Math

[anthrobabe]

OK-HELP!!!!!
I'm good at many things- I have talents- math is not one of them

I am stuck- I should probably get this- I will probably slap my forehead when I do get it-
this is basic elementary algebra stuff- (yes I'm a college senior- no laughing out loud)

linear inequality
I'm having trouble at the very start
I'll give you some examples
no exponents but will use / to denote a fraction
also < or > and =to, I don't have a combination key for less/greater than or equal to, so you'll have to picture it

1) 6x-5 < or = to -1/3x -4
2) -8w -5 > or =to 1/3w -6
3) -3/5t +6 < -6t -8
4) 8y +4 < 5/4y +6
5) 6x -5 < or =to -1/3x -4
6) -7x +4 > 8/3x -8

this stuff
now I get the reducing part the (PEMDAS) or whatever
but how do I know which/where to begin

look at #1 because I looked at the answer(I'm using a computer math program) I know that 1st I have to add 1/3w to each side

for #2 how do I decide/know I have to add 8w to each side to begin

for #3 how do I decide/know I have to add 6t to each side

do you get what I'm asking?
completing the problems makes sense to me--- it's getting started that is driving me crazy for some reason!

look at #1 again
6x -5 < or =to -1/3x -4
first add 1/3x to each side(how do I know to add 1/3 instead of going with the opposit of positive 6x which would be -6x)
then combine the like terms
then get rid of the -5 by adding 5
then multiply
to get the answer of x<or =to 3/19
so on that level I get it but 

Oh I'm loosing my mind

If anyone knows about the Aleks math program and can email me through it -- or try to help here-- I will love you forever!

Thanks

[Darlica]

Unfortunatly my brain isn't wired for math either...
so I can just give you :hug:s 

/D

[beagle]

It doesn't really matter which side you eliminate the unknown variable from. It's just slightly more convenient to do it so as to have it remaining on the side where it is positive.

e.g. here's doing 1 the "long way"

6x-5 <= -(1/3)x -4

(subtract 6x)

-5 <= -(19/3)x-4

(add 4)

-1 <= -(19/3)x

(multiply by 3/19)

-(3/19) <= -x

(multiply by -1, remembering doing this reverses the inequality condition)

(3/19) >= x

(rewrite in equivalent form)

x <= 3/19



same answer as the other way of working...


6x-5 < or = to -(1/3x) -4

(add (1/3)x to both sides )

(19/3)x-5  <= -4

(add 5)

(19/3)x  <=  1

(multiply by 3/19 )

x <= 3/19


You get exactly the same result either way (maths is like that  ) , but the working is slightly easier. The important thing is to eliminate the unknown variable from one side of the inequality.


Does that answer the question?
(Be back in an hour or so).


P.S. Linus in Peanuts reckoned that in algebra x is almost always 9, and y almost always 11.  I've found this advice invaluable over the years.

[Alpaca]

Here's my approach to equations and inequalities in general ('cept for the more esoteric stuff), and especially for linear ones. Of course, the more mathematical weirdness you throw in, the more wrong this approach gets, but it's fine for linear things.

1. I have a bunch of stuff in (in)equality.
2. My goal is to get my variable on one side and all the other crap on the other.
3. What I do to one side, I have to do to the other.
4. If things get simpler, I know I'm making progress.

So, since beagle did such an admirable job of #1, I'll take #2.

-8w -5 ≥ (1/3)w -6

This conforms nicely to #1.

First, I'm going to add 6 to each side. I could add any number of any other things arbitrarily, but I happened to choose 6 because that would conveniently add with -6 on the right side of the inequality to give me a result with no constants. After doing the same thing to the left side as per #3, I'm left with:

-8w +1 ≥ (1/3)w

Now, since the ONLY thing on the right side is (1/3w, I know that I'm making progress, as per #4, because (1/3)w is simpler than (1/3)w -6, and -8w +1 is no more complex than -8w -5 was.

Since, as per #2, my goal is to get x alone on one side, I'm now electing to add 8w to each side. That way, on the left side, I only have a constant (as opposed to a constant and a variable, so it's simpler and better), and on the right side I only have a variable with coefficient, which is no more complex than a variable with a different coefficient. If I had chosen to subtract (1/3)w, I would have made no progress, because my left side would be just as complex as it was before, and my right side would now contain a 0 which, just like (1/3)w, is a single term. If I had chosen to subtract 1, then my left side would be simpler, but my right side would become more complex, so I wouldn't have accomplished much. Anywho, I'm left with:

1 ≥ (25/3)w

Now, I just have to divide by (25/3), which is the same thing as multiplying by (3/25), and I'm left with:

3/25 ≥ w or w ≤ 25/3, whichever you prefer.

