Solve for n 1/(n^2-2n)-(n+6)/(n-2)=6/(n^2-2n)

$\frac{1}{n^{2} - 2 n} - \frac{n + 6}{n - 2} = \frac{6}{n^{2} - 2 n}$

Step 1

Factor each term.

Factor $n$ out of $n^{2} - 2 n$ .

Factor $n$ out of $n^{2}$ .

$\frac{1}{n \cdot n - 2 n} - \frac{n + 6}{n - 2} = \frac{6}{n^{2} - 2 n}$

Factor $n$ out of $- 2 n$ .

$\frac{1}{n \cdot n + n \cdot - 2} - \frac{n + 6}{n - 2} = \frac{6}{n^{2} - 2 n}$

Factor $n$ out of $n \cdot n + n \cdot - 2$ .

$\frac{1}{n (n - 2)} - \frac{n + 6}{n - 2} = \frac{6}{n^{2} - 2 n}$

Factor $n$ out of $n^{2} - 2 n$ .

Factor $n$ out of $n^{2}$ .

$\frac{1}{n (n - 2)} - \frac{n + 6}{n - 2} = \frac{6}{n \cdot n - 2 n}$

Factor $n$ out of $- 2 n$ .

$\frac{1}{n (n - 2)} - \frac{n + 6}{n - 2} = \frac{6}{n \cdot n + n \cdot - 2}$

Factor $n$ out of $n \cdot n + n \cdot - 2$ .

$\frac{1}{n (n - 2)} - \frac{n + 6}{n - 2} = \frac{6}{n (n - 2)}$

Step 2

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$n (n - 2), n - 2, n (n - 2)$

Since $n (n - 2), n - 2, n (n - 2)$ contain both numbers and variables, there are four steps to find the LCM. Find LCM for the numeric, variable, and compound variable parts. Then, multiply them all together.

Steps to find the LCM for $n (n - 2), n - 2, n (n - 2)$ are:

1. Find the LCM for the numeric part $1, 1, 1$ .

2. Find the LCM for the variable part $n^{1}, n^{1}$ .

3. Find the LCM for the compound variable part $n - 2, n - 2, n - 2$ .

4. Multiply each LCM together.

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

The factor for $n^{1}$ is $n$ itself.

$n^{1} = n$

$n$ occurs $1$ time.

The LCM of $n^{1}, n^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$n$

The factor for $n - 2$ is $n - 2$ itself.

$(n - 2) = n - 2$

$(n - 2)$ occurs $1$ time.

The LCM of $n - 2, n - 2, n - 2$ is the result of multiplying all factors the greatest number of times they occur in either term.

$n - 2$

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$n (n - 2)$

Step 3

Multiply each term in $\frac{1}{n (n - 2)} - \frac{n + 6}{n - 2} = \frac{6}{n (n - 2)}$ by $n (n - 2)$ to eliminate the fractions.

Multiply each term in $\frac{1}{n (n - 2)} - \frac{n + 6}{n - 2} = \frac{6}{n (n - 2)}$ by $n (n - 2)$ .

$\frac{1}{n (n - 2)} (n (n - 2)) - \frac{n + 6}{n - 2} (n (n - 2)) = \frac{6}{n (n - 2)} (n (n - 2))$

Simplify the left side.

Simplify each term.

Cancel the common factor of $n (n - 2)$ .

Cancel the common factor.

$\frac{1}{n (n - 2)} (n (n - 2)) - \frac{n + 6}{n - 2} (n (n - 2)) = \frac{6}{n (n - 2)} (n (n - 2))$

Rewrite the expression.

$1 - \frac{n + 6}{n - 2} (n (n - 2)) = \frac{6}{n (n - 2)} (n (n - 2))$

Cancel the common factor of $n - 2$ .

Move the leading negative in $- \frac{n + 6}{n - 2}$ into the numerator.

$1 + \frac{- (n + 6)}{n - 2} (n (n - 2)) = \frac{6}{n (n - 2)} (n (n - 2))$

Factor $n - 2$ out of $n (n - 2)$ .

$1 + \frac{- (n + 6)}{n - 2} ((n - 2) n) = \frac{6}{n (n - 2)} (n (n - 2))$

Cancel the common factor.

$1 + \frac{- (n + 6)}{n - 2} ((n - 2) n) = \frac{6}{n (n - 2)} (n (n - 2))$

Rewrite the expression.

$1 - (n + 6) n = \frac{6}{n (n - 2)} (n (n - 2))$

Apply the distributive property.

$1 + (- n - 1 \cdot 6) n = \frac{6}{n (n - 2)} (n (n - 2))$

Multiply $- 1$ by $6$ .

$1 + (- n - 6) n = \frac{6}{n (n - 2)} (n (n - 2))$

Apply the distributive property.

$1 - n \cdot n - 6 n = \frac{6}{n (n - 2)} (n (n - 2))$

Multiply $n$ by $n$ by adding the exponents.

Move $n$ .

$1 - (n \cdot n) - 6 n = \frac{6}{n (n - 2)} (n (n - 2))$

Multiply $n$ by $n$ .

$1 - n^{2} - 6 n = \frac{6}{n (n - 2)} (n (n - 2))$

Simplify the right side.

Cancel the common factor of $n (n - 2)$ .

Cancel the common factor.

$1 - n^{2} - 6 n = \frac{6}{n (n - 2)} (n (n - 2))$

Rewrite the expression.

$1 - n^{2} - 6 n = 6$

Step 4

Solve the equation.

Subtract $6$ from both sides of the equation.

$1 - n^{2} - 6 n - 6 = 0$

Subtract $6$ from $1$ .

$- n^{2} - 6 n - 5 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- n^{2} - 6 n - 5$ .

Factor $- 1$ out of $- n^{2}$ .

$- (n^{2}) - 6 n - 5 = 0$

Factor $- 1$ out of $- 6 n$ .

$- (n^{2}) - (6 n) - 5 = 0$

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

$- (n^{2}) - (6 n) - 1 \cdot 5 = 0$

Factor $- 1$ out of $- (n^{2}) - (6 n)$ .

$- (n^{2} + 6 n) - 1 \cdot 5 = 0$

Factor $- 1$ out of $- (n^{2} + 6 n) - 1 (5)$ .

$- (n^{2} + 6 n + 5) = 0$

Factor.

Factor $n^{2} + 6 n + 5$ 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 $5$ and whose sum is $6$ .

$1, 5$

Write the factored form using these integers.

$- ((n + 1) (n + 5)) = 0$

Remove unnecessary parentheses.

$- (n + 1) (n + 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$n + 1 = 0$

$n + 5 = 0$

Set $n + 1$ equal to $0$ and solve for $n$ .

Set $n + 1$ equal to $0$ .

$n + 1 = 0$

Subtract $1$ from both sides of the equation.

$n = - 1$

Set $n + 5$ equal to $0$ and solve for $n$ .

Set $n + 5$ equal to $0$ .

$n + 5 = 0$

Subtract $5$ from both sides of the equation.

$n = - 5$

The final solution is all the values that make $- (n + 1) (n + 5) = 0$ true.

$n = - 1, - 5$

Solve for n 1/(n^2-2n)-(n+6)/(n-2)=6/(n^2-2n)

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