Solve 3x-1/2x+4geq3

$$\frac { 3 x - 1 } { 2 x + 4 } \geq 3$$

Solve for x

$$x\in [-\frac{13}{3},-2)$$

Solution Steps

Denominator 2x+4 cannot be zero since division by zero is not defined. There are two cases.

$$2x+4>0 2x+4<0$$

Consider the case when 2x+4 is positive. Move 4 to the right hand side.

$$2x>-4$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x>-2$$

The initial inequality does not change the direction when multiplied by 2x+4 for 2x+4>0.

$$3x-1\geq 3\left(2x+4\right)$$

Multiply out the right hand side.

$$3x-1\geq 6x+12$$

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

$$3x-6x\geq 1+12$$

Combine like terms.

$$-3x\geq 13$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

$$x\leq -\frac{13}{3}$$

Consider condition x>-2 specified above.

$$x\in \emptyset$$

Now consider the case when 2x+4 is negative. Move 4 to the right hand side.

$$2x<-4$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x<-2$$

The initial inequality changes the direction when multiplied by 2x+4 for 2x+4<0.

$$3x-1\leq 3\left(2x+4\right)$$

Multiply out the right hand side.

$$3x-1\leq 6x+12$$

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

$$3x-6x\leq 1+12$$

Combine like terms.

$$-3x\leq 13$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

$$x\geq -\frac{13}{3}$$

Consider condition x<-2 specified above.

$$x\in [-\frac{13}{3},-2)$$

The final solution is the union of the obtained solutions.

$$x\in [-\frac{13}{3},-2)$$