 = 5) | |
What if we have an x2?
|Start with: |x2 + 2 = 27 |
|To remove the plus 2 use minus 2 (because 2-2=0) |x2 + 2 -2 = 27 -2 |
|A little arithmetic (2-2 = 0 and 27-2 = 25) becomes: |x2 = 25 |
|Take the square root of both sides |x = ±5 |
What if we have an x2 and an x term? There are several ways to solve problems like this. One of the most important ways is the Quadratic Equation
|This is a Quadratic Equation: |
|[pic] |
|(a, b, and c can have any value, except that a can't be 0.) |
Examples of Quadratic Equations:
|2x2 + 5x + 3 =0 | |In this one a=2, b=5 and c=3 |
| | | |
|x2 – 3x=0 | |This one is a little more tricky: |
| | |Where is a? In fact a=1, because we don't usually write "1x2" |
| | |b=-3 |
| | |And where is c? Well, c=0, so is not shown. |
|5x – 3 = 0 | |Oops! This one is not a quadratic equation, because it is missing x2 (in other words a=0, and that |
| | |means it can't be quadratic) |
Why is it special?
Quadratic equations can be solved using a special formula called the Quadratic Formula:
[pic]
|[pic]|The "±" means you need to do a plus AND a minus, and therefore there are normally TWO solutions ! |
|[pic]|The blue part (b2 - 4ac) is called the discriminant, because it can "discriminate" between the possible types of answer: |
| |if it is positive, you will get two solutions |
| |if it is zero you get just ONE solution, |
| |and if it is negative you get two solutions that include Imaginary Numbers . |
Solving
To solve, just plug the values of a, b and c into the Quadratic Formula, and do the calculations.
Example: Solve 5x² + 6x + 1 = 0
Quadratic Formula: x = [ -b ± √(b2-4ac) ] / 2a
Coefficients are: a = 5, b = 6, c = 1
Substitute a,b,c: x = [ -6 ± √(62-4×5×1) ] / 2×5
Solve: x = [ -6 ± √(36-20) ]/10 = [ -6 ± √(16) ]/10 = ( -6 ± 4 )/10
Answer: x = -0.2 and -1
(Check:
5×(-0.2)² + 6×(-0.2) + 1 = 5×(0.04) + 6×(-0.2) + 1 = 0.2 -1.2 + 1 = 0
5×(-1)² + 6×(-1) + 1 = 5×(1) + 6×(-1) + 1 = 5 - 6 + 1 = 0)
Quadratic Equation In Disguise
Some equations may not look like quadratic equations, but with a little clever work they can be made into one:
|In disguise |What to do |In standard form |a, b and c |
|x2 = 3x -1 |Move all terms to left hand side |x2 - 3x + 1 = 0 |a=1, b=-3, c=1 |
|2(x2 - 2x) = 5 |Undo parentheses |2x2 - 4x - 5 = 0 |a=2, b=-4, c=-5 |
|x(x-1) = 3 |Undo parentheses |x2 - x - 3 = 0 |a=1, b=-1, c=-3 |
|5 + 1/x - 1/x2 = 0 |Multiply by x2 |5x2 + x - 1 = 0 |a=5, b=1, c=-1 |
IV. Solving Algebra Equations: Two Equations, two unknowns: modified from ()
There are several ways of solving a system where you have two equations AND two unknowns (say x and y). We’re going to do it by substitution. This method works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this into the other equation, "substituting" for the chosen variable and solving for the other. Then you back-solve for the first variable. Here is how it works:
Solve the following system by substitution.
2x – 3y = –2
4x + y = 24
We can start with either x or y, or either the first or second equation. It’s easiest to solve for a variable that is by itself, so let’s solve for “y” from the 2nd equation:
4x + y = 24
y = –4x + 24
Now I'll plug this in ("substitute it") for "y" in the first equation, and solve for x:
2x – 3(–4x + 24) = –2
2x + 12x – 72 = –2
14x = 70
x = 5
Now I can plug this back into either equation, and solve for y. But since I already have an expression for "y =", it will be simplest to just plug into this:
y = –4(5) + 24 = –20 + 24 = 4
So the answer is: x=5, y=4
If I had substituted into the same equation as I'd used to solve for "y =", I would have gotten a true, but useless, statement: 2007 All Rights Reserved
4x + (–4x + 24) = 24
4x – 4x + 24 = 24
24 = 24
V. Linear Equations: modified from ()
The GENERAL equation of a straight line is given in the form:
y = mx + b
What does the equation stand for?
