Inequalities with two different fractions which include x in the denominator.

You want to determine when [itex](x-2)/(x+3)\le (x+1)/(x)[/itex].

As I said before, since we clearly want to multiply both sides by x+ 3 and x, to clear the fractions, we need to consider when they are positive of negative. So consider 3 cases:
1) x< -3. Then both x+ 3 and x are negative. Multiplying the above inequality by x(x+ 3) is multiplying by a positive number (the product of two negatives) so the direction of inequality does not change. [itex](x- 2)(x)\le (x+ 3)(x+ 1)[/itex]. [itex]x^2- 2x\le x^2+ 4x+ 3[/itex]. Subtract [itex]x^2[/itex] from both sides to get [itex]-2x\le 4x+ 3[/itex] so that [itex]0\le 6x+ 3= 3(x+ 2)[/itex]. Dividing both sides by the positive number 3, we have [itex]0\le x+ 2[/itex] or [itex]x\ge -2[/itex] which can't happen when x< -3.

2) -3< x< 0. Now x+ 3 is positive but x is still negative. Multiplying both sides by x(x+ 3) is now multiplying by a negative number and changes the direction of the inequality: [itex](x- 2)(x)\ge (x+ 3)(x+ 1)[/itex]. The same calculations as before go through with the changed inequality sign: [itex]x\le -2[/itex]. That tells us that the orignal inequality is true for [itex]-3< x< -2[/itex].

3) x> 0. Now both x+ 3 and x are positive so j=multiplying both sides of the inequality by x(x+ 3) does not change the sign: [itex](x- 2)(x)\le (x+ 3)(x+ 1)[itex] and again we get [itex]x\ge -2[/itex]. Of course that is true for all x> 0 so we have the inequality true for all x> 0.

We have, so far, that the inequality is true for -3< x< -2 and x> 0. You should also check to see if it is true at x= -3, x= -2, and x= 0.

2) -3< x< 0. Now x+ 3 is positive but x is still negative. Multiplying both sides by x(x+ 3) is now multiplying by a negative number and changes the direction of the inequality: [itex](x- 2)(x)> (x+ 3)(x+ 1)[/itex]. The same calculations as before go through with the changed inequality sign: [itex]x< -2[/itex]. That tells us that the orignal inequality is true for [itex]-3< x< -2[/itex].