# How do you solve 3(n-1)-2=4(n-4)-(n-11)?

Dec 9, 2015

There are infinite solutions.

#### Explanation:

$3 \left(n - 1\right) - 2 = 4 \left(n - 4\right) - \left(n - 11\right)$

Distribute the $3$ on the left side.

$3 n - 3 - 2 = 4 \left(n - 4\right) - \left(n - 11\right)$

Distribute the $4$ on the right side.

$3 n - 3 - 2 = 4 n - 16 - \left(n - 11\right)$

Distribute the $- 1$ on the right side.

$3 n - 3 - 2 = 4 n - 16 - n + 11$

Combine like terms on both sides.

$3 n - 5 = 4 n - n - 16 + 11$

Simplify.

$3 n - 5 = 3 n - 5$

There infinite solutions.