Solve 2x-3/x-5<3

$$\frac { 2 x - 3 } { x - 5 } < 3$$

Solve for x

$$x\in \left(-\infty,5\right)\cup \left(12,\infty\right)$$

Solution Steps

Denominator x-5 cannot be zero since division by zero is not defined. There are two cases.

$$x-5>0 x-5<0$$

Consider the case when x-5 is positive. Move -5 to the right hand side.

$$x>5$$

The initial inequality does not change the direction when multiplied by x-5 for x-5>0.

$$2x-3<3\left(x-5\right)$$

Multiply out the right hand side.

$$2x-3<3x-15$$

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

$$2x-3x<3-15$$

Combine like terms.

$$-x<-12$$

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

$$x>12$$

Consider condition x>5 specified above. The result remains the same.

$$x>12$$

Now consider the case when x-5 is negative. Move -5 to the right hand side.

$$x<5$$

The initial inequality changes the direction when multiplied by x-5 for x-5<0.

$$2x-3>3\left(x-5\right)$$

Multiply out the right hand side.

$$2x-3>3x-15$$

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

$$2x-3x>3-15$$

Combine like terms.

$$-x>-12$$

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

$$x<12$$

Consider condition x<5 specified above.

$$x<5$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,5\right)\cup \left(12,\infty\right)$$