Solve 3x+1/2x-1<2

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

Solve for x

$$x\in \left(-\infty,\frac{1}{2}\right)\cup \left(3,\infty\right)$$

Solution Steps

Denominator 2x-1 cannot be zero since division by zero is not defined. There are two cases.

$$2x-1>0 2x-1<0$$

Consider the case when 2x-1 is positive. Move -1 to the right hand side.

$$2x>1$$

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

$$x>\frac{1}{2}$$

The initial inequality does not change the direction when multiplied by 2x-1 for 2x-1>0.

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

Multiply out the right hand side.

$$3x+1<4x-2$$

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

$$3x-4x<-1-2$$

Combine like terms.

$$-x<-3$$

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

$$x>3$$

Consider condition x>\frac{1}{2} specified above. The result remains the same.

$$x>3$$

Now consider the case when 2x-1 is negative. Move -1 to the right hand side.

$$2x<1$$

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

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

The initial inequality changes the direction when multiplied by 2x-1 for 2x-1<0.

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

Multiply out the right hand side.

$$3x+1>4x-2$$

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

$$3x-4x>-1-2$$

Combine like terms.

$$-x>-3$$

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

$$x<3$$

Consider condition x<\frac{1}{2} specified above.

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

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,\frac{1}{2}\right)\cup \left(3,\infty\right)$$