Help please with algebra assignment....?

Question

I need help finding the solution sets to this equations:

1. [32/(x^2-3x+2)]-3 = (x-3)/(x+1)

2. CubeRoot(3x+1) = x-1

3. [5x/(x^2+2x-8)] => 3
note: => means equal or greater than

4. |4x-3|-3>6
note: |4x-3| means the absolute value of 4x-3


Thats all.....thanks

Answer 1

1)After eliminating denominators,multiplying both sides by
(x^2-3x+2)(x+1) (too difficult to type all steps) you get
x^3-3x^2-6x+8=0which can be factored as (x-1)(x-4)(x+2)=0
so x= 1 ,4 and -2
Factoring x^2-3x+2 =(x-2)(x-1) so x=1 is not a solution as the denominator is 0(starting equation)
2) (3x+1) = x^3-3x^2+3x-1
x^3-3x^2-2=0 ( are you sure of the question?)x=3.1958233
if it were cube root (3x-1) you get
x^3-3x^2=0 so x=0 and x= 3
3)5x/(x^2+2x-8)-3>=0 // (5x-3x^2-6x+24)/(x-2)(x+4)>=0
factoring
-3(x+3)(x-8)/(x-2)(x+4)>=0 solution -4 4)
-9<4x-3 <9 so sum 3 and divide by 4
-6/4 -3/2
Thats enough