The new forums will be named Coin Return (based on the most recent
vote)! You can check on the status and timeline of the transition to the new forums
here.
We now return to our regularly scheduled PA Forums. Please let me (Hahnsoo1) know if something isn't working. The Holiday Forum will remain up until January 10, 2025.
Posts
EDIT: Beaten by Demerdar. It's so easy to take this stuff for granted. It always kills me that I haven't the slightest idea how to explain it to people who have difficulties with it.
I think you may be getting confused about the difference between fractions and division. You can manipulate fractions as much as you want, as long as you are only multiplying or dividing the whole fraction by something that equals 1.
So if you have : 2q/3
You can do something this (if you really wanted to):
2*q/3 * 3/3
because 3/3 equals 1 (anything divided by itself equals one).
If you multiply or divide anything by 1, it doesn't change it.
If you have something like this: 3/2
You can do this:
3/2 * 5/3
which equals:
5/2
However, you are multiplying by 5/3, which does not equal 1, so when you are doing that you are changing the value of 3/2 to something else.
Take an equation: 2q + 4 = q - 2
You can multiply any one of those terms by anything that equals 1. However, say you want to do this:
Take 2q and multiply it by 1/q to get 2.
1/q does not equal 1 (unless you absolutely know that q = 1), so if you do that, then the left side of the equation no longer equals the right side, since you've changed something on the left (2q) without changing the right side the exact same amount.
So, you need to use other methods of isolating your variable.
Say you've got the following:
3x+5=7x-3
You can divide everything by 3 if you want, it just doesn't really get you anywhere:
(3x+5)/3=(7x-3)/3
x+5/3=7/3x-1
So we have to use addition and subtraction:
(x+5/3)-x=(7/3x-1)-x
5/3=4/3x-1
(5/3)+1=(4/3x-1)+1
8/3=4/3x
x=2
We can always plug x back into the original equation to test it:
3(2)+5=7(2)-3
11=11
Now, let's look at the original problem, again, but we'll substitute the actual answer in for the variables.
3(2)+5=7(2)-3
So we want to divide everything by 3. What's 3(2) divided by 3? 2. 2=x. So when we divide 3x by 3, we get x.
I've only used that trick to obtain a form equation for derivatives, integrals, and transforms.
Everything makes sense and then you take calculus and all of a sudden 0 and infinity become completely different things.
Also, keep in mind that you can rearrange the terms on either side if that makes it easier for you -- I know I like to put the terms with the variable first on either side even if that makes the equation look less "pretty".*
So for solving that last equation you put up there I would work it out like this:
-3x + 1 = 5 - 2x
Rearrange it for ease of looking at:
-3x + 1 = -2x + 5
Now I add 2x to both sides in order to cancel out the -2x on the right.
-3x + 2x + 1 = -2x + 2x + 5
Simplify:
-x + 1 = 5
Subtract 1 from both sides
-x + 1 - 1 = 5 - 1
Simplify:
-x = 4
Multiply both sides by -1
-x * -1 = 4 * -1
Simplify:
x = - 4
ETA: Don't forget to plug whatever you get back in to your original equation to make sure you got the right answer!
-3(-4) + 1 = 5 - 2(-4)
12 + 1 = 5 + 8
13 = 13
YAY!
*Also, I think your teacher may not be emphasizing the actual rules behind this enough, so pardon the longwindedness -- the reason this is possible is because addition and multiplication are commutative within their own operations. (Subtraction is considered the same operation as addition -- because for subtraction, you are just adding by a negative number (ie x - 2 is equal to x + -2). Similarly, division is considered the same operation as multiplication -- because for division you are multiplying by 1 over the number (ie x/2 is equal to x * 1/2).)
Fine, so they're "commutative". What the heck does that mean? It means that it doesn't matter what order you do them in, you'll still get the same result.
So: you can rearrange things that are being multiplied together: 1 * 2 * 3 is the same as 2 * 3 * 1. Or, with variables: -2 * x is equal to x * -2. ALSO: x / 2 is equal to 1/2 * x .
And: you can rearrange things that are being added: 1 + 2 + 3 is the same as 2 + 3 + 1. With variables: 2 + x is the same as x + 2. ALSO: x - 2 is the same as -2 + x.
However you can't reorder things between the two. So: 3x + 2 is the same as 2 + 3x or (x * 3) + 2 or 2 + (x * 3). I'm allowed to do that because I'm only reordering things within the addition or the multiplication (first one, just switching addition order, second one, just switching multiplication, last one switching both separately). But 3x + 2 isn't the same as 3 + 2x or x + (2 *3) or whatever other combinations, because I can't reorder things across the addition and the multiplication.
