$\displaystyle \frac{1+y}{1-y}>0$

I get y > -1 which is wrong since if y = 2, the expression is negative.

I never learned how to solve them properly :P

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- Oct 25th 2009, 11:41 PMscorpion007Solve Inequality.
$\displaystyle \frac{1+y}{1-y}>0$

I get y > -1 which is wrong since if y = 2, the expression is negative.

I never learned how to solve them properly :P - Oct 26th 2009, 12:29 AMGusbob
There are multiple ways of doing this. I'll just show you one of the simpler ones for now.

Multiply both sides of the inequality by the denominator squared. In this case (1-y)². This ensures that the denominator is not neglected.

You will get (1+y)(1-y) > 0

Now solve for y. These are the 'starting points' of the range. Check both sides of your solutions on the number line to see which side satisfies the inequality (whether the the inequality holds true for y values < the solution or > solution)

Combine the two solutions to form one that satisfies both of your solutions. For example if you get y > 6 and y < 10, your combined solution is 6 < y < 10. There are some times when you won't be able to combine them. For example y < 6 and y > 10. In this case just write the two solutions down. - Oct 26th 2009, 01:30 AMscorpion007
Awesome, thanks!

Harder than I thought. - Oct 26th 2009, 03:33 AMHallsofIvy
One simple way to do this is to realize that a fraction is positive if and only if the numerator and denominator have the

**same**sign.

That is, $\displaystyle \frac{1+y}{1-y}> 0$ if and only if either 1+y> 0**and**1-y> 0 or 1+y< 0**and**1-y< 0.

Case 1) 1+y> 0 and 1- y> 0. The first inequality gives y> -1 and the second gives 1> y. Therefore, -1< y< 1 is a solution.

Case 2) 1+y< 0 and 1- y< 0. The first inequality gives y< -1 and the second gives 1< y. Since those cannot both be true, this gives no solution: there is no case in which the numerator and denominator are both negative.

$\displaystyle \frac{1+y}{1-y}> 0$ if and only if -1< y< 1. - Oct 26th 2009, 04:57 AMKrizalid