Solve x-3/3x+1leq1

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

Solve for x

$$x\in (-\infty,-2]\cup (-\frac{1}{3},\infty)$$

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.

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

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.

$$-2x\leq 4$$

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

$$x\geq -2$$

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

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

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.

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

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.

$$-2x\geq 4$$

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

$$x\leq -2$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-2]\cup (-\frac{1}{3},\infty)$$