Form: |ax + b| = c
Condition: c ≥ 0
Solution: |ax + b| = c ⟺ ax + b = c or ax + b = -c
|x| = 5
Form: |ax + b| = |cx + d|
Solution: Consider two cases:
- ax + b = cx + d
- ax + b = -(cx + d)
|x| = |x - 2|
Left side: |ax + b|
Right side: |cx + d|
Form: |ax + b| + |cx + d| = e
Solution: Create a sign table and solve for each interval
|x + 1| + |x - 2| = 5