Solve 1/z^2-14z+48-1/z-8=1/9z-54

$$\frac { 1 } { z ^ { 2 } - 14 z + 48 } - \frac { 1 } { z - 8 } = \frac { 1 } { 9 z - 54 }$$

Solve for z

$$z = \frac{71}{10} = 7\frac{1}{10} = 7.1$$

Steps for Solving Linear Equation

Variable z cannot be equal to any of the values 6,8 since division by zero is not defined. Multiply both sides of the equation by 9\left(z-8\right)\left(z-6\right), the least common multiple of z^{2}-14z+48,z-8,9z-54.

$$9-\left(9z-54\right)=z-8$$

To find the opposite of 9z-54, find the opposite of each term.

$$9-9z+54=z-8$$

Add 9 and 54 to get 63.

$$63-9z=z-8$$

Subtract z from both sides.

$$63-9z-z=-8$$

Combine -9z and -z to get -10z.

$$63-10z=-8$$

Subtract 63 from both sides.

$$-10z=-8-63$$

Subtract 63 from -8 to get -71.

$$-10z=-71$$

Divide both sides by -10.

$$z=\frac{-71}{-10}$$

Fraction \frac{-71}{-10} can be simplified to \frac{71}{10} by removing the negative sign from both the numerator and the denominator.

$$z=\frac{71}{10}$$