Solve 2leqx+3/x-1

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

Solve for x

$$x\in (1,5]$$

Solution Steps

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

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

Denominator x-1 cannot be zero since division by zero is not defined. There are two cases.

$$x-1>0 x-1<0$$

Consider the case when x-1 is positive. Move -1 to the right hand side.

$$x>1$$

The initial inequality does not change the direction when multiplied by x-1 for x-1>0.

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

Multiply out the right hand side.

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

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

$$x-2x\geq -3-2$$

Combine like terms.

$$-x\geq -5$$

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

$$x\leq 5$$

Consider condition x>1 specified above.

$$x\in (1,5]$$

Now consider the case when x-1 is negative. Move -1 to the right hand side.

$$x<1$$

The initial inequality changes the direction when multiplied by x-1 for x-1<0.

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

Multiply out the right hand side.

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

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

$$x-2x\leq -3-2$$

Combine like terms.

$$-x\leq -5$$

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

$$x\geq 5$$

Consider condition x<1 specified above.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in (1,5]$$