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avatar+377 

And the 3rd one:

 

\({3x(2x-3)-(3-2x) \over 3-2x}-{18x^3 \over 9x^3-27x^2+x-3}\)

 

This is a lenghty and complicated sum, to try and add here the steps I took, will take a looong time, if someone out there could show me the way with this one also, I would really appreciate it!!

 Nov 1, 2017
 #1
avatar+99117 
+2

I do not think this simplifies very nicely

 

\({3x(2x-3)-(3-2x) \over 3-2x} - {18x^3 \over 9x^3-27x^2+x-3}\\ ={-3x(3-2x)-1(3-2x) \over 3-2x} - {18x^3 \over 9x^2(x-3)+1(x-3)}\\ ={(-3x-1)(3-2x) \over 3-2x} - {18x^3 \over (9x^2+1)(x-3)}\\ ={(-3x-1)(3-2x) \over 3-2x} - {18x^3 \over (9x^2+1)(x-3)}\\ ={-(3x+1)} - {18x^3 \over (9x^2+1)(x-3)}\\ ={-(3x+1) (9x^2+1)(x-3)-18x^3 \over (9x^2+1)(x-3)}\\\)

 

You can expand the numerator and denominator out if you want but nothing will cancel.

 

I should have added that x cannot equal 3 or 3/2

 Nov 1, 2017
edited by Melody  Nov 1, 2017
 #2
avatar+99117 
+1

The others were equations .. this one is a subtraction.

Maybe it was meant to be an equals sign ?

Melody  Nov 1, 2017
 #3
avatar+377 
+1

Hi Melody,

 

the left side I also got, however this is what I did with the right side:

\({18x^3} \over9x^3+x-27x^2-3\)

\(18x^3 \over(9x^3+x)+(-27x^2-3)\)

\(18x^3 \over x(9x^2+1)-3(9x^2+1)\)

\(18x^3 \over (x-3)(9x^2+1)\)

\(18x^2 \over (x-3)(3x+1)(3x+1)\)

 

so now it's;

 

\(-(3x+1)-{18x^3 \over(x-3)(3x+1)(3x+1)}\)

 

and this is where it ends for me...yes, I do not know, either = or * would have worked nicely, but subtraction?..I don't know..

 

Melody, anyways, thank you kindly once again for your help...I think I'll let this one pass..

juriemagic  Nov 1, 2017
 #4
avatar+99117 
+1

Hi Juriemagic,

Your factorisation on the right is not correct.

 

the first few lines are good but

 

\(9x^2+1 \ne (3x+1)^2\\~\\ (3x+1)^2=9x^2+6x+1\)

 

 

Other than that little error our answers are the same I think :)

Melody  Nov 1, 2017
 #5
avatar+377 
+2

Melody,

Yep, I see now...Thank you!!

juriemagic  Nov 2, 2017

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