Intermediate Algebra, Chapter 5, 5.2, Section 5.2, Problem 86

Solve $5k - (5k - [2k - (4k - 8k)]) + 11k - (9k - 12k)$
Since $4k$ and $-8k$ are liek terms, add $-8k$ to $4k$ to get $-4k$.
$5k−(5k−(2k−(−4k)))+11k−(9k−12k)$


Multiply $−1$ by each term inside the parentheses.

$5k−(5k−(2k+4k))+11k−(9k−12k)$


Since $2k$ and $4k$ are like terms, add $4k$ to $2k$ to get $6k$.

$5k−(5k−(6k))+11k−(9k−12k)$


Multiply $−1$ by each term inside the parentheses.

$5k−(5k−6k)+11k−(9k−12k)$


Since $5k$ and $−6k$ are like terms, add $−6k$ to $5k$ to get $−k$.

$5k−(−k)+11k−(9k−12k)$


Multiply $−1$ by each term inside the parentheses.

$5k+k+11k−(9k−12k)$


Since $9k$ and $−12k$ are like terms, add $−12k$ to $9k$ to get $−3k$.

$5k+k+11k−(−3k)$


Multiply $−1$ by each term inside the parentheses.

$5k+k+11k+3k$


Combine all similar terms in the polynomial $5k+k+11k+3k$.

$20k$