Solve 2x+1/x-3leq3

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

Solve for x

$$x\in (-\infty,3)\cup [10,\infty)$$

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.

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

Multiply out the right hand side.

$$2x+1\leq 3x-9$$

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

$$2x-3x\leq -1-9$$

Combine like terms.

$$-x\leq -10$$

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

$$x\geq 10$$

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.

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

Multiply out the right hand side.

$$2x+1\geq 3x-9$$

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

$$2x-3x\geq -1-9$$

Combine like terms.

$$-x\geq -10$$

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

$$x\leq 10$$

Consider condition x<3 specified above.

$$x<3$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,3)\cup [10,\infty)$$