Solve 3x+3/9x+8geq-9

$$\frac { 3 x + 3 } { 9 x + 8 } \geq - 9$$

Solve for x

$$x\in (-\infty,-\frac{25}{28}]\cup (-\frac{8}{9},\infty)$$

Solution Steps

Denominator 9x+8 cannot be zero since division by zero is not defined. There are two cases.

$$9x+8>0 9x+8<0$$

Consider the case when 9x+8 is positive. Move 8 to the right hand side.

$$9x>-8$$

Divide both sides by 9. Since 9 is positive, the inequality direction remains the same.

$$x>-\frac{8}{9}$$

The initial inequality does not change the direction when multiplied by 9x+8 for 9x+8>0.

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

Multiply out the right hand side.

$$3x+3\geq -81x-72$$

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

$$3x+81x\geq -3-72$$

Combine like terms.

$$84x\geq -75$$

Divide both sides by 84. Since 84 is positive, the inequality direction remains the same.

$$x\geq -\frac{25}{28}$$

Consider condition x>-\frac{8}{9} specified above.

$$x>-\frac{8}{9}$$

Now consider the case when 9x+8 is negative. Move 8 to the right hand side.

$$9x<-8$$

Divide both sides by 9. Since 9 is positive, the inequality direction remains the same.

$$x<-\frac{8}{9}$$

The initial inequality changes the direction when multiplied by 9x+8 for 9x+8<0.

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

Multiply out the right hand side.

$$3x+3\leq -81x-72$$

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

$$3x+81x\leq -3-72$$

Combine like terms.

$$84x\leq -75$$

Divide both sides by 84. Since 84 is positive, the inequality direction remains the same.

$$x\leq -\frac{25}{28}$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-\frac{25}{28}]\cup (-\frac{8}{9},\infty)$$