Solve x-3/2x-3<1

$$\frac { x - 3 } { 2 x - 3 } < 1$$

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(\frac{3}{2},\infty\right)$$

Solution Steps

Denominator 2x-3 cannot be zero since division by zero is not defined. There are two cases.

$$2x-3>0 2x-3<0$$

Consider the case when 2x-3 is positive. Move -3 to the right hand side.

$$2x>3$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x>\frac{3}{2}$$

The initial inequality does not change the direction when multiplied by 2x-3 for 2x-3>0.

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

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

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

Combine like terms.

$$-x<0$$

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

$$x>0$$

Consider condition x>\frac{3}{2} specified above.

$$x>\frac{3}{2}$$

Now consider the case when 2x-3 is negative. Move -3 to the right hand side.

$$2x<3$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

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

The initial inequality changes the direction when multiplied by 2x-3 for 2x-3<0.

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

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

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

Combine like terms.

$$-x>0$$

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

$$x<0$$

Consider condition x<\frac{3}{2} specified above. The result remains the same.

$$x<0$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,0\right)\cup \left(\frac{3}{2},\infty\right)$$