Solve 1-x/1+x<1

$$\frac { 1 - x } { 1 + x } < 1$$

Solve for x

$$x\in \left(-\infty,-1\right)\cup \left(0,\infty\right)$$

Solution Steps

Denominator 1+x cannot be zero since division by zero is not defined. There are two cases.

$$1+x>0 1+x<0$$

Consider the case when 1+x is positive. Move 1 to the right hand side.

$$x>-1$$

The initial inequality does not change the direction when multiplied by 1+x for 1+x>0.

$$1-x<1+x$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$-x-x<-1+1$$

Combine like terms.

$$-2x<0$$

Divide both sides by -2. Since -2 is negative, the inequality direction is changed.

$$x>0$$

Consider condition x>-1 specified above. The result remains the same.

$$x>0$$

Now consider the case when 1+x is negative. Move 1 to the right hand side.

$$x<-1$$

The initial inequality changes the direction when multiplied by 1+x for 1+x<0.

$$1-x>1+x$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$-x-x>-1+1$$

Combine like terms.

$$-2x>0$$

Divide both sides by -2. Since -2 is negative, the inequality direction is changed.

$$x<0$$

Consider condition x<-1 specified above.

$$x<-1$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,-1\right)\cup \left(0,\infty\right)$$