|[pic] | |[pic] |
| | | |
| | |Slope |
| | |Y Intercept |
| | | |
| | | |
|y = how far up |
|x = how far along |
|m = Gradient or Slope (how steep the line is) |
|b = the Y Intercept (where the line crosses the Y axis) |
Slope of a Straight Line:
The slope of a straight line shows how steep a straight line is. The method to calculate the slope is:
|Slope = |Change in Y |
| | |
| |————— |
| | |
| |Change in X |
| | |
An example of calculating slope:
|[pic] |
|The slope of the above line = |3 | = 1 |
| | | |
| |— | |
| | | |
| |3 | |
| | | |
|The slope of the above line is equal to 1 |
Different Types of slopes: There are different types of slopes of a straight line. Below are listed most of the ones that you will come up against. Remember, starting from the left end of the line going across to the right is positive, but going across to the left is negative. And up is positive, but down is negative:
| | | | |
| | | | |
| |slope = |3 | = 0.6 |
| | | | |
| | |— | |
| | | | |
| | |5 | |
| | | | |
| | | | |
| | | | |
| | | | |
| | | | |
| |slope = |4 | = 2 |
| | | | |
| | |— | |
| | | | |
| | |2 | |
| | | | |
| | | | |
| | | | |
| | | | |
| | | | |
| |slope = |-4 | = –2 |
| | | | |
| | |— | |
| | | | |
| | |2 | |
| | | | |
| | | | |
| | | | |
| | | | |
| | | | |
| |slope = |0 | = 0 |
| | | | |
| | |— | |
| | | | |
| | |5 | |
| | | | |
| | | | |
| | | | |
Y Intercept of a Straight Line: The Y intercept of a straight line is simply where the line crosses the Y axis. This is “b” in the linear equation.
Example:
[pic]
In the above diagram the line crosses the Y axis at 1.
So, the Y intercept is equal to 1.
Knowing how to calculate slope and the y intercept, we can work out the equation of a straight line:
Example 1
|m | = |2 | = |2 |
| | | | | |
| | |— | | |
| | | | | |
| | |1 | | |
| | | | | |
|b = 1 | | | | |
|Therefore |y = 2x + 1 |
Example 2
|m | = |3 | = |–3 |
| | | | | |
| | |— | | |
| | | | | |
| | |-1 | | |
| | | | | |
|b = 0 | | | | |
|Therefore |y = –3x |
This gives us y = –3x + 0
We don’t need the zero!
Material taken and modified from the following sources:
I. Basic Operations
A. Order of Operations ()
1. Pierce, Rod. "Order of Operations - PEMDAS" Math Is Fun. Ed. Rod Pierce. 20 Sep 2007. 5 Feb 2008
B. Associative, Commutative and Distributive Laws
1. Pierce, Rod. "Associative, Commutative and Distributive Laws" Math Is Fun. Ed. Rod Pierce. 16 Sep 2006. 5 Feb 2008
C. Log functions, powers, exponents
1. Pierce, Rod. "Laws of Exponents" Math Is Fun. Ed. Rod Pierce. 26 Sep 2007. 5 Feb 2008
2. Log functions -
col_alg_tut44_logprop.htm
II. Simplifying Algebraic expressions
A. Expanding:
Pierce, Rod. "Algebra - Expanding" Math Is Fun. Ed. Rod Pierce. 12 Jul 2007. 5 Feb 2008
B. Fractions in Algebra:
Pierce, Rod. "Fractions in Algebra" Math Is Fun. Ed. Rod Pierce. 24 Nov 2007. 5 Feb 2008
C. Adding and Subtracting Polynomials
Pierce, Rod. "Adding and Subtracting Polynomials" Math Is Fun. Ed. Rod Pierce. 23 Jan 2008. 5 Feb 2008
D. Multiplying the Polynomials
Pierce, Rod. "Multiplying Polynomials" Math Is Fun. Ed. Rod Pierce. 7 Nov 2007. 5 Feb 2008
III. Solving Algebra Equations: One Equation, one unknown
Pierce, Rod. "Introduction to Algebra" Math Is Fun. Ed. Rod Pierce. 16 Sep 2006. 5 Feb 2008
Quadratic Equation
Pierce, Rod. "Quadratic Equation" Math Is Fun. Ed. Rod Pierce. 14 Feb 2007. 5 Feb 2008
IV. Solving Algebra Equations: Two Equations, two unknowns:
V. Linear Equations:
Pierce, Rod. "Equation of a Straight Line" Math Is Fun. Ed. Rod Pierce. 24 May 2007. 5 Feb 2008
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