PLZ Help!!! I'm lost!!

[tbryant]

Could someone help me simplify these....

1 - 3/x(squared) divided by 2 + 4/x

and this is the doosie

9x(squared) - 6x divided by 3x(squared)+4x-4 multiplied by x(cubed)-x(squared)-12 divided by x(to the 3rd power) - 2x+3x-6.

I'm stumped PLEASE HELP!!!

 

[Gene]

Please look at Karl's Notes to learn how to type questions. I thought you meant
(1 - (3/x))/ (2 + (4/x))
Multiply top & bottom by x
(x-3)/(2x+4)
But that doesn't look to promising so you may mean
((1 - 3)/x)/ ((2 + 4)/x)
Multiply top & bottom by x
-2/6 = -1/3

The second may start
(9x^2-6x)/ (3x^2+4x-4) =
3x(3x-2)/((3x-2)(x+2)) =
3x/(x+2)
but the second half is too confusing to even try. Check for typos and Karl.

 

[soroban]

Hello, tbryant!

\(\displaystyle \L1)\;\;\frac{1\,-\,\frac{3}{x^2}}{2\,+\,\frac{4}{x}}\)

\(\displaystyle \text{Multiply top and bottom by }x^2:\L\;\;\frac{x^2\left(1\,-\,\frac{3}{x^2}\right)}{x^2\left(2\,+\,\frac{4}{x}\right)} \;=\;\frac{x^2\,-\,3}{2x^2\,+\,4x}\)

\(\displaystyle \;\;\text{which factors: }\L\,\frac{x^3\,-\,3}{2x(x\,+\,2)\) . . . \(\displaystyle \text{but doesn't simplify.}\)

\(\displaystyle \L2)\;\;\frac{9x^2\,-\,6x}{3x^2\,+\,4x\,-\,4}\,\times\,\frac{x^3\,-\,x^2\,-\,12}{x^3\,-\,2x^2\,+\,3x\,-\,6}\)

You already had a typo in the problem . . . are there any others?
\(\displaystyle \;\;\)Only a few parts can be factored and there's very little reducing.

\(\displaystyle \L\;\;\frac{3x(3x\,-\,2)}{(3x\,-\,2)(x\,+\,2)}\,\times\,\frac{x^3\,-\,x^2\,-\,12}{(x\,-\,2)(x^2\,+\,3)\) . . . not very satisfying, is it?

 

[Gene]

Gee, re #1: I lost an x from 3/x² but Soroban picked one up in x³-3
The answer is (x²-3)/(2x(x+2))
---------------
Gene