Question

determine the solution for each equation. show your work.

1. $|6 - 4k| - 30 = 0$
2. $|8x - 11| + 45 = 6$

$k = $
$x = $

1. $|n + 6| = 19$
2. $-3|y + 5| + 27 = 0$

Explanation:

Problem 3: \(|6 - 4k| - 30 = 0\)

Step 1: Isolate the absolute value

Add 30 to both sides:
\(|6 - 4k| = 30\)

Step 2: Set up two equations

Case 1: \(6 - 4k = 30\)
Subtract 6: \(-4k = 24\)
Divide by -4: \(k = -6\)

Case 2: \(6 - 4k = -30\)
Subtract 6: \(-4k = -36\)
Divide by -4: \(k = 9\)

Problem 4: \(|8x - 11| + 45 = 6\)

Step 1: Isolate the absolute value

Subtract 45: \(|8x - 11| = -39\)

The absolute value of a number is always non - negative, so no solution.

Problem 5: \(|n + 6| = 19\)

Step 1: Set up two equations

Case 1: \(n + 6 = 19\)
Subtract 6: \(n = 13\)

Case 2: \(n + 6 = -19\)
Subtract 6: \(n = -25\)

Problem 6: \(-3|y + 5| + 27 = 0\)

Answer:

s:

1. \(k = -6\) or \(k = 9\)
2. No solution
3. \(n = 13\) or \(n = -25\)
4. \(y = 4\) or \(y = -14\)