Find Where Undefined/Discontinuous 1/(x-5)-5/(x+3)=8/(x^2-2x-15)

$\frac{1}{x - 5} - \frac{5}{x + 3} = \frac{8}{x^{2} - 2 x - 15}$

Step 1

Subtract $\frac{8}{x^{2} - 2 x - 15}$ from both sides of the equation.

$\frac{1}{x - 5} - \frac{5}{x + 3} - \frac{8}{x^{2} - 2 x - 15} = 0$

Step 2

Simplify $\frac{1}{x - 5} - \frac{5}{x + 3} - \frac{8}{x^{2} - 2 x - 15}$ .

Factor $x^{2} - 2 x - 15$ 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 $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$\frac{1}{x - 5} - \frac{5}{x + 3} - \frac{8}{(x - 5) (x + 3)} = 0$

To write $\frac{1}{x - 5}$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

$\frac{1}{x - 5} \cdot \frac{x + 3}{x + 3} - \frac{5}{x + 3} - \frac{8}{(x - 5) (x + 3)} = 0$

To write $- \frac{5}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{x - 5}{x - 5}$ .

$\frac{1}{x - 5} \cdot \frac{x + 3}{x + 3} - \frac{5}{x + 3} \cdot \frac{x - 5}{x - 5} - \frac{8}{(x - 5) (x + 3)} = 0$

Write each expression with a common denominator of $(x - 5) (x + 3)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{x - 5}$ by $\frac{x + 3}{x + 3}$ .

$\frac{x + 3}{(x - 5) (x + 3)} - \frac{5}{x + 3} \cdot \frac{x - 5}{x - 5} - \frac{8}{(x - 5) (x + 3)} = 0$

Multiply $\frac{5}{x + 3}$ by $\frac{x - 5}{x - 5}$ .

$\frac{x + 3}{(x - 5) (x + 3)} - \frac{5 (x - 5)}{(x + 3) (x - 5)} - \frac{8}{(x - 5) (x + 3)} = 0$

Reorder the factors of $(x - 5) (x + 3)$ .

$\frac{x + 3}{(x + 3) (x - 5)} - \frac{5 (x - 5)}{(x + 3) (x - 5)} - \frac{8}{(x - 5) (x + 3)} = 0$

Combine the numerators over the common denominator.

$\frac{x + 3 - 5 (x - 5)}{(x + 3) (x - 5)} - \frac{8}{(x - 5) (x + 3)} = 0$

Combine the numerators over the common denominator.

$\frac{x + 3 - 5 (x - 5) - 8}{(x + 3) (x - 5)} = 0$

Simplify each term.

Apply the distributive property.

$\frac{x + 3 - 5 x - 5 \cdot - 5 - 8}{(x + 3) (x - 5)} = 0$

Multiply $- 5$ by $- 5$ .

$\frac{x + 3 - 5 x + 25 - 8}{(x + 3) (x - 5)} = 0$

Subtract $5 x$ from $x$ .

$\frac{- 4 x + 3 + 25 - 8}{(x + 3) (x - 5)} = 0$

Add $3$ and $25$ .

$\frac{- 4 x + 28 - 8}{(x + 3) (x - 5)} = 0$

Subtract $8$ from $28$ .

$\frac{- 4 x + 20}{(x + 3) (x - 5)} = 0$

Factor $4$ out of $- 4 x + 20$ .

Factor $4$ out of $- 4 x$ .

$\frac{4 (- x) + 20}{(x + 3) (x - 5)} = 0$

Factor $4$ out of $20$ .

$\frac{4 (- x) + 4 (5)}{(x + 3) (x - 5)} = 0$

Factor $4$ out of $4 (- x) + 4 (5)$ .

$\frac{4 (- x + 5)}{(x + 3) (x - 5)} = 0$

Cancel the common factor of $- x + 5$ and $x - 5$ .

Factor $- 1$ out of $- x$ .

$\frac{4 (- (x) + 5)}{(x + 3) (x - 5)} = 0$

Rewrite $5$ as $- 1 (- 5)$ .

$\frac{4 (- (x) - 1 (- 5))}{(x + 3) (x - 5)} = 0$

Factor $- 1$ out of $- (x) - 1 (- 5)$ .

$\frac{4 (- (x - 5))}{(x + 3) (x - 5)} = 0$

Rewrite $- (x - 5)$ as $- 1 (x - 5)$ .

$\frac{4 (- 1 (x - 5))}{(x + 3) (x - 5)} = 0$

Cancel the common factor.

$\frac{4 (- 1 (x - 5))}{(x + 3) (x - 5)} = 0$

Rewrite the expression.

$\frac{4 \cdot (- 1)}{x + 3} = 0$

Multiply $4$ by $- 1$ .

$\frac{- 4}{x + 3} = 0$

Move the negative in front of the fraction.

$- \frac{4}{x + 3} = 0$

Step 3

Set the denominator in $\frac{4}{x + 3}$ equal to $0$ to find where the expression is undefined.

$x + 3 = 0$

Step 4

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

Find Where Undefined/Discontinuous 1/(x-5)-5/(x+3)=8/(x^2-2x-15)

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