For You Math Wiz-Kids

[trust]

Help a brotha out.

Long Division:

5X^3 - 3X^2 - 4
---------------------
          X^2


Solve:

5x-4    = 2/3
5x+4                     

1                3                4
___     +    __        =    __
x-2            x+6          (x+3)(x-2)


Simplify:

1) (9x+81)/(x^2-9) * (2x^2-6x)/(x^2+18x+81)

2) (2/x^2-7x) + (3/x-7) - (3/x)

3) (1/x) + (1/y)
---------------------
(2/x) - (2/y)


Thanks.

[Sidoh]

Get MathType.  My head hurts from reading that, damnit.

Haha, just kidding.  You seriously should check out MathType, though.  It's pretty neat.

Actually, I think Microsoft Word comes equipped with an "Equasion Editor," which is basically a light version of MathType.

[iago]

What exactly are you doing to those equations?  I can change them into different equations, if that's what you want...

5X^3 - 3X^2 - 4
---------------------
          X^2

5x - 3 - 4/x²

After that, it gets too long and boring.  I have better things to do

[MyndFyre]

Long Division:

5X^3 - 3X^2 - 4
---------------------
          X^2

Commutative property: you can split this into two fractions:
5x³ - 3x²       4
----------  -  -----
    x²               x²

Factor out x², leaving you with :
5x - 3 - 4/x²

That's the most-simplified as that can get, and arguably the first form is even better.

Solve:

5x-4    = 2/3

Multiply both sides by 3:
15x - 12 = 2
Add 12 to both sides
15x = 14
Divide both sides by 15
x = 14/15

5x+4                     
1                3                4
___     +    __        =    __
x-2            x+6          (x+3)(x-2)

I don't know what you want out of those....  The second one, hrm.  Multiply both sides by (x+3)(x-2)
1(x+3) + [3(x² - 5x - 6) / (x+6)] = 4
Make into common terms by multiplying 1(x+3) and 4 by (x+6)/(x+6)
[(x² + 9x + 18)/(x+6)] + [3(x² - 5x - 6)/(x+6)] = [(4x + 24)/(x+6)]
Multiply both sides by (x+6) to get a quadratic equation:
x² + 9x + 18      +      3x² - 5x - 6       =     4x + 24
Simplify
4x² + 4x + 12 = 4x + 24
Divide both sides by 4:
x² + x + 3 = x + 6
Subtract (x+6) from both sides:
x² - 3 = 0

Solution: x² = 3, so x = +/- sqrt(3)


Simplify:

1) (9x+81)/(x^2-9) * (2x^2-6x)/(x^2+18x+81)

Separate each polynomial term into factors:
[9(x+9)] / [(x+3)(x-3)] * [2x(x-3)]/(x+9)²
Cross-cancel: one factor of (x+9), one factor of (x-3):
9/(x+3) * 2x/(x+9)
Multiply:
18x/[(x+3)(x+9)]
= 18x/(x²+12x+27)

2) (2/x^2-7x) + (3/x-7) - (3/x)

2 / [x(x-7)] + 3/(x-7) - 3/x
Get everything in common terms: multiply the 2nd term by x/x and the third term by (x-7)/(x-7)
2/[x(x-7)] + 3x/[x(x-7)] - (3x-21)/[x(x-7)]
Combine like terms:
-19 / [x(x-7)], or -19/(x² - 7x)

3) (1/x) + (1/y)
---------------------
(2/x) - (2/y)

Separate them out.
[(1/x) / [(2/x)-(2/y)]]    +   [(1/y) / [(2/x)-(2/y)]]
Realize that 1/x means you can multiply the denominator by x
1 / (x[(2/x)-(2/y)])     +     1 / (y[(2/x)-(2/y)])
Distribute:
(1 / [2 - 2x/y]) + (1 / [2y/x - 2])
Whether that's actually more simple is anyone's argument

[d&q]

  Reminds me of last year's math homework...

[Sidoh]

Reminds me of last year's math homework...

Haha, reminds me of Geometry.

[Towelie]

Reminds me of last year's math homework...

me too, what math are you in?

[trust]

Thanks MyndFyre!

It's for analysis.

[trust]

Mynd would you mind checking that again? I think you got some of the problems mixed up (like you took the denominator from one and put it with another... on a few?)

[MyndFyre]

OK here's me checking it over, this time not trying to do it with ASCII fractions.  I started out with what I understood from your problems and worked step-by-step similarly (or identically) to the way I did it before:

1.)

2.)

3.)

Simplification problems:

1.)

2.)

3.)