+1124 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!!
+118609 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
+118609 The others were equations .. this one is a subtraction.
Maybe it was meant to be an equals sign ?
+1124 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..
