|x² - 3x + 1| = x - 1 |x² - 3x + 1| = ±1(x - 1) |x² - 3x + 1| = 1(x - 1) or |x² - 3x + 1| = -1(x - 1) |x² - 3x + 1| = 1(x) - 1(1) or |x² - 3x + 1| = -1(x) + 1(1) |x² - 3x + 1| = x - 1 or |x² - 3x + 1| = -x + 1 x² - 3x + 1 = x - 1 or x² - 3x + 1 = -x + 1 - x - x + x + x x² - 4x + 1 = -1 or x² - 2x + 1 = 1 + 1 + 1 - 1 - 1 x² - 4x + 1 = 0 or x² - 2x + 0 = 0 x = -(-4) ± √((-4)² - 4(1)(1)) or x = -(-2) ± √((-2)² - 4(1)(0)) 2(1) 2(1) x = 4 ± √(16 - 4) or x = 2 ± √(4 - 0) 2 2 x = 4 ± √(12) or x = 2 ± √(4) 2 2 x = 4 ± 2√(3) or x = 2 ± 2 2 2 x = 2 ± √(3) or x = 1 ± 1 x = 2 + √(3) or x = 2 - √(3) or x = 1 + 1 or x = 1 - 1 x = 2 or x = 0 y = x - 1 or y = x - 1 or y = x - 1 or y = x - 1 y = (2 + √(3)) - 1 or y = (2 - √(3)) - 1 or y = 2 - 1 or y = 0 - 1 y = 2 - 1 + √(3) or y = 2 - 1 - √(3) or y = 1 or y = -1 y = 1 + √(3) or y = 1 - √(3) (x, y) = (2, 1) or (x, y) = (0, -1) (x, y) = (2 ± √(3), 1 ± √(3))
The solution (0, -1) can be made by one function (y = x - 1) while the solution (2 ± √(3), 1 ± √(3)) can be made by another function (y = |x² - 3x + 1|). So the solution (2, 1) can be made by both functions, making the two solutions equal.
