Solve 2x-1/3-4x>1

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

Solve for x

$$x\in \left(\frac{2}{3},\frac{3}{4}\right)$$

Solution Steps

Denominator 3-4x cannot be zero since division by zero is not defined. There are two cases.

$$3-4x>0 3-4x<0$$

Consider the case when 3-4x is positive. Move 3 to the right hand side.

$$-4x>-3$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x<\frac{3}{4}$$

The initial inequality does not change the direction when multiplied by 3-4x for 3-4x>0.

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

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

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

Combine like terms.

$$6x>4$$

Divide both sides by 6. Since 6 is positive, the inequality direction remains the same.

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

Consider condition x<\frac{3}{4} specified above.

$$x\in \left(\frac{2}{3},\frac{3}{4}\right)$$

Now consider the case when 3-4x is negative. Move 3 to the right hand side.

$$-4x<-3$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x>\frac{3}{4}$$

The initial inequality changes the direction when multiplied by 3-4x for 3-4x<0.

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

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

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

Combine like terms.

$$6x<4$$

Divide both sides by 6. Since 6 is positive, the inequality direction remains the same.

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

Consider condition x>\frac{3}{4} specified above.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{2}{3},\frac{3}{4}\right)$$