Solve x+1/2-xgeq-2

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

Solve for x

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

Solution Steps

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

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

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

$$-x>-2$$

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

$$x<2$$

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

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

Multiply out the right hand side.

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

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.

$$x<2$$

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

$$-x<-2$$

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

$$x>2$$

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

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

Multiply out the right hand side.

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

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.

$$x\geq 5$$

The final solution is the union of the obtained solutions.

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