maths inequalities help please!?

This is different from an equation. You cannot cross-multiply.

Pull everything to one side and get a common denominator.

X/(x+3)-1/(x-2) >0. The lcd is (x+3)(x-2)

[X(x-2)-1(x+3)]/(x+3)(x-2)>0

(x^2-2x-x-3)= x^2-3x-3 is the numerator, but does not factor

So (x^2-3x-3)/(x+3)(x-2)>0

you now have to find the zeroes of all the factors

X= -3, or 2 are zeroes of the denominator. Unfortunately you have to use the quadratic formula for the numerator.

X= (3+/-sqr(21))/2=3.9or-.8

Now put all these numbers on a number line

.......-3.......-.8........2..........3.9......

So check all five regions to see which ones work!

+...-3.....(-)......-8......(+).....2.....(-).....3.9....+.....

so the solution is ( -infinity,-3) and( -.8,2) and (3.9, infinity)

This is sooo much easier with a graphing calculator...just graph both sides and see where the first graph is above (>) the other.

Good luck!

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

multiply both sides by (x + 3)(x - 2)

now the inequality reads

x^2 - 2x / (x + 3)(x - 2) > x + 3 / (x + 3)(x - 2)

multiply both sides by (x + 3)(x - 2)

now the inequality reads

x^2 - 2x > x + 3

x^2 - 3x - 3 > 0

now use the quadratic formula.

There are two inequalities: 2w > eight AND 2w < 12 You can remedy it now. The first one says w > four. The moment one says w < 6. So the reply is "w > four AND w < 6". You can write this as "four < w < 6".