Q25 of 50 Page 1

Solve for x:

; x ≠ 3, 5


Given equation





Cross-multiplying,


⇒ 3 (2x2 – 14x + 22) = 10 (x2 – 8x + 15)


⇒ 6x2 – 42x + 66 = 10x2 – 80x + 150


⇒ 4x2 – 38x + 84 = 0


⇒ 2x2 – 19x + 42 = 0


⇒ 2x2 – 12x – 7x + 42 = 0


⇒ 2x (x – 6) – 7 (x – 6) = 0


⇒ (x – 6) (2x – 7) = 0


⇒ x = 6 and x = 7/2


∴ The solutions of x are 6 and 7/2.


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