In the following question two equations I and II are given. You have to solve both the equations and Give answer

I. \( \Large 8x^{2} \) + 3x = 38

II. \( \Large 6y^{2} \) + 34 = 29y

A) x > y

B) x \( \Large \leq \) y

C) x < y

D) x \( \Large \geq \) y

Correct answer:
B) x \( \Large \leq \) y

Description for Correct answer:
I. 8 \( \Large x^{2} \) + 3x - 38 = 0

=> 8 \( \Large x^{2} \) + 19x - 16x - 38 = 0

=> x ( 8x + 19 ) - 2 ( 8x + 19 ) = 0

=> ( 8x + 19 ) ( x - 2) = 0

=> x = 2 or - \( \Large \frac{19}{8} \)

II. 6 \( \Large y^{2} \) - 29y + 34 = 0

=> 6 \( \Large y^{2} \) - 17y - 12y + 34 = 0

=> y ( 6y - 17 ) - 2 ( 6y - 17 ) = 0

=> ( y - 2 ) ( 6y - 17 ) = 0

=> y = 2 or \( \Large \frac{17}{6} \)

Clearly, x \( \Large \leq \) y.

Please provide the error details in above question

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