Solve 3-x/x-4geq-4

$$\frac { 3 - x } { x - 4 } \geq - 4$$

Solve for x

$$x\in (-\infty,4)\cup [\frac{13}{3},\infty)$$

Solution Steps

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

$$x-4>0 x-4<0$$

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

$$x>4$$

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

$$3-x\geq -4\left(x-4\right)$$

Multiply out the right hand side.

$$3-x\geq -4x+16$$

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

$$-x+4x\geq -3+16$$

Combine like terms.

$$3x\geq 13$$

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

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

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

$$x<4$$

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

$$3-x\leq -4\left(x-4\right)$$

Multiply out the right hand side.

$$3-x\leq -4x+16$$

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

$$-x+4x\leq -3+16$$

Combine like terms.

$$3x\leq 13$$

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

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

Consider condition x<4 specified above.

$$x<4$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,4)\cup [\frac{13}{3},\infty)$$