solving inequalities?

(x-3)/2x > 1 ... you have two answers; if x>0 and if x<0

if x>0 then (x - 3) > 2x ... or ... -3 > x

thus a solution would be -3 > x > 0.

but -3 is not greater than 0, so this is not a valid solution set.

which leaves us with x<0

then the equation becomes (x-3) < 2x ... or ... -3 < x

( you change the direction of the inequality when multiplying or dividing by a negative )

or the solution set is: - 3 < x < 0

(x-3)/2x -1>0

((x-3) -2x )/2x >0

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

(x+3)/2x <0

take roots of ( x+3) and 2x

x+3=0 x=-3

2x=0 x=0

For -infinity < x < -3 x+3 <0 and 2x <0 so (x+3)/2x >0 NO /

For -3< x< 0 x+3 >0 and 2x <0 so (x+3)/2x <0 OK /

For 0< x<+infinity x+3 >0 and 2x >0 so (x+3)/2x >0 NO /

Only -3< x < 0 .- You can´t put -3<= x <=0 , because at x=-3 (exactly) the solution is 0 (exactly) , thus it is valid if (x+3)/2x <= 0 .- At x=0 the solution is undefined , not valid.-

(x-3)/2x >1

There are three cases: x = 0, x<0, and x>0.

Case 1: x=0 is impossible because the denominator would be zero and you cannot divide by 0.

Case 2:

If x>0, then

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

x-3 > 2x

-3 > x

This is impossible, since, x cannot be greater than 0 and less than -3 at the same time.

Case 3:

If x<0,

(x-3/2x) * 2x < 1*2x

x-3 < 2x

-3 < x

So -3<x<0

I hope this helps!

(x-3)/2x > 1

multiply both side by 2x

x-3>2x

put x's in the same side

-2x+1>3

-x>3

therefore X Element of {-infinity,-4}

(x - 3)/2x > 1

Let x>0.

x - 3 > 2x

-3 > 3x

x < -1, a contradiction, so x is not positive.

Let x < 0.

x - 3 < 2x

-3 < x

x = (-3, 0)

(x - 3)/2x > 1 if x > 0

x - 3 > 2x

x < - 3

and if x < 0

x - 3 < 2x

x > - 3

hi

x=>-1