Solve x-98/3x+18geq-4

$$\frac { x - 98 } { 3 x + 18 } \geq - 4$$

Solve for x

$$x\in (-\infty,-6)\cup [2,\infty)$$

Solution Steps

Denominator 3x+18 cannot be zero since division by zero is not defined. There are two cases.

$$3x+18>0 3x+18<0$$

Consider the case when 3x+18 is positive. Move 18 to the right hand side.

$$3x>-18$$

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

$$x>-6$$

The initial inequality does not change the direction when multiplied by 3x+18 for 3x+18>0.

$$x-98\geq -4\left(3x+18\right)$$

Multiply out the right hand side.

$$x-98\geq -12x-72$$

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

$$x+12x\geq 98-72$$

Combine like terms.

$$13x\geq 26$$

Divide both sides by 13. Since 13 is positive, the inequality direction remains the same.

$$x\geq 2$$

Now consider the case when 3x+18 is negative. Move 18 to the right hand side.

$$3x<-18$$

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

$$x<-6$$

The initial inequality changes the direction when multiplied by 3x+18 for 3x+18<0.

$$x-98\leq -4\left(3x+18\right)$$

Multiply out the right hand side.

$$x-98\leq -12x-72$$

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

$$x+12x\leq 98-72$$

Combine like terms.

$$13x\leq 26$$

Divide both sides by 13. Since 13 is positive, the inequality direction remains the same.

$$x\leq 2$$

Consider condition x<-6 specified above.

$$x<-6$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-6)\cup [2,\infty)$$