Solve for n a(n-3)+8=bn

$a (n - 3) + 8 = b n$

Step 1

Simplify each term.

Apply the distributive property.

$a n + a \cdot - 3 + 8 = b n$

Move $- 3$ to the left of $a$ .

$a n - 3 a + 8 = b n$

Step 2

Subtract $b n$ from both sides of the equation.

$a n - 3 a + 8 - b n = 0$

Step 3

Move all terms not containing $n$ to the right side of the equation.

Add $3 a$ to both sides of the equation.

$a n + 8 - b n = 3 a$

Subtract $8$ from both sides of the equation.

$a n - b n = 3 a - 8$

Step 4

Factor $n$ out of $a n - b n$ .

Factor $n$ out of $a n$ .

$n a - b n = 3 a - 8$

Factor $n$ out of $- b n$ .

$n a + n (- b) = 3 a - 8$

Factor $n$ out of $n a + n (- b)$ .

$n (a - b) = 3 a - 8$

Step 5

Divide each term in $n (a - b) = 3 a - 8$ by $a - b$ and simplify.

Divide each term in $n (a - b) = 3 a - 8$ by $a - b$ .

$\frac{n (a - b)}{a - b} = \frac{3 a}{a - b} + \frac{- 8}{a - b}$

Simplify the left side.

Cancel the common factor of $a - b$ .

Cancel the common factor.

$\frac{n (a - b)}{a - b} = \frac{3 a}{a - b} + \frac{- 8}{a - b}$

Divide $n$ by $1$ .

$n = \frac{3 a}{a - b} + \frac{- 8}{a - b}$

Simplify the right side.

Combine the numerators over the common denominator.

$n = \frac{3 a - 8}{a - b}$