abs(x-1)+abs(x+1)=1
1) Let's solve in ℝ: abs(x -1) + abs(x + 1) = 1
The absolute-value arguments are zero at { -1, 1 }. Split ℝ at these points and solve on each interval
On (-∞, -1]: x -1 ≤ 0, so abs(x -1) = -(x -1) ; x + 1 ≤ 0, so abs(x + 1) = -(x + 1)
On (-∞, -1], the equation becomes -(x -1) -(x + 1) = 1
2) Let's solve in ℝ: -(x -1) -(x + 1) = 1
-x + 1 -x -1 = 1
ⓘ Start from 1 and move back 1 on the number line, down to zero: 1 -1 = 0■Like terms are grouped by adding their coefficients: -x -x = -2x
-2x + 0 = 1
ⓘ Adding 0 changes nothing: -2x + 0 = -2x
-2x = 1
ⓘ Both sides are divided by -2
x = 1-2
ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator: 1-2 = -12
Solution found: x = -12
As -12 ∉ (-∞, -1], we cannot keep the solution -12
On [-1, 1]: x -1 ≤ 0, so abs(x -1) = -(x -1) ; x + 1 ≥ 0, so abs(x + 1) = x + 1
On [-1, 1], the equation becomes -(x -1) + (x + 1) = 1
3) Let's solve in ℝ: -(x -1) + (x + 1) = 1
ⓘ A minus sign before a parenthesis changes the sign of each of its terms: -(x -1) = -x + 1
-x + 1 + x + 1 = 1
ⓘ Start from 1 and move forward 1 on the number line: 1 + 1 = 2■Like terms are grouped by adding their coefficients: -x + x = 0
0 + 2 = 1
ⓘ Adding 0 changes nothing: 0 + 2 = 2
2 = 1
The statement 2 = 1 is false, whatever the value of x
Solution: x ∈ ∅
On [1, +∞): x -1 ≥ 0, so abs(x -1) = x -1 ; x + 1 ≥ 0, so abs(x + 1) = x + 1
On [1, +∞), the equation becomes (x -1) + (x + 1) = 1
4) Let's solve in ℝ: (x -1) + (x + 1) = 1
ⓘ Start from -1 and move forward 1 on the number line, up to zero: -1 + 1 = 0■Like terms are grouped by adding their coefficients: x + x = 2x
2x + 0 = 1
ⓘ Adding 0 changes nothing: 2x + 0 = 2x
2x = 1
ⓘ Both sides are divided by 2
Solution obtained: x = 12
As 12 ∉ [1, +∞), we cannot keep the solution 12
▶Solution: x ∈ ∅