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Maths Question
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SeenButNotHeard
Perkone
Caldari State
#1 - 2012-02-11 20:07:23 UTC  |  Edited by: SeenButNotHeard
I apologise in advance for the level of stupid in this thread.

I am utterly fail at mathematics. I just managed to scrape a C level at GCSE here in the UK.

I was recently inspired to re-visit equations and see if they made any more sense.

To my utter astonishment I found that two-step equations (don't laugh) started to make sense after about three hours of swatting up.

One that stumped me for a while was: 6a + 12 = 2a - 16

Now I appreciate that to some of you that may be laughably easy. It stumped me for about 25mins until I spotted the answer.

What I am asking for is the next step for me to start furthering my maths understanding. It is totally foreign to me so please be forgiving.

TL;DR What is my next step after two-step equations

Edit: Links to any good sites would be much appreciated - thanks!
Caleidascope
Republic Military School
Minmatar Republic
#2 - 2012-02-11 20:41:18 UTC
6a+12=2a-16
6a-2a=-16-12
(6-2)a=-28
4a=-28
a=-28/4
a=-7

Life is short and dinner time is chancy

Eat dessert first!
Micheal Dietrich
Kings Gambit Black
#3 - 2012-02-11 20:43:46 UTC
I suck at math as well. I downloaded Microsoft Mathematics for my Trig and Calculus and went to tutoring every chance I got. Now, I tried to solve each problem on my own first before plugging them in to the calculator. In any case the program is a life saver.

Microsoft Mathematics

Youtube example

Out of Pod is getting In the Pod - Join in game channel **IG OOPE **
SeenButNotHeard
Perkone
Caldari State
#4 - 2012-02-11 20:50:27 UTC  |  Edited by: SeenButNotHeard
Wow - Cal you got there a very different way to me. Not that I asked you to solve it!

Thanks Michael - I will have a look at those although from the you-tube vide it looks hideously unfriendly.

I was more asking about the type of maths puzzles I should be looking at learning and attempting next as a logical progression from where I am.

I wouldn't even know where to begin looking and google just brings up a repetitive list of the same stuff (no doubt due to my lack of knowing what to look for)

Cheers
Micheal Dietrich
Kings Gambit Black
#5 - 2012-02-11 21:05:42 UTC
It's not really as long as you know what to plug in to it. The best part is that it shows you how the problem is done step by step (especially great when the teacher asks you to show your work). Like I said though, try to figure out the problem before running it on the calc. My teacher didn't allow laptops during a quiz but we could use our notes so we still had to do all this manually. I didn't want this to be a crutch.

There are several math forums out there as well where people challenge each other. I caught one where a guy broke down the batman graph step by step and it made me envious of those math skills.

Out of Pod is getting In the Pod - Join in game channel **IG OOPE **
SeenButNotHeard
Perkone
Caldari State
#6 - 2012-02-11 21:13:35 UTC
It sounds like a useful tool then to use when you get stuck.

It was intriguing to see that Cal came to the same answer as me but via a very different method. It appears that maths has more lee-way than I had thought.

Can you suggest some learning I can do from here (bearing in mind I am proud to have mastered two-step equations!) Am not in a class so I have no teacher. It's just something that has started to interest me.

Appreciated
Caleidascope
Republic Military School
Minmatar Republic
#7 - 2012-02-11 22:31:24 UTC
I am not sure why the way I solved it surprises you. This is how I was taught to solve equations of this type, probably since 6th or 7th grade.

Also this type of setup is used to find the point where two lines intersect.

Life is short and dinner time is chancy

Eat dessert first!
Brisco County
Deep Core Mining Inc.
Caldari State
#8 - 2012-02-11 22:33:34 UTC
It's interesting how that kind of thing works. In my freshman year, I took Calculus, physics and chemistry all in the same semester. I never studied or did homework, skipped lectures, and still pulled all A's easily. The next semester, I took freshman English, a philosophy class, and music appreciation, studied my ass off and went to all lectures, even took supplemental instruction, and barely pulled C's.
SeenButNotHeard
Perkone
Caldari State
#9 - 2012-02-11 22:45:08 UTC
Caleidascope wrote:
I am not sure why the way I solved it surprises you. This is how I was taught to solve equations of this type, probably since 6th or 7th grade.

Also this type of setup is used to find the point where two lines intersect.


I didn't say it surprised me - just that it was different.

I went through:

6a + 12 = 2a - 16
6a + 28 = 2a
3a + 14 = a
3a = a - 14
2a = -14
a = -7

This might be completely wrong - and I am not asking you to beat the **** out of me for it. It's just how I got to the answer.

If it is mathematically wrong then I would ask you to help me with a sensible link. If it is ok then we arrived at the same answer via a different technique.

Am very new to maths so making me look like a male chicken won't help me.

I hope you can find the time to land on the right side. o7
SeenButNotHeard
Perkone
Caldari State
#10 - 2012-02-11 22:53:38 UTC
Brisco County wrote:
It's interesting how that kind of thing works. In my freshman year, I took Calculus, physics and chemistry all in the same semester. I never studied or did homework, skipped lectures, and still pulled all A's easily. The next semester, I took freshman English, a philosophy class, and music appreciation, studied my ass off and went to all lectures, even took supplemental instruction, and barely pulled C's.


Haha - I totally understand that one. None of my actual results matched my expectations after my exams.

