Solve 2x+1/3-xgeq1

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

Solve for x

$$x\in [\frac{2}{3},3)$$

Solution Steps

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

$$3-x>0 3-x<0$$

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

$$-x>-3$$

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

$$x<3$$

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

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

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

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

Combine like terms.

$$3x\geq 2$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

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

Consider condition x<3 specified above.

$$x\in [\frac{2}{3},3)$$

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

$$-x<-3$$

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

$$x>3$$

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

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

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

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

Combine like terms.

$$3x\leq 2$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

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

Consider condition x>3 specified above.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in [\frac{2}{3},3)$$