Solve x-1/1+x<-2

$$\frac{ x-1 }{ 1+x } < -2$$

Solve for x

$$x\in \left(-1,-\frac{1}{3}\right)$$

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.

$$x-1<-2\left(1+x\right)$$

Multiply out the right hand side.

$$x-1<-2-2x$$

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

$$x+2x<1-2$$

Combine like terms.

$$3x<-1$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

$$x<-\frac{1}{3}$$

Consider condition x>-1 specified above.

$$x\in \left(-1,-\frac{1}{3}\right)$$

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.

$$x-1>-2\left(1+x\right)$$

Multiply out the right hand side.

$$x-1>-2-2x$$

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

$$x+2x>1-2$$

Combine like terms.

$$3x>-1$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

$$x>-\frac{1}{3}$$

Consider condition x<-1 specified above.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(-1,-\frac{1}{3}\right)$$