Very annoying but it probably worked out ok. Am now just trying to expand my mathematics a bit, it has always been a thorn for me.

Mind you - if you believed the teachers I had I should be a crak addicted, card-board box loving, wastrel by now.
Caleidascope
Republic Military School
Minmatar Republic
#11 - 2012-02-11 22:55:04 UTC  |  Edited by: Caleidascope
SeenButNotHeard wrote:
Caleidascope wrote:
I am not sure why the way I solved it surprises you. This is how I was taught to solve equations of this type, probably since 6th or 7th grade.

Also this type of setup is used to find the point where two lines intersect.


I didn't say it surprised me - just that it was different.

I went through:

6a + 12 = 2a - 16
6a + 28 = 2a
3a + 14 = a
3a = a - 14
2a = -14
a = -7

This might be completely wrong - and I am not asking you to beat the **** out of me for it. It's just how I got to the answer.

If it is mathematically wrong then I would ask you to help me with a sensible link. If it is ok then we arrived at the same answer via a different technique.

Am very new to maths so making me look like a male chicken won't help me.

I hope you can find the time to land on the right side. o7

Nothing wrong. I just like keep things organized, I guess.

See, what I did is move all the letters on the left side, all the numbers on the right side, then proceeded to solve the thing.

You started to solve right away. The end result is the same. All the letters are on one side, all the numbers on another side: 2a=-14, then you divide both sides by two and get the final answer. The answer is the same, your solution is as good as mine.

Life is short and dinner time is chancy

Eat dessert first!
SeenButNotHeard
Perkone
Caldari State
#12 - 2012-02-11 23:02:58 UTC  |  Edited by: SeenButNotHeard
Thanks Cal.

I appreciate you explaining the way you work it out versus mine.

Given my obvious noobishness to all this, would you post me something to work on? Something that is roughly just above what I posted (in terms of difficulty - I don't mind buggering off and doing some research).

As you can tell I am just feeling out in all this so please don't screw me too hard :)

Edit: For clarity
Caleidascope
Republic Military School
Minmatar Republic
#13 - 2012-02-11 23:22:34 UTC
SeenButNotHeard wrote:
Thanks Cal.

I appreciate you explaining the way you work it out versus mine.

Given my obvious noobishness to all this, would you post me something to work on? Something that is roughly just above what I posted.

As you can tell I am just feeling out in all this so please don't screw me too hard :)

I hate to sound like pompous a$$, but this is basic algebra.

The next step is probably systems of equations. Where you have two unknowns or more unknowns. What happens in cases like that is you need as many equations as there are unknowns. Example:
2a+3b=6
3a-7b=10
If you try to solve each equations, you will never solve it. What you have to do is to use one equation to represent one unknown in terms of the second unknown. Say:
2a+3b=6
2a=6-3b
a=3-1.5b
Then you plug a into the second equation:
3a-7b=10
3(3-1.5b)-7b=10
9-4.5b-7b=10
-11.5b=10-9
-11.5b=1
b=-1/11.5=-0.087
Now you know b, with this you can find a. Going back to:
a=3-1.5b
a=3-1.5(-0.087)=3.1305

Like I said, the only way to solve equations with multiple unknowns is to have same number of equations as there are unknowns. In the example above I have two unknowns, so I provided two equations. One equations is not enough. Two equations is just right. Three is fine, but the extra equations can be simply ignored.

I think next area after algebra is calculus with its derivatives and integrals.


Life is short and dinner time is chancy

Eat dessert first!
SeenButNotHeard
Perkone
Caldari State
#14 - 2012-02-11 23:36:31 UTC  |  Edited by: SeenButNotHeard
Caleidascope wrote:
SeenButNotHeard wrote:
Thanks Cal.

I appreciate you explaining the way you work it out versus mine.

Given my obvious noobishness to all this, would you post me something to work on? Something that is roughly just above what I posted.

As you can tell I am just feeling out in all this so please don't screw me too hard :)

I hate to sound like pompous a$$, but this is basic algebra.

The next step is probably systems of equations. Where you have two unknowns or more unknowns. What happens in cases like that is you need as many equations as there are unknowns. Example:
2a+3b=6
3a-7b=10
If you try to solve each equations, you will never solve it. What you have to do is to use one equation to represent one unknown in terms of the second unknown. Say:
2a+3b=6
2a=6-3b
a=3-1.5b
Then you plug a into the second equation:
3a-7b=10
3(3-1.5b)-7b=10
9-4.5b-7b=10
-11.5b=10-9
-11.5b=1
b=-1/11.5=-0.087
Now you know b, with this you can find a. Going back to:
a=3-1.5b
a=3-1.5(-0.087)=3.1305

Like I said, the only way to solve equations with multiple unknowns is to have same number of equations as there are unknowns. In the example above I have two unknowns, so I provided two equations. One equations is not enough. Two equations is just right. Three is fine, but the extra equations can be simply ignored.

I think next area after algebra is calculus with its derivatives and integrals.




Ok - that has ****** me royally.

That is exactly the kind of thing I was looking for. Looking at it I kind of understand it - but working it out the way you have has fried my brain.

As I stated in my first post, I am not good at this. Would you post me a problem of a similar nature that I can go away and work on? No answers this time - just something I can bend my head around.

The amount of effort you have gone to, in order to reply to my musings is very much appreciated. Thanks.

PS - wasn't taught algebra ever - so two-step was pretty cool for me