Find the approximate solution of this system of equations.

y = |x^2 - 3x + 1|

y = x - 1

Y = |x² - 3x + 1|

y = x - 1

|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.