|x² - 3x + 2| + x = 1
|(x - 1)(x -2)| = 1 - x

y = x² - 3x + 2 is the expression for a parabola which opens upwards. It cuts the x-axis at x = 1 and x = 2 (where y = 0)

Between x = 1 and x = 2 inclusive, y is negative and the absolute/modulus would reflect that part of the curve upwards about the x-axis. For this range, it is equivalent to solving for the intersection between y = 1 - x and the reflected curve.

So,

For x ≤ 1 and x ≥ 2,

x² - 3x + 2 = 1 - x
x² - 2x + 1 = 0
(x - 1)² = 0
x - 1 = 0
x = 1



For 1 ≤ x ≤ 2,

-(x² - 3x + 2) = 1 - x
x² - 3x + 2 = x - 1
x² - 4x + 3 = 0
(x - 1)(x - 3) = 0
x - 1 = 0 or x = 3
x = 1 or x = 3 (rejected, as 1 ≤ x ≤ 2)


So there is only one point of intersection at x = 1