Solve x+4/2x-1leq3

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

Solve for x

$$x\in (-\infty,\frac{1}{2})\cup [\frac{7}{5},\infty)$$

Solution Steps

Denominator 2x-1 cannot be zero since division by zero is not defined. There are two cases.

$$2x-1>0 2x-1<0$$

Consider the case when 2x-1 is positive. Move -1 to the right hand side.

$$2x>1$$

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

$$x>\frac{1}{2}$$

The initial inequality does not change the direction when multiplied by 2x-1 for 2x-1>0.

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

Multiply out the right hand side.

$$x+4\leq 6x-3$$

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

$$x-6x\leq -4-3$$

Combine like terms.

$$-5x\leq -7$$

Divide both sides by -5. Since -5 is negative, the inequality direction is changed.

$$x\geq \frac{7}{5}$$

Now consider the case when 2x-1 is negative. Move -1 to the right hand side.

$$2x<1$$

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

$$x<\frac{1}{2}$$

The initial inequality changes the direction when multiplied by 2x-1 for 2x-1<0.

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

Multiply out the right hand side.

$$x+4\geq 6x-3$$

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

$$x-6x\geq -4-3$$

Combine like terms.

$$-5x\geq -7$$

Divide both sides by -5. Since -5 is negative, the inequality direction is changed.

$$x\leq \frac{7}{5}$$

Consider condition x<\frac{1}{2} specified above.

$$x<\frac{1}{2}$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,\frac{1}{2})\cup [\frac{7}{5},\infty)$$