Solve 4-x/x+1<1

$$\frac{ 4-x }{ x+1 } < 1$$

Solve for x

$$x\in \left(-\infty,-1\right)\cup \left(\frac{3}{2},\infty\right)$$

Solution Steps

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

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

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

$$x>-1$$

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

$$4-x<x+1$$

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

$$-x-x<-4+1$$

Combine like terms.

$$-2x<-3$$

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

$$x>\frac{3}{2}$$

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

$$x>\frac{3}{2}$$

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

$$x<-1$$

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

$$4-x>x+1$$

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

$$-x-x>-4+1$$

Combine like terms.

$$-2x>-3$$

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

$$x<\frac{3}{2}$$

Consider condition x<-1 specified above.

$$x<-1$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,-1\right)\cup \left(\frac{3}{2},\infty\right)$$