My steps in summary:
-8w -5 ≥ (1/3)w -6
-8w +1 ≥ (1/3)w
1 ≥ (25/3)w
3/25 ≥ w

In reverse, i.e. moving the "w" around before the constants, this would've been
-8w -5 ≥ (1/3)w -6
-5 ≥ (25/3)w -6
1 ≥ (25/3)w
3/25 ≥ w

If I had gone the "other way," i.e. adding 5 and subtracting (1/3)w, this would've been
-8w -5 ≥ (1/3)w -6
-8w ≥ (1/3)w -1
(-25/3)w ≥ -1
w ≤ 3/25 N.B. In this step, I had to flip the inequality sign, because I was dividing by a negative number.

Does that help at all?

Oh, for the record, I assumed that 1/3w meant "one third w," not "one divided by [three times w]."

[Scriblerus the Philosophe]

This stuff is the little that makes much sense to me.

3) -3/5t +6 < -6t -8
             -6           -6


-3/5t<-6t-14
            -3

[Griffin NoName]

-7x +4 > 8/3x -8

My philosophy is always go for the minus and always make it look nicer

so ..... .....work on the minus and make it look nicer

add 8 to both sides 

12 -7x > 8/3x

next step : get the x on one side only so that it looks even nicer


so..... add 7x to both sides

12 > 7x + 8/3x

next step:  reduce it to one instance of x so it looks nicer

so....... take the x outside some brackets

12 > x (7 + 8/3)

sort out the silly way the stuff inside brackets looks

12 > x ((7 x 3) + 8/3)
           --------
              3
is

12 > x (21+8)
           -------
             3

12 > x (29)
           ----
             3

brackets look silly now so make it look nicer

12 > 29x
       ---
         3

ugh it can look better than that, move the 3 by multiplying both sides

12 x 3 = 29x x 3
             ---
              3

ooo threes on right hand side cancel out so it looks even better

36 > 29x

want x alone - need 29 under to cancel out so do that both sides

36  > x
---
29

hmmm prefer x on right really so change signs and swap

x <= 35
        ---
        29

dont like fractions? get decimal equivalent of 35/29 !!




Method: eliminate minuses first, do both to same side, divide or multiply by thing you want to move sides or cancel out, and always think what would make it look nicer

[Alpaca]

and always think what would make it look nicer

Hehe... at this point, my definition of "nice" is so warped... 
I just go for simple. If I get into making things look nice, I'm liable to re-express things using trigonometric identities.

[Griffin NoName]

and always think what would make it look nicer

Hehe... at this point, my definition of "nice" is so warped... 
I just go for simple. If I get into making things look nice, I'm liable to re-express things using trigonometric identities.

back in my younger days I'd have gone for a Fourier look good factor - now I like the quiet life 

[goat starer]

OK-HELP!!!!!
I'm good at many things- I have talents- math is not one of them

Nor is English. It is maths (or better still mathematics). For goodness sakes can we not have a bit more respect for the mother tongue from our former colonies.

yours pedantically and imperialistically,

Duke Goat

[Sibling Chatty]

Goatie, we in the New World simplify it and reduce it to math. Not maths. The full word is mathematic, so the shortened form could well be math (no extraneous s). With all the s's saved, you could combine them with all the u's ya'll overuse, and spell us a LOT...

By the way, the mathematical language expressed here might as well be Swahili to me. I don't understand it, and what's worse, I no longer want to!!

Old, lazy and proud of it!

[goat starer]

the full word is mathematics (note s)

[Griffin NoName]

I have to agree with Goat. The full word is mathematics.

Mathmatic would be, well,  like asthmatic, or something.

It would help if American English was not called English

[Scriblerus the Philosophe]

Perhaps we ought to chalk it up to differences in dialects? 

[Sibling Chatty]

Mathematics-al formulas?

Hmmph...

[Sibling Zono (anon1mat0)]

It is the plural in Spanish too: Matemáticas

(mathematica is/was a math software).