Math help, please...?

[Jeepgyrl]

I have this packet full of 40 wonderful Algebra problems I need to turn in when I go back to school after Spring Break. I'm stuck on a problem....can anyone solve this for me please?
Any help is appreciated. :)

Solve for X:
a(x+y)=b(x-y)

[SharpTJ]

a(x+y)=b(x-y)
ax+ay=bx+by

divide both sides by "a" and "y"
x=(b+b)/y

I really dont have much of an idea...Is it just basic Algebra/is there a word problem?

[Jeepgyrl]

this is how far I can get in the problem:

a(x+y)=b(x-y)
ax+ay=bx-by

then I don't know what to do since you have to solve for X...if I divide both sides by X, since they both have that variable??

I hate math.

[SharpTJ]

To isolate X, divide by evrything except X inthis Y and A

I had a brain fart earlier sorry...
Here is the answer....I think

(x+y)=b(x-y)
ax+ay=bx-by

divide both sides by A then by Y to get:

X=(BX-B)/A


Maybe?!?!

[EP]

Geez, I took calculous so I thought this would be easy. I haven't taken a math class in over three years so I had some trouble. I don't even know if I'm right but this is what I came up with.

x=y(a-b)/(a+b)

Am I right?

Edit: Are the answers in the back? :D

Edit2: You should ask Jon Brule. He would know for sure. I think he's going by a different name on this board. The Man with the Plan?

[Jeepgyrl]

Ok so how do u divide both sides by A then by Y...I thought both sides of the equals sign had to have a common variable to be able to divide...
Oh forget it. Thank you both for giving me possible answers...I'll just write somethin down!

[EP]

Hey check my 2nd edit. This is how I got to my answer though.

Ax+Ay=Bx-By

Ax-Bx+Ay=-By

Ax-Bx=Ay-By

x(A+B)=y(A-B)

x=y(A-B)/(A+B)

[Jeepgyrl]

Well hey that looks good. You got the A on both sides...so there's the common variable. Thanks again :)

[SharpTJ]

You dont have to have the same variable on both sides because in my Chemistry classes we transpose formulas all the time w/o the same variablw on both sides.

E=mc^2

m=c^2/m

[Jeepgyrl]

Oh...lol
Well, whatever! :)

[TOOLman]

a(x+y) = b(x-y)
ax + ay = bx - by
ay + by = bx - ax
y(a+b) = x(b-a)

y(a+b)/(b-a) = x

EP - you got the sign wrong at one place there.

[EP]

Toolman-

Oops. I did it right on my notepad though. Just didn't copy it down right. I always got points deducted because I didn't double-check my answers too. Still haunting me. :D

[The Man with the Plan]

TOOLman got it.

[Jeepgyrl]

Can I give you guys one more problem?
Factor:
8x3 + 27
(8x cubed)

How can this be factored?
I didn't think there was a solution for this problem.

[John Strenk]

I'm not good at math but isn't that the same as:
2^3X^3+3^3

[Jeepgyrl]

Hmm.
I don't think so....
But thanks!
:D

[SharpTJ]

here goes...

8x^3+27
x^3+(27/8)

Take the cube root of 27/8 for the answer.

Am I right this time

[Jeepgyrl]

lol
Well I dunno....I didn't know how to simplify that problem, I thought it had no solution...but your answer could be right. Maybe I'll just write that down.
Thanks for helping.
Do you actually like math??

[SharpTJ]

I really hate Algebra...its juct because I have to use it alot in Chemistry that I know any of it. If there was one subject in High Scholl I hated it would definitely be Algebra.

[ACERBIC]

Interestingly there is a second answer to the first question:

ax+ay = bx-by

ax-bx=-ay-by

x(a-b)=y(-a-b)

x=y(-a-b)/(a-b)

which if you plug any two numbers in for a and b works out to be the same as if you did so to the first answer!

Yes I was an Engi-nerd in college and I like this stuff!

As for the second question, there is a way to do that but I forget the trick.
I know something like 9x^2+16 breaks down to (3x+4)(3x-4) but there is another trick for cubes that makes it more of a PITA to do.

[Jeepgyrl]

Yea, as you can tell, I don't like it either.
I don't understand why we need to learn all of it...I bet I'll never use 90% of it ever again.

[SharpTJ]

Acerbic is right!!! I just dont remember how to do it either.

[Guest]

Q1: x = y(a+b)/(b-a)
Q2: (3+2x)(9-6x+4x^2)

You can factorize further if you use complex variables.

[ACERBIC]

I remebered right after I posted the last time but the forum died on me! :mad:

8x^3+27

breaks down to

(2x+3)(4x^2-6x+9)

mulitply it out it works!

the rule is take the cuberoot of the two numbers and use them in the single x part, then use the square of them in the x^2 forumula with the multiplied number of the two roots for the x term. It only works for a plus sign with two cubes, so it wouldn't work for 8x^3-27.

In other words the rule is

A^3x^3+B^3

roots to

(Ax+B)(A^2x-ABx+B^2)

:D
Al


Jeep Trick beat me by 2 mins, bastard! :D Forget any help on the complex variables, I'd need the book for that one!

[Ryan4x4XJ]

Yea, as you can tell, I don't like it either.
I don't understand why we need to learn all of it...I bet I'll never use 90% of it ever again.

Yes you will, trust me.

If you plan on taking any type of Calculus course in the future, know your algebra!! Im in Cal 3 right now in college and Im still using that algebra, it never goes away in any math class.

Besides it helps develop analytical skills, at least thats what my teachers say :)

[Jeepgyrl]

Thanks :D

[Guest]

... and then you graduate, buy a copy of Mathematica, and never look back. (As I did in the above solution I provided.... heh heh heh)