....I hope that made some sort of sense.
Edit:
If you multiply a variable (or anything else) by itself, it is squared.
That is, x * x = x^2.
If you're just multiplying it by a number, it's just the number times x (ie x * 2 = 2x).
Anytime you multiply a variable once by itself it becomes squared. For example:
y * y = y^2
You can extend this to cube something or even go to an arbitrary exponent:
y * y * y = y^3
y * y * y * y = y^4
Also, you can do stuff like this:
y^2 * y^2 = y^4
because:
y^2 = y * y
so:
y^2 * y^2 = y * y * y *y = y^4
When you add or subtract variables with different exponents, you keep them separate.
y^2 + y does not equal 2y^2 or 2y or y^3 or anything like that. y^2 + y is about as simple as you can get.
You can put them together if they are the same exponent.
y^2 + y^2 = 2 * y^2 or 2y^2
If you've got two different variables going to the same exponent, you leave them apart, though.
y^2 + x^2 does not really simplify any further.
This may be beyond what you're doing right now though. You'll probably learn this stuff in a few months.
Here are a couple basic things rules that should get you through all of these questions you are having difficulty with:
1) BEDMAS
The order of operations for ANY math problem that ever existed
B - Brackets
E - Exponents
D and M - Division/Multiplication in the order that they occur (Ex. 2 x 3 / 4 First multiply then divide because that is the order in which they occur)
A and S - Addition/Subtraction in the order that they occur (Ex. 2 + 3 - 1 First add then subtract because that is the order in which they occur)
2) Balance/ Equality
Whatever you do to one side in a mathematical equation (Meaning there is an = sign somewhere in the problem) you must do to the other in order to keep it balanced.
So say we wanted to simplify the following:
3a - 6 = 12a + 9
In order to simplify properly we must divide ALL terms (Any time you see an add/subtract they are seperating two terms, however division and multiplication do not seperate terms) on BOTH sides by the same thing.
So we'll divide by 3...
a - 2 = 4a +3
Now we want to isolate the variable, so we move it all to one side (making sure to change the signs along the way).
-2 - 3 = 4a - a You could also write it like this: 4a - a = -2 -3 OR this: -4a + a = 3 + 2 although that
-5 = 3a gives you a dumb
negative
Now we still need to isolate our variable so we just divide our -5 by the 3 in order to take it out and weve got our answer: -5/3 = a
Thats just one example of where you need to keep the equation balanced, but just remember that when dividing out/simplifying that you always do it to both sides. Simplifying is not always a valid move to make as it will sometimes give you fractions if the terms dont have any common factors and no one likes fractions :P
3) Sign changes
As I mentioned in brackets above ALWAYS remember to change your sign values when you move a term across the equals sign (This is incredibly important especially when proving identities later on)
For example:
2a = 5 - 3a
2a + 3a = 5
5a = 5
Now dividing out the five to isolate the variable (which in this case just HAPPENS to be a common factor) we get a = 1 of course.
4) A note on Equality
Remember that when you are changing the base of a fraction so that you can add it to another fraction you DO NOT multiply all terms because when you change the base the value of the fraction stays the same, it has merely changed its disguise if you will :P
However, if you are changing the base of a more advanced fraction that has multiple terms on top then you must be sure to multiply all the terms on top of the fraction like so:
4a + 6 2a + 5
______ + _______ = 0 We want the bases the same so we can add the terms on top
2 4
So, we multiply the base of the fraction on the left by 2 of course to make them the same, but we must do this also to all the terms above the fraction line too which gives us:
8a + 12 2a + 5
_______ + _______ = 0
4 4
Writing the problem like this however is poor form, not incorrect, just inneficient and poor, and depending on how much of a hardass your teacher is he may deduct a half mark or something so be sure to write it properly like so:
8a + 12 + 2a + 5
______________ = 0 Since they have the same base you just move them together essentially and collect
like terms
4
Now we have:
10a + 17
________ = 0
4
Not an example with a smooth finish I know but its not a big deal
(10/4)a + 17/4 = 0 (simplify the 10a/4)
(5/2)a = -17/4 Yes the zero disappears :P
And then you would divide out the 5/2 to isolate your variable once more, however because you cannot divide two fractions you must flip the fraction (called a reciprocal) that you are dividing out and MULTIPLY it instead, so you would essentially have this:
a = -17/4 X 2/5
Note that in this case you do not have to change the sign of the fraction because you aren't actually moving it, you're dividing it out and it just happens to have moved as a result, so you dont change the sign!
