Solve 2x+1/3x-6geq3

$$\frac{ 2x+1 }{ 3x-6 } \geq 3$$

Solve for x

$$x\in (2,\frac{19}{7}]$$

Solution Steps

Denominator 3x-6 cannot be zero since division by zero is not defined. There are two cases.

$$3x-6>0 3x-6<0$$

Consider the case when 3x-6 is positive. Move -6 to the right hand side.

$$3x>6$$

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

$$x>2$$

The initial inequality does not change the direction when multiplied by 3x-6 for 3x-6>0.

$$2x+1\geq 3\left(3x-6\right)$$

Multiply out the right hand side.

$$2x+1\geq 9x-18$$

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

$$2x-9x\geq -1-18$$

Combine like terms.

$$-7x\geq -19$$

Divide both sides by -7. Since -7 is negative, the inequality direction is changed.

$$x\leq \frac{19}{7}$$

Consider condition x>2 specified above.

$$x\in (2,\frac{19}{7}]$$

Now consider the case when 3x-6 is negative. Move -6 to the right hand side.

$$3x<6$$

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

$$x<2$$

The initial inequality changes the direction when multiplied by 3x-6 for 3x-6<0.

$$2x+1\leq 3\left(3x-6\right)$$

Multiply out the right hand side.

$$2x+1\leq 9x-18$$

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

$$2x-9x\leq -1-18$$

Combine like terms.

$$-7x\leq -19$$

Divide both sides by -7. Since -7 is negative, the inequality direction is changed.

$$x\geq \frac{19}{7}$$

Consider condition x<2 specified above.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in (2,\frac{19}{7}]$$