How do you solve # x/2>1+(4/x)#?
x is outside
x cannot be 0.
Multiply throughout by 2x and rearrange. The inequality sign is
reversed, when x is negative. Accordingly,
#x^2-2x-8 > < 0, respectively.
In the first case, x > 0, #and so, x > 4; the other inequality is
When x is negative,