how do i solve these maths (inequalities and partial fractions)?

Question

how do i solve these maths (inequalities and partial fractions)

1. Solve x + 3/x - 2 > x + 1/x - 3

2. Resolve [x - 1]/x + 3 > into partial fractions

Answer 1

1. x + 3/x - 2 > x + 1/x - 3
x^2 - 2x + 3 > x^2 - 3x + 1
x^2 - x^2 + 3 - 1 > -3x + 2x
2 > -x
x > -2
Dunno about 2, though

Answer 2

1. x+3/x -2 > x + 1/x - 3

x is a denominator, so it cannot be 0

Separate into x<0 , x>0

If x<0
i. Multiply by x and change ">" to "<"
x^2 + 3 - 2x > x^2 + 1 - 3x // - x^2

3 - 2x > 1 - 3x // + 3x
3 + x > 1 // - x
x > -2

-2 < x < 0

ii x>0, Multiply by x
x^2 + 3 - 2x > x^2 + 1 - 3x // - x^2
3 - 2x > 1 - 3x // + 3x
3 + x > 1 // -3
x>-2
True for every x>0

Unite both, and get (x > 0) or (-2 < x < 0)
In other words, x>-2 and x =/= 0

2. [x - 1]/x + 3 = 1 - 1/x + 3 = 4 - 1/x

Answer 3

1) first you should simplify x from the both sides
if x>0
multiplay by x
3-2x>1-3x
x>-2 it's always possible coz x>0
if x<0
when u multiply by x u should reverse the sign
3-2x<1-3x
x<-2
the solutions are {x/x<-2 or x>0}

2) all u have to do id separate the nomirator
(x-1)/x+3 = 1-1/x+3 = 4-1/x

Answer 4

x^2+3-2x>x^2+1-3x ( the 2 x's cancel out)
x^2-x^2-2x+3x>1-3
x>-2