## solving inequalities with three variables

Can you help me solve this inequality for x?

$$\frac{1+(\gamma+x(r-\alpha)-1)t}{1+\frac{\gamma+x(r-\alpha)-1}{2}}>0$$

where $$\gamma>1, 0<t<1, 0<r<3\alpha, \alpha>0$$
I really don't know where to start ...

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 Mentor I would multiply both sides of the inequality by what's in the denominator, and then make two cases: one in which the denominator is assumed to be positive, and the other in which the denominator is assumed to be negative. The reason for the two cases is that if you multiply both sides of an inequality by a variable quantity, the direction of the inequality symbol changes if what you multiplied by is negative. If the quantity you multiply by is positive, the inequality symbol doesn't change direction. For example, 2 < 3, and 2(2) < 2(3), but -1(2) > -1(3).
 Well, my way of thinking of solving such inequalities is similiar to the one presented by Mark44, I just get to it in a different way. Since it's fairly obvious, that $$\frac{x}{y} >0 \iff xy>0$$, you can simply write: $$(1+(\gamma+x(r-\alpha)-1)t)(1+\frac{\gamma+x(r-\alpha)-1}{2})>0$$ insted of the fraction. And the product of 2 numbers is >0 only if both of them are of the same sign. Just choose whatever interpretation you prefer.