Solve 3x-5/1-xleq-2

$$\frac { 3 x - 5 } { 1 - x } \leq - 2$$

Solve for x

$$x\in (-\infty,1)\cup [3,\infty)$$

Solution Steps

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

$$1-x>0 1-x<0$$

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

$$-x>-1$$

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

$$x<1$$

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

$$3x-5\leq -2\left(1-x\right)$$

Multiply out the right hand side.

$$3x-5\leq -2+2x$$

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

$$3x-2x\leq 5-2$$

Combine like terms.

$$x\leq 3$$

Consider condition x<1 specified above.

$$x<1$$

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

$$-x<-1$$

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

$$x>1$$

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

$$3x-5\geq -2\left(1-x\right)$$

Multiply out the right hand side.

$$3x-5\geq -2+2x$$

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

$$3x-2x\geq 5-2$$

Combine like terms.

$$x\geq 3$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,1)\cup [3,\infty)$$