Solve z/z-2-9/z+6=z^2/z^2+4z-12

$$\frac { z } { z - 2 } - \frac { 9 } { z + 6 } = \frac { z ^ { 2 } } { z ^ { 2 } + 4 z - 12 }$$

Solve for z

$$z=6$$

Steps for Solving Linear Equation

Variable z cannot be equal to any of the values -6,2 since division by zero is not defined. Multiply both sides of the equation by \left(z-2\right)\left(z+6\right), the least common multiple of z-2,z+6,z^{2}+4z-12.

$$\left(z+6\right)z-\left(z-2\right)\times 9=z^{2}$$

Use the distributive property to multiply z+6 by z.

$$z^{2}+6z-\left(z-2\right)\times 9=z^{2}$$

Use the distributive property to multiply z-2 by 9.

$$z^{2}+6z-\left(9z-18\right)=z^{2}$$

To find the opposite of 9z-18, find the opposite of each term.

$$z^{2}+6z-9z+18=z^{2}$$

Combine 6z and -9z to get -3z.

$$z^{2}-3z+18=z^{2}$$

Subtract z^{2} from both sides.

$$z^{2}-3z+18-z^{2}=0$$

Combine z^{2} and -z^{2} to get 0.

$$-3z+18=0$$

Subtract 18 from both sides. Anything subtracted from zero gives its negation.

$$-3z=-18$$

Divide both sides by -3.

$$z=\frac{-18}{-3}$$

Divide -18 by -3 to get 6.

$$z=6$$