Am I solving this kind of inequality right?

Whenever I'm faced with an inequality like this one:

1+x/1-x > 0

I don't really use a set plan. I basically just take the numerator and denominator and solve for them separately. Example:

1+x>0 ----> x>-1
1-x>0 -----> x<1

Is it OK to do this for all inequalites with numerators and denominators that are just set against 0 as opposed to a quantity, or am I missing something? Thanks.

Linty Fresh said:

1+x/1-x > 0

Click to expand...

First, upgrade your notation a little.

1+x/1-x = 1 + (1/x) - x

Is that what you had in mind?

Maybe (1+x)/(1-x)?

Second, get a set plan. If you are going to multiply or divide by expressions containing variables, you MUST know if (or where) they are negative.

In this case, 1-x < 0 ==> 1 < x, so

(1+x)/(1-x) > 0 is equivalent to (1+x) > (1-x)*0 or 1+x > 0 WHEN x < 1
(1+x)/(1-x) > 0 is equivalent to (1+x) < (1-x)*0 or 1+x < 0 WHEN x > 1

Having arrived at:

1+x > 0 WHEN x < 1
1+x < 0 WHEN x > 1

We have

x > -1 WHEN x < 1 <== Solutions!
x < -1 WHEN x > 1 <== No Solutions