Solve for n (n-6)/(3n)=(n-5)/(3n+1)

$\frac{n - 6}{3 n} = \frac{n - 5}{3 n + 1}$

Step 1

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$(n - 6) (3 n + 1) = 3 n (n - 5)$

Step 2

Solve the equation for $n$ .

Simplify $(n - 6) (3 n + 1)$ .

Start simplifying.

$0 + 0 + (n - 6) (3 n + 1) = 3 n (n - 5)$

Simplify by adding zeros.

$(n - 6) (3 n + 1) = 3 n (n - 5)$

Expand $(n - 6) (3 n + 1)$ using the FOIL Method.

Apply the distributive property.

$n (3 n + 1) - 6 (3 n + 1) = 3 n (n - 5)$

Apply the distributive property.

$n (3 n) + n \cdot 1 - 6 (3 n + 1) = 3 n (n - 5)$

Apply the distributive property.

$n (3 n) + n \cdot 1 - 6 (3 n) - 6 \cdot 1 = 3 n (n - 5)$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$3 n \cdot n + n \cdot 1 - 6 (3 n) - 6 \cdot 1 = 3 n (n - 5)$

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

Move $n$ .

$3 (n \cdot n) + n \cdot 1 - 6 (3 n) - 6 \cdot 1 = 3 n (n - 5)$

Multiply $n$ by $n$ .

$3 n^{2} + n \cdot 1 - 6 (3 n) - 6 \cdot 1 = 3 n (n - 5)$

Multiply $n$ by $1$ .

$3 n^{2} + n - 6 (3 n) - 6 \cdot 1 = 3 n (n - 5)$

Multiply $3$ by $- 6$ .

$3 n^{2} + n - 18 n - 6 \cdot 1 = 3 n (n - 5)$

Multiply $- 6$ by $1$ .

$3 n^{2} + n - 18 n - 6 = 3 n (n - 5)$

Subtract $18 n$ from $n$ .

$3 n^{2} - 17 n - 6 = 3 n (n - 5)$

Simplify $3 n (n - 5)$ .

Apply the distributive property.

$3 n^{2} - 17 n - 6 = 3 n \cdot n + 3 n \cdot - 5$

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

Move $n$ .

$3 n^{2} - 17 n - 6 = 3 (n \cdot n) + 3 n \cdot - 5$

Multiply $n$ by $n$ .

$3 n^{2} - 17 n - 6 = 3 n^{2} + 3 n \cdot - 5$

Multiply $- 5$ by $3$ .

$3 n^{2} - 17 n - 6 = 3 n^{2} - 15 n$

Move all terms containing $n$ to the left side of the equation.

Subtract $3 n^{2}$ from both sides of the equation.

$3 n^{2} - 17 n - 6 - 3 n^{2} = - 15 n$

Add $15 n$ to both sides of the equation.

$3 n^{2} - 17 n - 6 - 3 n^{2} + 15 n = 0$

Combine the opposite terms in $3 n^{2} - 17 n - 6 - 3 n^{2} + 15 n$ .

Subtract $3 n^{2}$ from $3 n^{2}$ .

$- 17 n - 6 + 0 + 15 n = 0$

Add $- 17 n - 6$ and $0$ .

$- 17 n - 6 + 15 n = 0$

Add $- 17 n$ and $15 n$ .

$- 2 n - 6 = 0$

Add $6$ to both sides of the equation.

$- 2 n = 6$

Divide each term in $- 2 n = 6$ by $- 2$ and simplify.

Divide each term in $- 2 n = 6$ by $- 2$ .

$\frac{- 2 n}{- 2} = \frac{6}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 n}{- 2} = \frac{6}{- 2}$

Divide $n$ by $1$ .

$n = \frac{6}{- 2}$

Simplify the right side.

Divide $6$ by $- 2$ .

$n = - 3$

Solve for n (n-6)/(3n)=(n-5)/(3n+1)

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