Solve 2x-3/x+1geq1

$$\frac{ 2x-3 }{ x+1 } \geq 1$$

Solve for x

$$x\in (-\infty,-1)\cup [4,\infty)$$

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.

$$2x-3\geq x+1$$

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

$$2x-x\geq 3+1$$

Combine like terms.

$$x\geq 4$$

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.

$$2x-3\leq x+1$$

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

$$2x-x\leq 3+1$$

Combine like terms.

$$x\leq 4$$

Consider condition x<-1 specified above.

$$x<-1$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-1)\cup [4,\infty)$$