Question
I. 2x² - 15x  + 13 = 0
II. 3y² - 6y + 3 = 0
Solution
I. 2 x ² - 15 x  + 13=0 2 x ² - 2 x - 13 x  + 13= 0 2 x  ( x  – 1) – 13 ( x – 1) (2 x – 13) ( x  – 1) = 0 x = 1, 13/2 II. 3 y ² - 6 y  + 3 = 0 3 y ² - 3 y  - 3 y + 3 = 0 3 y ( y – 1) - 3 ( y  – 1) (3 y – 3) ( y – 1) = 0  y = 1, 1  Hence, x ≥ y
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