In each of these questions two equations are given. You have to solve these equations and give answer.
I. $3 x^{2} + 7 x = 6$
II. $6 (2 y^{2} + 1) = 17 y$

x < y

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x > y

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x = y

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x \geq y

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x \leq y

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Solution

The correct option is E If $x \leq y$
I. $3 x^{2} + 7 x - 6 = 0$
or, $3 x^{2} + 9 x - 2 x - 6 = 0$
or, $(x + 3) (3 x - 2) = 0$
or, $x = - 3, \frac{2}{3}$

II. $6 (2 y^{2} + 1) = 17 y$
or, $12 y^{2} + 6 - 17 y = 0$
or, $12 y^{2} - 9 y - 8 y + 6 = 0$
or, $(4 y - 3) (3 y - 2) = 0$
or, $y = \frac{3}{4}, \frac{2}{3}$
Hence, $y \geq x$

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In each of these questions two equations are given. You have to solve these equations and give answer. I. 3x2+7x=6 II. 6(2y2+1)=17y

The correct option is E If x≤yI. 3x2+7x−6=0 or, 3x2+9x−2x−6=0 or, (x+3)(3x−2)=0 or, x=−3,23 II. 6(2y2+1)=17y or, 12y2+6−17y=0 or, 12y2−9y−8y+6=0 or, (4y−3)(3y−2)=0 or, y=34,23 Hence, y≥x

In each of these questions two equations are given. You have to solve these equations and give answer.
I. $3 x^{2} + 7 x = 6$
II. $6 (2 y^{2} + 1) = 17 y$