Solve x/x-7<5

$$\frac{ x }{ x-7 } < 5$$

Solve for x

$$x\in \left(-\infty,7\right)\cup \left(\frac{35}{4},\infty\right)$$

Solution Steps

Denominator x-7 cannot be zero since division by zero is not defined. There are two cases.

$$x-7>0 x-7<0$$

Consider the case when x-7 is positive. Move -7 to the right hand side.

$$x>7$$

The initial inequality does not change the direction when multiplied by x-7 for x-7>0.

$$x<5\left(x-7\right)$$

Multiply out the right hand side.

$$x<5x-35$$

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

$$x-5x<-35$$

Combine like terms.

$$-4x<-35$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x>\frac{35}{4}$$

Consider condition x>7 specified above. The result remains the same.

$$x>\frac{35}{4}$$

Now consider the case when x-7 is negative. Move -7 to the right hand side.

$$x<7$$

The initial inequality changes the direction when multiplied by x-7 for x-7<0.

$$x>5\left(x-7\right)$$

Multiply out the right hand side.

$$x>5x-35$$

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

$$x-5x>-35$$

Combine like terms.

$$-4x>-35$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x<\frac{35}{4}$$

Consider condition x<7 specified above.

$$x<7$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,7\right)\cup \left(\frac{35}{4},\infty\right)$$