Solve x+1/x-2leq2

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

Solve for x

$$x\in (-\infty,2)\cup [5,\infty)$$

Solution Steps

Denominator x-2 cannot be zero since division by zero is not defined. There are two cases.

$$x-2>0 x-2<0$$

Consider the case when x-2 is positive. Move -2 to the right hand side.

$$x>2$$

The initial inequality does not change the direction when multiplied by x-2 for x-2>0.

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

Multiply out the right hand side.

$$x+1\leq 2x-4$$

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

$$x-2x\leq -1-4$$

Combine like terms.

$$-x\leq -5$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x\geq 5$$

Now consider the case when x-2 is negative. Move -2 to the right hand side.

$$x<2$$

The initial inequality changes the direction when multiplied by x-2 for x-2<0.

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

Multiply out the right hand side.

$$x+1\geq 2x-4$$

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

$$x-2x\geq -1-4$$

Combine like terms.

$$-x\geq -5$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x\leq 5$$

Consider condition x<2 specified above.

$$x<2$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,2)\cup [5,\infty)$$