Alright, now we have...(drum roll) a = -34/20! YAY! But not really yay, its a really lame answer, though it is correct. However, this answer is also not in proper form either, clearly it is not in lowest terms :O so we must do that as our last step! Finally we have reached our answer after simply reducing the fraction by 2:
a = -17/10
Whew, that's quite the silly fraction if I do say so myself, but it is correct, so keep in mind that if you get something weird like that it is correct so dont lose confidence!
Now just to be sure...we will check our answer just in case as it is rather odd:
4(-17/10) + 6 2(-17/10) + 5
____________ + ____________ = 0 *When using BEDMAS to figure your order of ops here
keep in mind always that you must figure out the top THEN
2 4 divide the bottoms as it applies to ALL the terms on top.
You can think of it this way: ( 4(-17/10) + 6 ) / 2 This shows
more clearly the workings of BEDMAS and that you must
simplify the brackets first
For my own sake, I am going to use a calculator to check this haha
-0.4 + 0.4 = 0
0 = 0
At last! We have now reassured ourselves that a = -17/10 Good job!
5) Some quick vocabulary for functions
Range: All possible y-axis values in the function
Domain: All possible x-axis values in the function
Degree: The highest exponent on any given term within the function Ie. x^2 + 2x + 4 The degree of this trinomial ( A function with three terms ) is 2. Note (in case you dont know) that ^ means exponent
Thats it for now at any rate, I hope you understand everything I wrote as I tried my very best to write it in very simple terms. I would advise practicing your problems by writing out each and EVERY step along the way and thinking about what you are doing each time and maybe remind yourself of the rule if you need to. Eventually it will become second nature to you and you can start skipping some of the more meticulous steps. When taking notes in class I would advise asking questions if you dont understand a step on teh board even if they make you feel a little dumb and always write down EVERY step during lessons as nothing is worse than reviewing your notes and getting hung up on a step that you dont understand. You will probably start doing functions eventually and I can give you an overview of those as well if you would like one just to get a look at them. Best of luck to you! I would also recommend that since you have some math geniuses in your family that you ask them for help if they can, however, if they look at it and are just as confused as you or are trying to figure it out themselves because it's been so long since High School then just kindly ask them to leave because they will only confuse you with methods that they may have been taught that are different than the ones being taught to you. Oh and last but not least, ask any of your friends who are good at math to come and help you for an hour or two each week and give you an overview of what you learned that week and run through some problems that you find challenging with them so you aren't left behind
Well that did not take long, I went to the next section and I can not even begin to imagine how this stuff works. I tried putting in real numbers but it still does not work, even with the books answer.
I = Prt
solve for P
So I/p = rt
Where the hell do I go from there?
thus
1. (I) /r = (Prt) /r
2. I/r = Pt
3. (I/r) /t = (Pt) /t
4. I/rt = P
It's all the exact same operations you've been doing until now except there are no "numbers" to cancel each other out.
I=Prt
I=P(rt)
I/(rt)=P(rt)/(rt)
I/rt=P
P=I/rt
From I/P=rt
(I/P)/rt=rt/rt
I/Prt=1
P(I/Prt)=P(1)
I/rt=P
Alternatively, you can invert each side:
1/(I/P)=1/(rt)
P/I=1/rt
I(P/I)=I(1/rt)
P=I/rt
x/2 + 3 = x -1/2
The first thing I think of doing is +1/2 so that gives me:
x+3=x
right there that seems wrong, that does not even look possible so I do not know what I am doing wrong.
I have no idea why I am getting this wrong and continue to get everything wrong. I am just not retaining what I am learning, its like I forget it all right away as soon as I get the hang of it.
x/2+3=x-1/2
x/2+3+(1/2)=x-1/2+(1/2)
x/2+3 1/2=x
I got the answer, it is yet another rule to remember.
2 (x/2+3) = 2 (x-1/2)
x+6 = 2x-1
x+7 = 2x
7=x
It is going to help you immensely to show what you're doing to each sides as you do the problem. Also just a few tips I give to my kids (I teach 9th grade algebra all the mistakes you do happen ALL THE TIME):
When you're moving all of the variables to one side, and everything else to the other, move the smaller amount of variables. This is not in any way necessary but it seems to help get rid of negatives (not always) which tends to reduce the amount of sign errors. For example in the second problem there when you have x+6 and 2x-1 move the x+6 because if you move the 2x you'll get -x which will probably fuck you up.
Some people like to talk about working through the order of operations backwards to solve equations, first get rid of your addition and subtraction, and then get rid of your multiplication, then any powers. The only problem I ever run into is you usually have to dump the parenthesis first which isn't consistent.
