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I am horrible at math
Fizban140
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I do not get math, I am horrible at it. I am taking an intro to college math class and I am barely understanding the class. I will struggle doing practice problems every time I do them, I can do the same set on two different days I get the same questions wrong on both days even though I go over them a bunch of times. I can not remember any of these rules, they seem to all change for the slightest of reasons none of which seem to matter.
Right now I am working on equations which I thought I understood but tomorrow I will probably do the same set of problems and get 10 wrong. How can I remember all these rules and how can I remember when they do and do not apply to certain situations? It seems like there are hundreds of rules with thousands of different conditions where they can be applied and thousands more where they do no apply.
My only goal right now is to be good at math, I want to be really good at it. It is all I spend my free time doing during the week and at work when I have nothing else to do. Some days I will spend six hours out of class working on this stuff.
As an example I do not see why this does not work.
2/3a -5 = 1/3 a +5
to get rid of 2/3a I multiply by 3/2 which equals -5 = 1/2a + 5
-10 = 1/2a
-5=a
Right now I am working on equations which I thought I understood but tomorrow I will probably do the same set of problems and get 10 wrong. How can I remember all these rules and how can I remember when they do and do not apply to certain situations? It seems like there are hundreds of rules with thousands of different conditions where they can be applied and thousands more where they do no apply.
My only goal right now is to be good at math, I want to be really good at it. It is all I spend my free time doing during the week and at work when I have nothing else to do. Some days I will spend six hours out of class working on this stuff.
As an example I do not see why this does not work.
2/3a -5 = 1/3 a +5
to get rid of 2/3a I multiply by 3/2 which equals -5 = 1/2a + 5
-10 = 1/2a
-5=a
Fizban140 on
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It is just something I want to be good at and I do not see any jobs I would enjoy that do not involve at least a little math.
The idea that there are a lot of rules and they never stay the same is definitely going to be a hurdle for you. To be honest, there are only a few hard and fast rules that you need to stick by; the difficulty is all in knowing when and how to use them.
Unless you're memorising shortcuts, the rules of mathematics tend to be pretty consistent. For the most part, pre-college mathematics only blow up when you do things like try to divide by zero, or venture into places where complex mathematics (terms involving the square root of -1) are required.
Could you give an example of a rule and its conditions?
Oh mate. Two things wrong.
1.) You need to multiply everything by 3/2, including the -5 and +5. For whatever reason, you've multiplied the 1/3a to give 1/2a, which is correct, but ignored -5 and +5.
2.) 2/3a * 3/2 = a. I'm not quite sure why you're under the impression that it disappears.
So it actually equals: a - 15/2 = a/2 + 15/2
It solves to a = 15.
edit:
Actually, a = 30.
a - 15/2 = a/2 + 15/2
a = a/2 + 30/2
a/2 = 30/2
a = 30
I think I am memorizing shortcuts, which I have no fucking interest in doing but my teacher keeps teaching us all these stupid rhymes and shit (its one of those classes) to memorize stuff to do the easy way. I hate the class actually, everyone in the class is perfectly happy to memorize all that bullshit and never question why it works.
Also I can not get a tutor, the teacher only shows up a few minutes early and does not have much time after class. I am in the military so I am lucky to even be able to take this class, I have had to drop two classes already because I had no time.
Haha, so it is. =P I fail, clearly.
(2/3)a -5 = (1/3)a +5
(2/3)a = (1/3)a +10
(2/3-1/3)a = 10
(1/3)a = 10
a = 10*3
a = 30
What you did wrong there was only multiplying half of the problem.
All you know is that the left half of the equation is equal to the right half of the equation, so if you were to multiply just the 2/3a and the 1/3a by 3/2, you will no longer know if the sides are equal- in fact, you could pretty easily show that they would not be. Pretend the sides of the equation have parentheses around them, that could make it easier.
Whenever you use multiplication or division on an algebra equation, you need to apply it to every term, because otherwise you're dealing with unequal portions of the same quantity.
Also, when you multiply 3/2 and 2/3a, you get a, because 2/3 times 3/2 equals 1.
Were I to solve this equation, i would first add five to each side, and then subtract 1/3a from each side.
2/3a -5 +5=1/3a +5 +5
2/3a =1/3a +10
1/3a = 10
3(1/3 a)=3(10)
a=30 (or 1/30 if I misunderstood your notation, and you intended one over 2a, as opposed to one half of a)
Edit: Beaten while writing my novel.
1. Whatever you do to one side of the equation, you must do EXACTLY THE SAME THING to the other side.
2. Your goal is to isolate the variable you're solving for, meaning get it by itself on one side of the equation.
3. In order to do that, you must undo whatever operations are there. If there's addition, you have to subtract. If there's multiplication, you need to divide.
Actually, make it 4.
4. If you are multiplying or dividing both sides of an equation by a number, you must make sure that all the terms on both sides are multiplied or divided. That's called the Distributive Property.
Also, one of the most important things you can do to improve your algebra is get in the habit of checking your answer, every single question. If you identify when you've made a mistake, you can then work on finding and fixing it. Never, ever assume your answer is right.
IOS Game Center ID: Isotope-X
That is right.
We also had a rote memory quiz that he drilled us with to deal with exponents, so helpful. And the way I got it was
a = 30
2/3a - 5 = 1/3a +5
3[2/3a - 5 = 1/3a + 5]
2a - 15 = a +15
a = 30
PSN Hypacia
Xbox HypaciaMinnow
Discord Hypacia#0391
When you multiply an equation with an equals sign, you have to multiply everything on both sides of the equation by the same thing. If you multiply the 'a' terms by 3/2, then you're only manipulating a part of each side. Since you don't really know what 'a' is yet, you can't really know how much you're increasing each side. So, what happens is, you're increasing one side by more than the other, so it's no longer equal.
