Solve 3x/6x+1geq1

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

Solve for x

$$x\in [-\frac{1}{3},-\frac{1}{6})$$

Solution Steps

Denominator 6x+1 cannot be zero since division by zero is not defined. There are two cases.

$$6x+1>0 6x+1<0$$

Consider the case when 6x+1 is positive. Move 1 to the right hand side.

$$6x>-1$$

Divide both sides by 6. Since 6 is positive, the inequality direction remains the same.

$$x>-\frac{1}{6}$$

The initial inequality does not change the direction when multiplied by 6x+1 for 6x+1>0.

$$3x\geq 6x+1$$

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

$$3x-6x\geq 1$$

Combine like terms.

$$-3x\geq 1$$

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

$$x\leq -\frac{1}{3}$$

Consider condition x>-\frac{1}{6} specified above.

$$x\in \emptyset$$

Now consider the case when 6x+1 is negative. Move 1 to the right hand side.

$$6x<-1$$

Divide both sides by 6. Since 6 is positive, the inequality direction remains the same.

$$x<-\frac{1}{6}$$

The initial inequality changes the direction when multiplied by 6x+1 for 6x+1<0.

$$3x\leq 6x+1$$

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

$$3x-6x\leq 1$$

Combine like terms.

$$-3x\leq 1$$

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

$$x\geq -\frac{1}{3}$$

Consider condition x<-\frac{1}{6} specified above.

$$x\in [-\frac{1}{3},-\frac{1}{6})$$

The final solution is the union of the obtained solutions.

$$x\in [-\frac{1}{3},-\frac{1}{6})$$