|2x-1|+3=|5-x|
⚠ |2x-1| is read as the absolute value abs(2x-1)
⚠ |5-x| is read as the absolute value abs(5-x)
1) Let's solve in ℝ: abs(2x -1) + 3 = abs(5 -x)
The absolute-value arguments vanish at { 12, 5 }. Split ℝ at these points and solve on each interval
On (-∞, 12]: 2x -1 <= 0, so abs(2x -1) = -(2x -1) ; 5 -x >= 0, so abs(5 -x) = 5 -x
On (-∞, 12], the equation becomes -(2x -1) + 3 = 5 -x
2) Let's solve in ℝ: -(2x -1) + 3 = 5 -x
-2x + 1 + 3 = 5 -x
-2x + 4 = 5 -x
4 -5 = -x + 2x
ℹ4 -5 = -1‖-x + 2x = x
-1 = x
Solution obtained: x = -1
Within this interval, keep -1
On [12, 5]: 2x -1 >= 0, so abs(2x -1) = 2x -1 ; 5 -x >= 0, so abs(5 -x) = 5 -x
On [12, 5], the equation becomes (2x -1) + 3 = 5 -x
3) Let's solve in ℝ: (2x -1) + 3 = 5 -x
2x + 2 = 5 -x
2x = 3 -x
2x + x = 3
3x = 3
x = 33
Solution found: x = 1
Within this interval, keep 1
On [5, +∞): 2x -1 >= 0, so abs(2x -1) = 2x -1 ; 5 -x <= 0, so abs(5 -x) = -(5 -x)
On [5, +∞), the equation becomes (2x -1) + 3 = -(5 -x)
4) Let's solve in ℝ: (2x -1) + 3 = -(5 -x)
ℹ-1 + 3 = 2‖-(5 -x) = -5 + x
2x + 2 = -5 + x
2x = -7 + x
2x -x = -7
Solution found: x = -7
Within this interval, keep ∅
▶Solutions found: x = { -1, 1 }