If I ever see parenthesis it helps to draw sort of a rainbow from the number outside the parenthesis to EVERY number inside, if you do it every time it will help you remember when you distribute you have to take the sign with you, and you have to multiply times every term. Here are a couple problems I did in class the other day (systems are a little tougher but there are a lot of equations here...just look at the way I do like +4 on each side of the equal sides to get to the next step). And yes I realize there is a lot of crap on these pages, I assure you explanations as we go make more sense
http://www.mgwyatt.com/systemsSample.pdf
Trust us when we say there really aren't that many rules for the level of Algebra you're learning. I wouldn't trust myself to enumerate them all since I'd probably forget a couple I take for granted, but there can't be more than 20 or so. Probably even less, honestly.
So, when you make an observation like that, your next thought should be "okay, so how do I get rid of the fractions?" This is where the "tricks" come in. Multiplying both sides by 2 is simply the most obvious one, since it gets rid of the fractions on both sides of the equation. Then it just becomes the same kind of stuff you've been doing already.
Many equations can be solved in this way. If one side is under a square root, it might be easier to solve by squaring both sides. If your x is inside a function such as e^x, you can simplify it by taking the natural log of both sides (don't worry about this now, you'll get there later). In all of these cases, the one big "rule" to remember is that whatever you change on one side of the equation, you need to make the same change on the other side. In your case above, it worked out well because you had a fraction on both sides with a 2 in the denominator, so multiplying both sides by 2 got rid of both fractions. However, even if there had been only one fraction on one side of the equation, it still would have been a good move to make. For example:
(1/4)x - 5 = 3x + 6
There's only one fraction, but we still would like to get rid of it for simplicity. So, since there's a 4 in the denominator, we'll multiply by 4. BOTH SIDES.
4*[(1/4)x - 5] = 4*[3x + 6]
x - 20 = 12x + 24
x - 44 = 12x
-44 = 11x
x = -4
Or, you might have a case where there is a fraction on both sides of the equation, but the number in the denominators isn't the same. For example:
8x - (1/4) = (1/3)x + (1/2)
ZOMG! But it's not really that bad. You still use the exact same trick, which is to multiply both sides by some number to simplify the fractions. The only catch here is that you need to pick a number which will get rid of all three fractions at once. In this case, if you multiply both sides by 12, you simplify all three fractions to whole numbers all at once:
12*[8x - (1/4)] = 12*[(1/3)x + (1/2)]
96x - 3 = 4x + 6
96x = 4x + 9
92x = 9
x = (9/92)
So you see, it's the same general rule, you just have to use your noggin to figure out the best way to apply it. I hope this helps some, good luck.
Oculus: TheBigDookie | XBL: Dook | NNID: BigDookie
You don't HAVE to do it that way either fizban. Like I said, you can multiply anything by anything as long as you do it to both sides. Lets solve this WITHOUT multiplying both sides by 2 at first, k?
x/2 + 3 = x - 1/2
add 1/2 to both sides.
x/2 + 3 + 1/2 = x
x/2 + 7/2 = x
now subtract an x/2 from both sides.. (to get all x's on the right).
7/2 = x - x/2
7/2 = x/2 (because 1 - 1/2 is 1/2.. right?)
now multiply each side by 2 (see, you can do this at any step fizban
7 = x
Not another rule to remember, except maybe the rules of addition, subtraction, multiplication, and division with maybe some fractions mixed in. Stuff you've been doing all along.
Essentially, yes you can. But remember that you don't have to. You can do it to make it look nice (and in the problem shown above, multiplying everything by 2 from the get-go seems to make it the "prettiest").
Essentially frizban, algebra boils down to one rule, and one rule only:
Anything you do on the left side MUST be done on the right side.
Try some of them out, and when you get stuck, let us know so we can help you
Also, I looked at the PDF just now. That's not exactly what you are doing, but they are related. Those are systems of linear algebra equations, and I just took a whole course on them and their applications. Don't worry about those right now.
p.s. they aren't falling asleep because I'll jump their shit
That's right, but be careful - you don't 'get rid' of the fraction, just the denominator. e.g. 2x/3, multiply by three, now you have 2x.
Get the variable on one side, solve for that variable. It's that easy.
Seriously, if it were really that easy, would the OP be having so many problems? Or the millions of people around the world who have trouble working through algebraic equations every single day? Yes, that may be essentially what you're doing in the end, but getting to that point is not always clear or easy.
Oculus: TheBigDookie | XBL: Dook | NNID: BigDookie
Once it clicks, math becomes really easy.
They only find it hard because they make it out to be hard.
And that's what I'm explaining to the OP, without being an asshole.