+1124 And the 3rd one:
3x(2x−3)−(3−2x)3−2x−18x39x3−27x2+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!!
+118703 I do not think this simplifies very nicely
3x(2x−3)−(3−2x)3−2x−18x39x3−27x2+x−3=−3x(3−2x)−1(3−2x)3−2x−18x39x2(x−3)+1(x−3)=(−3x−1)(3−2x)3−2x−18x3(9x2+1)(x−3)=(−3x−1)(3−2x)3−2x−18x3(9x2+1)(x−3)=−(3x+1)−18x3(9x2+1)(x−3)=−(3x+1)(9x2+1)(x−3)−18x3(9x2+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
+118703 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:
18x39x3+x−27x2−3
18x3(9x3+x)+(−27x2−3)
18x3x(9x2+1)−3(9x2+1)
18x3(x−3)(9x2+1)
18x2(x−3)(3x+1)(3x+1)
so now it's;
−(3x+1)−18x3(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..
