Solve x-1/x+3geq5

$$\frac{ x-1 }{ x+3 } \geq 5$$

Solve for x

$$x\in [-4,-3)$$

Solution Steps

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

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

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

$$x>-3$$

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

$$x-1\geq 5\left(x+3\right)$$

Multiply out the right hand side.

$$x-1\geq 5x+15$$

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

$$x-5x\geq 1+15$$

Combine like terms.

$$-4x\geq 16$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x\leq -4$$

Consider condition x>-3 specified above.

$$x\in \emptyset$$

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

$$x<-3$$

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

$$x-1\leq 5\left(x+3\right)$$

Multiply out the right hand side.

$$x-1\leq 5x+15$$

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

$$x-5x\leq 1+15$$

Combine like terms.

$$-4x\leq 16$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x\geq -4$$

Consider condition x<-3 specified above.

$$x\in [-4,-3)$$

The final solution is the union of the obtained solutions.

$$x\in [-4,-3)$$