Solve (2x-5)div(3x-1)<1

$$(2x-5) \div (3x-1) < 1$$

Solve for x

$$x\in \left(-\infty,-4\right)\cup \left(\frac{1}{3},\infty\right)$$

Solution Steps

Denominator 3x-1 cannot be zero since division by zero is not defined. There are two cases.

$$3x-1>0 3x-1<0$$

Consider the case when 3x-1 is positive. Move -1 to the right hand side.

$$3x>1$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

$$x>\frac{1}{3}$$

The initial inequality does not change the direction when multiplied by 3x-1 for 3x-1>0.

$$2x-5<3x-1$$

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

$$2x-3x<5-1$$

Combine like terms.

$$-x<4$$

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

$$x>-4$$

Consider condition x>\frac{1}{3} specified above.

$$x>\frac{1}{3}$$

Now consider the case when 3x-1 is negative. Move -1 to the right hand side.

$$3x<1$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

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

The initial inequality changes the direction when multiplied by 3x-1 for 3x-1<0.

$$2x-5>3x-1$$

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

$$2x-3x>5-1$$

Combine like terms.

$$-x>4$$

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

$$x<-4$$

Consider condition x<\frac{1}{3} specified above. The result remains the same.

$$x<-4$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,-4\right)\cup \left(\frac{1}{3},\infty\right)$$