Solve by Factoring z/(z-3)+3/(z+3)=18/(z^2-9)

$\frac{z}{z - 3} + \frac{3}{z + 3} = \frac{18}{z^{2} - 9}$

Step 1

Subtract $\frac{18}{z^{2} - 9}$ from both sides of the equation.

$\frac{z}{z - 3} + \frac{3}{z + 3} - \frac{18}{z^{2} - 9} = 0$

Step 2

Simplify $\frac{z}{z - 3} + \frac{3}{z + 3} - \frac{18}{z^{2} - 9}$ .

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = z$ and $b = 3$ .

$\frac{z}{z - 3} + \frac{3}{z + 3} - \frac{18}{(z + 3) (z - 3)} = 0$

To write $\frac{z}{z - 3}$ as a fraction with a common denominator, multiply by $\frac{z + 3}{z + 3}$ .

$\frac{z}{z - 3} \cdot \frac{z + 3}{z + 3} + \frac{3}{z + 3} - \frac{18}{(z + 3) (z - 3)} = 0$

To write $\frac{3}{z + 3}$ as a fraction with a common denominator, multiply by $\frac{z - 3}{z - 3}$ .

$\frac{z}{z - 3} \cdot \frac{z + 3}{z + 3} + \frac{3}{z + 3} \cdot \frac{z - 3}{z - 3} - \frac{18}{(z + 3) (z - 3)} = 0$

Write each expression with a common denominator of $(z - 3) (z + 3)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{z}{z - 3}$ by $\frac{z + 3}{z + 3}$ .

$\frac{z (z + 3)}{(z - 3) (z + 3)} + \frac{3}{z + 3} \cdot \frac{z - 3}{z - 3} - \frac{18}{(z + 3) (z - 3)} = 0$

Multiply $\frac{3}{z + 3}$ by $\frac{z - 3}{z - 3}$ .

$\frac{z (z + 3)}{(z - 3) (z + 3)} + \frac{3 (z - 3)}{(z + 3) (z - 3)} - \frac{18}{(z + 3) (z - 3)} = 0$

Reorder the factors of $(z - 3) (z + 3)$ .

$\frac{z (z + 3)}{(z + 3) (z - 3)} + \frac{3 (z - 3)}{(z + 3) (z - 3)} - \frac{18}{(z + 3) (z - 3)} = 0$

Combine the numerators over the common denominator.

$\frac{z (z + 3) + 3 (z - 3)}{(z + 3) (z - 3)} - \frac{18}{(z + 3) (z - 3)} = 0$

Combine the numerators over the common denominator.

$\frac{z (z + 3) + 3 (z - 3) - 18}{(z + 3) (z - 3)} = 0$

Simplify each term.

Apply the distributive property.

$\frac{z \cdot z + z \cdot 3 + 3 (z - 3) - 18}{(z + 3) (z - 3)} = 0$

Multiply $z$ by $z$ .

$\frac{z^{2} + z \cdot 3 + 3 (z - 3) - 18}{(z + 3) (z - 3)} = 0$

Move $3$ to the left of $z$ .

$\frac{z^{2} + 3 \cdot z + 3 (z - 3) - 18}{(z + 3) (z - 3)} = 0$

Apply the distributive property.

$\frac{z^{2} + 3 z + 3 z + 3 \cdot - 3 - 18}{(z + 3) (z - 3)} = 0$

Multiply $3$ by $- 3$ .

$\frac{z^{2} + 3 z + 3 z - 9 - 18}{(z + 3) (z - 3)} = 0$

Add $3 z$ and $3 z$ .

$\frac{z^{2} + 6 z - 9 - 18}{(z + 3) (z - 3)} = 0$

Subtract $18$ from $- 9$ .

$\frac{z^{2} + 6 z - 27}{(z + 3) (z - 3)} = 0$

Factor $z^{2} + 6 z - 27$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 27$ and whose sum is $6$ .

$- 3, 9$

Write the factored form using these integers.

$\frac{(z - 3) (z + 9)}{(z + 3) (z - 3)} = 0$

Cancel the common factor of $z - 3$ .

Cancel the common factor.

$\frac{(z - 3) (z + 9)}{(z + 3) (z - 3)} = 0$

Rewrite the expression.

$\frac{z + 9}{z + 3} = 0$

Step 3

Set the numerator equal to zero.

$z + 9 = 0$

Step 4

Subtract $9$ from both sides of the equation.

$z = - 9$

Solve by Factoring z/(z-3)+3/(z+3)=18/(z^2-9)

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