In the following question, two equations numbered I and II are given. Solve both the equations and mark the appropriate option:

$I . 4 y^{2} + 8 y = 4 y + 8$
$I I . x^{2} + 9 x = 2 x - 12$

x = y

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x > y

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x < y

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Solution

The correct option is C x < y
$F r o m I, 4 y^{2} + 8 y = 4 y + 8 y^{2} + 2 y = y + 2 y^{2} + 2 y - y - 2 = 0 (y - 1) (y + 2) = 0 y = 1 o r y = - 2 \dots (1) F r o m I I, x^{2} + 9 x = 2 x - 12 x^{2} + 7 x + 12 = 0 (x + 4) (x + 3) = 0 x = - 4 o r x = - 3 \dots (2)$
From (1) and (2), we have y > x.

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In the following question, two equations numbered I and II are given. Solve both the equations and mark the appropriate option: I. 4y2+8y = 4y+8 II. x2+9x = 2x−12

The correct option is C x < yFrom I,4y2+8y=4y+8y2+2y=y+2y2+2y−y−2=0(y−1)(y+2)=0y=1 or y=−2⋯(1)From II,x2+9x=2x−12x2+7x+12=0(x+4)(x+3)=0x=−4 or x=−3⋯(2) From (1) and (2), we have y > x.

In the following question, two equations numbered I and II are given. Solve both the equations and mark the appropriate option:

$I . 4 y^{2} + 8 y = 4 y + 8$
$I I . x^{2} + 9 x = 2 x - 12$

The correct option is C x < y
$F r o m I, 4 y^{2} + 8 y = 4 y + 8 y^{2} + 2 y = y + 2 y^{2} + 2 y - y - 2 = 0 (y - 1) (y + 2) = 0 y = 1 o r y = - 2 \dots (1) F r o m I I, x^{2} + 9 x = 2 x - 12 x^{2} + 7 x + 12 = 0 (x + 4) (x + 3) = 0 x = - 4 o r x = - 3 \dots (2)$
From (1) and (2), we have y > x.