Solve x+3/3x+1geq1

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

Solve for x

$$x\in (-\frac{1}{3},1]$$

Solution Steps

Denominator 3x+1 cannot be zero since division by zero is not defined. There are two cases.

$$3x+1>0 3x+1<0$$

Consider the case when 3x+1 is positive. Move 1 to the right hand side.

$$3x>-1$$

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

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

The initial inequality does not change the direction when multiplied by 3x+1 for 3x+1>0.

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

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

$$x-3x\geq -3+1$$

Combine like terms.

$$-2x\geq -2$$

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

$$x\leq 1$$

Consider condition x>-\frac{1}{3} specified above.

$$x\in (-\frac{1}{3},1]$$

Now consider the case when 3x+1 is negative. Move 1 to the right hand side.

$$3x<-1$$

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

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

The initial inequality changes the direction when multiplied by 3x+1 for 3x+1<0.

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

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

$$x-3x\leq -3+1$$

Combine like terms.

$$-2x\leq -2$$

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

$$x\geq 1$$

Consider condition x<-\frac{1}{3} specified above.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in (-\frac{1}{3},1]$$