Memorizing shortcuts and learning rhymes is a stupid way to learn math. In my opinion, the best way to learn is to do word problems with physical real world objects. That's how I did it and it makes it really easy to see how everything gets manipulated.
Yes, except for the part of "only doing it once." Multiplying by each term is only doing it once, because you'll do that every time. That's just how I'd think of it.
Edit: In defense of memorizing little rhymes and things, you will want to know the quadratic formula, and it's a little faster to use the formula for squaring a binomial, and there are all sorts of trigonometric identities and integrations that it's only reasonable to just memorize. Some things will require it, or become easier because of it, even if many of the little rhymes aren't necessary.
2/3a -5 = 1/3 a +5 multiply everything by 3 as it makes everything into whole numbers (easier to deal with)
2a - 15 = a + 15 take the 15 from the left side add it to the right side
2a = a + 30 take the a from the right side and minus it from the left side
a = 30
PSN = Wicker86 ________ Gamertag = Wicker86
wtf?
The problem is 2(y-1/2) = 4(y-1/4) +y
Well, try it. You should be able to do this pretty much exactly the same way as any method that solved the last one.
2y-1 = 5y -1
2y = 5y
so since 2y and 5 y are equal to each other they must be 0
so y = 0
PSN = Wicker86 ________ Gamertag = Wicker86
2y - 1 = 5y = 1
2y = 5y
-3y = 0
y = 0
PSN Hypacia
Xbox HypaciaMinnow
Discord Hypacia#0391
The easiest way to do any algebra equation is to get the variable you want to solve for (y, in this case) and get all copies of it on one side.
So, subtract 2y from each side:
-1 = 3y - 1
Then, you want to get everything else on the opposite side from your variable.
Add 1 to each side.
0 = 3y
Now that your variable is isolated, you can multiply/divide/square/whatever to get it down to just that variable.
Divide by 3.
0 = y
It's not as hard to figure out what a is when -3a=0
And again, beaten. Twice.
2q+12=q
2q/1q= 2q
12=2q
q=6
but the answer is suppose to be -12 and it works when I put it in.
Yes, any number multiplied by x will equal 0.
You can use the same method I put above to solve that.
http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/equationsrev1.shtml
It'll help.
PSN = Wicker86 ________ Gamertag = Wicker86
2q+12 = q
12 = q-2q
12 = -1q
q = 12/-1
q = -12
Motherfucker, I swear there was a problem earlier where I wanted to do subtraction but it would not work and I have to divide, this is the shit that kills me.
I don't know what your first step there is supposed to be. Where did the 12 go?
2q+12=q
subtract 2q from both sides
12 = q-2q
12= -q
q=-12
Oops, didn't realize it had gone to a second page.
-3x=4-2x
-5x=4
Do I subtract or divide? I think I would divide because they are not like terms? But if they were like terms I would subtract? More rules to remember...
-3x = 4-2x
-3x + 2x = 4
-1x = 4
x = 4/-1
x = -4
Once all your x terms are combined on one side, divide by the number in front (called the coefficient).
Easiest way of doing algebra is to get your numbers on one side of the equal sign, and your variables on the other side like this:
2q+12 = q
q + 12 = 0
q=-12
Remember, you can ALWAYS check your answer by pluging it back into the equation, try it:
2q + 12 =q
if q = -12...
2(-12) + 12 = -12
-24 + 12 = -12
-12 = -12
And it works.
What you were trying to do was multiply the equation in whichever fashion you saw fit. You can't do this. If you were going to divide the entire equation by q you would get this:
2q + 12 = q .. dividing BOTH sides by q (remember you HAVE to divide BOTH sides)
2 + 12/q = 1 (remember, 2q/1q = 2, not 2q. Those variables go AWAY when you divide them out.. E.G. q^2/q = q)
But that really doesn't make a whole hell of a lot of sense to do it that way.. however... if you are to plug in -12 (the answer we got before) it will work out.. lets see...
2 + 12/-12 = 1
2 - 1 = 1 YES this does work.
Just remember, you aren't CHANGING the equation at all by preforming operations on it, you are just changing the way it looks, but the equation is still the same. This is the very crux of algebraic manipulation and computation.
I suggest getting a tutor.
It's really not as difficult as you are trying to make it. Just make sure you deliberately write out each step of what you are doing, as to catch simple arithemetic errors (such as subtracting when you are supposed to be adding).
Also I am having trouble understand why I can not divide like terms.
So for example, here is a extremely verbose version of 2q+12=q:
2q + 12 = q
(2q +12) - q = (q) - q
(2q - q) + 12 = 0
(2 - 1) * q + 12 = 0
(1) * q + 12 = 0
q + 12 = 0
(q + 12) - 12 = 0 - 12
q + (12 - 12) = -12
q + 0 = -12
q = -12
Now, you could probably skip a bunch of the steps in there, but you might want to start from the basics and skip nothing.
2q + 12 = q
If I divide everything by q, I get:
2 + 12/q = 1
All I accomplished was having to multiply both sides by q again later:
12/q = 1-2
12/q = -1
12 = -1q
q = 12/-1
q = -12
You can. You can divide ANYTHING BY ANYTHING fizban. You just have to make sure you are doing it RIGHT and not skipping steps or not following rules. You can add anything to anything. You can do whatever the hell you want, as long as you are following the rules.
Take for instance:
2q + 12 = q
You can add 5 to both sides.
2q + 17 = q + 5
now lets divide everything by q.
2 + 17/q = 1 + 5/q
And guess what, you'll still get the same answer