Solve x-1/2x-1geq2/3

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

Solve for x

$$x\in [-1,\frac{1}{2})$$

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.

$$x-1\geq \frac{2}{3}\left(2x-1\right)$$

Multiply out the right hand side.

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

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

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

Combine like terms.

$$-\frac{1}{3}x\geq \frac{1}{3}$$

Divide both sides by -\frac{1}{3}. Since -\frac{1}{3} is negative, the inequality direction is changed.

$$x\leq -1$$

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

$$x\in \emptyset$$

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.

$$x-1\leq \frac{2}{3}\left(2x-1\right)$$

Multiply out the right hand side.

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

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

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

Combine like terms.

$$-\frac{1}{3}x\leq \frac{1}{3}$$

Divide both sides by -\frac{1}{3}. Since -\frac{1}{3} is negative, the inequality direction is changed.

$$x\geq -1$$

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

$$x\in [-1,\frac{1}{2})$$

The final solution is the union of the obtained solutions.

$$x\in [-1,\frac{1}{2})$$