Solve x+1/x-1+2geq0

$$\frac{ x+1 }{ x-1 } +2 \geq 0$$

Solve for x

$$x\in (-\infty,\frac{1}{3}]\cup (1,\infty)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{x-1}{x-1}.

$$\frac{x+1}{x-1}+\frac{2\left(x-1\right)}{x-1}\geq 0$$

Since \frac{x+1}{x-1} and \frac{2\left(x-1\right)}{x-1} have the same denominator, add them by adding their numerators.

$$\frac{x+1+2\left(x-1\right)}{x-1}\geq 0$$

Do the multiplications in x+1+2\left(x-1\right).

$$\frac{x+1+2x-2}{x-1}\geq 0$$

Combine like terms in x+1+2x-2.

$$\frac{3x-1}{x-1}\geq 0$$

For the quotient to be ≥0, 3x-1 and x-1 have to be both ≤0 or both ≥0, and x-1 cannot be zero. Consider the case when 3x-1\leq 0 and x-1 is negative.

$$3x-1\leq 0 x-1<0$$

The solution satisfying both inequalities is x\leq \frac{1}{3}.

$$x\leq \frac{1}{3}$$

Consider the case when 3x-1\geq 0 and x-1 is positive.

$$3x-1\geq 0 x-1>0$$

The solution satisfying both inequalities is x>1.

$$x>1$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{1}{3}\text{; }x>1$$