Solve 3x+1/x+4geq1

$$\frac { 3 x + 1 } { x + 4 } \geq 1$$

Solve for x

$$x\in (-\infty,-4)\cup [\frac{3}{2},\infty)$$

Solution Steps

Denominator x+4 cannot be zero since division by zero is not defined. There are two cases.

$$x+4>0 x+4<0$$

Consider the case when x+4 is positive. Move 4 to the right hand side.

$$x>-4$$

The initial inequality does not change the direction when multiplied by x+4 for x+4>0.

$$3x+1\geq x+4$$

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

$$3x-x\geq -1+4$$

Combine like terms.

$$2x\geq 3$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x\geq \frac{3}{2}$$

Now consider the case when x+4 is negative. Move 4 to the right hand side.

$$x<-4$$

The initial inequality changes the direction when multiplied by x+4 for x+4<0.

$$3x+1\leq x+4$$

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

$$3x-x\leq -1+4$$

Combine like terms.

$$2x\leq 3$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x\leq \frac{3}{2}$$

Consider condition x<-4 specified above.

$$x<-4$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-4)\cup [\frac{3}{2},\infty)$$