Solve x+8/2-3xgeq2

$$\frac { x + 8 } { 2 - 3 x } \geq 2$$

Solve for x

$$x\in [-\frac{4}{7},\frac{2}{3})$$

Solution Steps

Denominator 2-3x cannot be zero since division by zero is not defined. There are two cases.

$$2-3x>0 2-3x<0$$

Consider the case when 2-3x is positive. Move 2 to the right hand side.

$$-3x>-2$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

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

The initial inequality does not change the direction when multiplied by 2-3x for 2-3x>0.

$$x+8\geq 2\left(2-3x\right)$$

Multiply out the right hand side.

$$x+8\geq 4-6x$$

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

$$x+6x\geq -8+4$$

Combine like terms.

$$7x\geq -4$$

Divide both sides by 7. Since 7 is positive, the inequality direction remains the same.

$$x\geq -\frac{4}{7}$$

Consider condition x<\frac{2}{3} specified above.

$$x\in [-\frac{4}{7},\frac{2}{3})$$

Now consider the case when 2-3x is negative. Move 2 to the right hand side.

$$-3x<-2$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

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

The initial inequality changes the direction when multiplied by 2-3x for 2-3x<0.

$$x+8\leq 2\left(2-3x\right)$$

Multiply out the right hand side.

$$x+8\leq 4-6x$$

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

$$x+6x\leq -8+4$$

Combine like terms.

$$7x\leq -4$$

Divide both sides by 7. Since 7 is positive, the inequality direction remains the same.

$$x\leq -\frac{4}{7}$$

Consider condition x>\frac{2}{3} specified above.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in [-\frac{4}{7},\frac{2}{3})$$