Find Where Undefined/Discontinuous 1/(x-5)-4/(x+7)=12/(x^2+2x-35)

$\frac{1}{x - 5} - \frac{4}{x + 7} = \frac{12}{x^{2} + 2 x - 35}$

Step 1

Subtract $\frac{12}{x^{2} + 2 x - 35}$ from both sides of the equation.

$\frac{1}{x - 5} - \frac{4}{x + 7} - \frac{12}{x^{2} + 2 x - 35} = 0$

Step 2

Simplify $\frac{1}{x - 5} - \frac{4}{x + 7} - \frac{12}{x^{2} + 2 x - 35}$ .

Factor $x^{2} + 2 x - 35$ 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 $- 35$ and whose sum is $2$ .

$- 5, 7$

Write the factored form using these integers.

$\frac{1}{x - 5} - \frac{4}{x + 7} - \frac{12}{(x - 5) (x + 7)} = 0$

To write $\frac{1}{x - 5}$ as a fraction with a common denominator, multiply by $\frac{x + 7}{x + 7}$ .

$\frac{1}{x - 5} \cdot \frac{x + 7}{x + 7} - \frac{4}{x + 7} - \frac{12}{(x - 5) (x + 7)} = 0$

To write $- \frac{4}{x + 7}$ as a fraction with a common denominator, multiply by $\frac{x - 5}{x - 5}$ .

$\frac{1}{x - 5} \cdot \frac{x + 7}{x + 7} - \frac{4}{x + 7} \cdot \frac{x - 5}{x - 5} - \frac{12}{(x - 5) (x + 7)} = 0$

Write each expression with a common denominator of $(x - 5) (x + 7)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{x - 5}$ by $\frac{x + 7}{x + 7}$ .

$\frac{x + 7}{(x - 5) (x + 7)} - \frac{4}{x + 7} \cdot \frac{x - 5}{x - 5} - \frac{12}{(x - 5) (x + 7)} = 0$

Multiply $\frac{4}{x + 7}$ by $\frac{x - 5}{x - 5}$ .

$\frac{x + 7}{(x - 5) (x + 7)} - \frac{4 (x - 5)}{(x + 7) (x - 5)} - \frac{12}{(x - 5) (x + 7)} = 0$

Reorder the factors of $(x - 5) (x + 7)$ .

$\frac{x + 7}{(x + 7) (x - 5)} - \frac{4 (x - 5)}{(x + 7) (x - 5)} - \frac{12}{(x - 5) (x + 7)} = 0$

Combine the numerators over the common denominator.

$\frac{x + 7 - 4 (x - 5)}{(x + 7) (x - 5)} - \frac{12}{(x - 5) (x + 7)} = 0$

Combine the numerators over the common denominator.

$\frac{x + 7 - 4 (x - 5) - 12}{(x + 7) (x - 5)} = 0$

Simplify each term.

Apply the distributive property.

$\frac{x + 7 - 4 x - 4 \cdot - 5 - 12}{(x + 7) (x - 5)} = 0$

Multiply $- 4$ by $- 5$ .

$\frac{x + 7 - 4 x + 20 - 12}{(x + 7) (x - 5)} = 0$

Subtract $4 x$ from $x$ .

$\frac{- 3 x + 7 + 20 - 12}{(x + 7) (x - 5)} = 0$

Add $7$ and $20$ .

$\frac{- 3 x + 27 - 12}{(x + 7) (x - 5)} = 0$

Subtract $12$ from $27$ .

$\frac{- 3 x + 15}{(x + 7) (x - 5)} = 0$

Factor $3$ out of $- 3 x + 15$ .

Factor $3$ out of $- 3 x$ .

$\frac{3 (- x) + 15}{(x + 7) (x - 5)} = 0$

Factor $3$ out of $15$ .

$\frac{3 (- x) + 3 (5)}{(x + 7) (x - 5)} = 0$

Factor $3$ out of $3 (- x) + 3 (5)$ .

$\frac{3 (- x + 5)}{(x + 7) (x - 5)} = 0$

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

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

$\frac{3 (- (x) + 5)}{(x + 7) (x - 5)} = 0$

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

$\frac{3 (- (x) - 1 (- 5))}{(x + 7) (x - 5)} = 0$

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

$\frac{3 (- (x - 5))}{(x + 7) (x - 5)} = 0$

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

$\frac{3 (- 1 (x - 5))}{(x + 7) (x - 5)} = 0$

Cancel the common factor.

$\frac{3 (- 1 (x - 5))}{(x + 7) (x - 5)} = 0$

Rewrite the expression.

$\frac{3 \cdot (- 1)}{x + 7} = 0$

Multiply $3$ by $- 1$ .

$\frac{- 3}{x + 7} = 0$

Move the negative in front of the fraction.

$- \frac{3}{x + 7} = 0$

Step 3

Set the denominator in $\frac{3}{x + 7}$ equal to $0$ to find where the expression is undefined.

$x + 7 = 0$

Step 4

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 5

Find Where Undefined/Discontinuous 1/(x-5)-4/(x+7)=12/(x^2+2x-35)

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