Solve 2x+1/x-5leq3

$$\frac { 2 x + 1 } { x - 5 } \leq 3$$

Solve for x

$$x\in (-\infty,5)\cup [16,\infty)$$

Solution Steps

Denominator x-5 cannot be zero since division by zero is not defined. There are two cases.

$$x-5>0 x-5<0$$

Consider the case when x-5 is positive. Move -5 to the right hand side.

$$x>5$$

The initial inequality does not change the direction when multiplied by x-5 for x-5>0.

$$2x+1\leq 3\left(x-5\right)$$

Multiply out the right hand side.

$$2x+1\leq 3x-15$$

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

$$2x-3x\leq -1-15$$

Combine like terms.

$$-x\leq -16$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x\geq 16$$

Now consider the case when x-5 is negative. Move -5 to the right hand side.

$$x<5$$

The initial inequality changes the direction when multiplied by x-5 for x-5<0.

$$2x+1\geq 3\left(x-5\right)$$

Multiply out the right hand side.

$$2x+1\geq 3x-15$$

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

$$2x-3x\geq -1-15$$

Combine like terms.

$$-x\geq -16$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x\leq 16$$

Consider condition x<5 specified above.

$$x<5$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,5)\cup [16,\infty)$$