Solve 4-3x/1-x>2

$$\frac { 4 - 3 x } { 1 - x } > 2$$

Solve for x

$$x\in \left(-\infty,1\right)\cup \left(2,\infty\right)$$

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.

$$4-3x>2\left(1-x\right)$$

Multiply out the right hand side.

$$4-3x>2-2x$$

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

$$-3x+2x>-4+2$$

Combine like terms.

$$-x>-2$$

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

$$x<2$$

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.

$$4-3x<2\left(1-x\right)$$

Multiply out the right hand side.

$$4-3x<2-2x$$

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

$$-3x+2x<-4+2$$

Combine like terms.

$$-x<-2$$

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

$$x>2$$

Consider condition x>1 specified above. The result remains the same.

$$x>2$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,1\right)\cup \left(2,\infty\right)$$