Solve -8x-9/3x-8leq-2

$$\frac { - 8 x - 9 } { 3 x - 8 } \leq - 2$$

Solve for x

$$x\in (-\infty,-\frac{25}{2}]\cup (\frac{8}{3},\infty)$$

Solution Steps

Denominator 3x-8 cannot be zero since division by zero is not defined. There are two cases.

$$3x-8>0 3x-8<0$$

Consider the case when 3x-8 is positive. Move -8 to the right hand side.

$$3x>8$$

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

$$x>\frac{8}{3}$$

The initial inequality does not change the direction when multiplied by 3x-8 for 3x-8>0.

$$-8x-9\leq -2\left(3x-8\right)$$

Multiply out the right hand side.

$$-8x-9\leq -6x+16$$

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

$$-8x+6x\leq 9+16$$

Combine like terms.

$$-2x\leq 25$$

Divide both sides by -2. Since -2 is negative, the inequality direction is changed.

$$x\geq -\frac{25}{2}$$

Consider condition x>\frac{8}{3} specified above.

$$x>\frac{8}{3}$$

Now consider the case when 3x-8 is negative. Move -8 to the right hand side.

$$3x<8$$

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

$$x<\frac{8}{3}$$

The initial inequality changes the direction when multiplied by 3x-8 for 3x-8<0.

$$-8x-9\geq -2\left(3x-8\right)$$

Multiply out the right hand side.

$$-8x-9\geq -6x+16$$

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

$$-8x+6x\geq 9+16$$

Combine like terms.

$$-2x\geq 25$$

Divide both sides by -2. Since -2 is negative, the inequality direction is changed.

$$x\leq -\frac{25}{2}$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-\frac{25}{2}]\cup (\frac{8}{3},\infty)$$