Algebra

See the English forum description? I think that you get this.
User avatar
K
Posts in topic: 1
Posts: 648
Joined: Wed Oct 06, 2010 8:33 pm
Guild:

Algebra

Post by K » Mon May 09, 2011 7:15 pm

Let's start from the ground up, since algebra is just a series of concepts.

Keep in mind the basic relations of mathematical operations:
Subtraction is negative addition.
Division is "negative" (reverse, really) multiplication.
Multiplication is repeated addition (or, when negative, repeated subtraction).
Division is repeated, proportional subtraction.
Exponents are repeated multiplication (or, when negative, repeated division).
Roots are divisions of exponents.
Fractions are division.

If you keep these relations in mind, everything else will become common sense.

Algebra is all about looking at an equation or inequality and moving things around until the equation shows you what you want to know. That's pretty much it. With word problems, you first figure out how to put the problem in math terms, then solve it. You may make two or more problems with common variables and substitute the answers on down until you get what you want to know. It's really simple, once you get the feel for it.

Now, you may be wondering what I'm talking about when I say that you just move things around. It's a simple matter of negating what you don't want to know in order to say that the opposite is what you want to know. Let's run this on a simple example to see what I mean:

2 = x - 2

So, we're told to solve for x. It seems like it's always either x or y that we're solving for... There's a reason for that, but it's irrelevant, right now. Anyway, if we want to know what x equals ("Equals" and "is" are interchangeable; in either case, we're speaking of an equation. Everything that "is" something "equals" just that.), we have to get it alone on one side of the equal sign. After all, we don't want to know what x - 2 equals; we want to know what x equals.
So, we have x - 2, but we want x. Well, why not just add 2?
2 = x
There! Not quite, though. We got x alone, but we didn't finish what we started. We added 2, but where did that 2 come from? Well, okay, we can't just go randomly adding stuff in until we get what we want; that's not logical. So, let's say that we then negate the added 2 by subtracting 2...
2 = x -2
Well, damn. That did a whole lot of nothing. All right, then adding 2 and immediately subtracting it shows that we can negate things, but we can do that all day and get nowhere. Wait, what if we subtract it from the other side?
0 = x
Not bad! Still wrong, though. Okay, how did it get to the other side? We just . . . kinda' put it there, huh? Not really a logical way to go about it, right? Right. Well, wait, what if we do that negatively? In stead of subtracting 2 from the left side after adding it to the right, let's add it to both sides.
4 = x
Is that right? It is?! Aha! Wait, how can we prove that this works? We proved that we can negate an operation by doing its reverse to the same side; that's how we got back to the original problem, so it must have been a logical thing to do. This is different, though.
Well, look at this:
2 + 2 = 4
Subtract 2 from both sides.
2 = 2
It's right, isn't it? Here's another:
10 - 3 = 7
Add 3 to both sides.
10 = 10
It works every time! What about with multiplication and division?
10 * 10 = 100
Divide both by 10.
10 = 10
Okay, now the other way:
1,000 / 5 = 200
Multiply both by 5 . . .
1,000 = 1,000
Exponents and roots?
24 = 2 * 2 * 2 * 2 = 16
Let's get rid of the middle part, since it's just there to show people how exponents work, then put both terms to a power of 1/4 (which is the same as putting both terms under a fourth root sign).
21 = 2
Haha, more success. You can take my word for it that it works the other way, too.
All right, so we've proved that we can do anything to a problem as long as we either do its inverse operation to the same side or doing the same thing to the other side. So, we look back on our problem and how we solved it:
2 = x - 2
+ 2 to both sides
4 = x
Now, it looks to be more obvious, doesn't it? That's all that it takes; enough practice, and all algebra becomes so obvious.

Of course, those are all such basic examples. What happens when there's more than one variable?

y = 3x + 5

Well, if they ask you to solve for x, it's not much different.
When you're doing some basic math, you follow the order of operations as follows:
Please Excuse My Dear Aunt Sally = Parentheses, Exponents, Multiplication and Division, Addition and Subtraction
That's when you're trying to evaluate basic math, like (3 + 4)3 / (6 - 1), but when you're moving things around (i.e. performing algebra), you need to go in reverse.

So, we subtract 5 from both sides.
y - 5 = 3x
Then, we divide both sides by 3.
(y - 5) / 3 = x
That's it. You solved for x.
Yes, really.
No, I mean it. That's the answer.
I know that it doesn't feel complete, but if you're just given an equation like that, that's all that you can do.
Yup.

Okay, so what if they then say that y = 5? Or, what if they give you another equation, and you use algebra to make it say that y = 5? Substitute it in, then use the same process.

(5) = 3x + 5
0 = 3x
0/3 = x
0/3 = 0, so x = 0

There. See, when we have one variable and an equation to relate it to another, we can use the solution of one to solve the other. However, if we have at least one unknown variable, we're dead in the water. Remember this, because that's a rule that you'll have to keep in mind throughout all math from here on out:
If there is at least one unknown factor, we have to either bring in another equation to solve for it, or stop once we get whatever they asked for. If they ask for a more accurate solution, we're forgetting an equation that we can bring in (usually something of geometry).

The only other thing that you'll have to keep in mind about your order of operations (for now) is that exponents come before parentheses unless they're in the parentheses. If you have
y = (x + 2)3 - 6,
you need to add the six, take the cubic root , then subtract 2.
y1/3 - 2 = x
However, if you have
y = (x3 + 2) - 6,
you must first add the six, then you can get rid of the parentheses, so you can subtract the 2, THEN take the cubic root.
(y + 4)1/3

Okay to be sure that you're all with me on the fundamental concepts, here's some homework for everyone who's interested in this lesson:

1) Where y = (7x -16)2, solve for x.

2) Where t = 7 - (3n + 9), solve for n.

3) Where 2x - 4 = r, and
5r + 9 = y,
solve for y.

Please begin your post by stating that it is "Homework 1" or the like, then give your answers in tags. It should look something like this:

Code: Select all

First Homework
[note]1) answer
2) answer
3) answer[/note]
Showing your work is optional. It may help us to see where you went wrong if you get one wrong.

Return to “Math”

×

Who is online

Users browsing this forum: No registered users and 1 guest