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

 Quote by HallsofIvy You want to determine when $(x-2)/(x+3)\le (x+1)/(x)$. 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. $(x- 2)(x)\le (x+ 3)(x+ 1)$. $x^2- 2x\le x^2+ 4x+ 3$. Subtract $x^2$ from both sides to get $-2x\le 4x+ 3$ so that $0\le 6x+ 3= 3(x+ 2)$. Dividing both sides by the positive number 3, we have $0\le x+ 2$ or $x\ge -2$ 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: $(x- 2)(x)\ge (x+ 3)(x+ 1)$. The same calculations as before go through with the changed inequality sign: $x\le -2$. That tells us that the orignal inequality is true for $-3< x< -2$. 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: $(x- 2)(x)\le (x+ 3)(x+ 1)[itex] and again we get [itex]x\ge -2$. 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.
I think you made a mistake in the firsf case where you ended up with 3(x+2)

Also the question didn't include any inequality symbols with equal signs. (No line under the inequality is what I mean.

 Recognitions: Gold Member Science Advisor Staff Emeritus Oh, bother! Thanks. I have gone back and edited my post.

And...
 Quote by HallsofIvy 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: $(x- 2)(x)> (x+ 3)(x+ 1)$. The same calculations as before go through with the changed inequality sign: $x< -2$. That tells us that the orignal inequality is true for $-3< x< -2$.
## (x−2)(x) > (x+3)(x+1) ##

## x^2-2x > x^2 +4x +3 ##

## -2x > 4x +3 ##

## 0 > 6x +3 ##

## x < -\frac 36 = -\frac 12 ##