**Directions(1-10):** Establish the correct relation between x and y and choose a correct option.

- I.6x – 2 = (x)^1/2

II.2y = (3)^1/2y + 3x=y or relation cannot be established.x=<yx<yx>yx>=yOption C

From I: 6x – 2 = (x)^1/2

Squaring on both the sides, we get

36x^2 – 25x + 4 = 0

=> 36x^2 – 16x – 9x +4 = 0

=> 4x(9x-4)-1(9x-4) = 0

=> x= 4/9,1/4

From II: 2y = (3)^1/2y + 3

Squaring on both the sides, we get

4y^2 – 15y + 9 = 0

=> 4y^2 – 12y -3y + 9 = 0

=> y(4y-3)-3(4y-3) = 0

=> y = ¾,3

x<y - I.6x^2 -31x+18 = 0

II.2y^2 -21y+54 = 0x>yx<yx=<yx=y or relation cannot be established.x>=yOption C

From I: 6x^2 -31x+18 = 0

=> 6x^2 -27x -4x+18 = 0

=>3x(2x-9)-2(2x-9) = 0

=> (2x-9)(3x-2) = 0

=> x = 9/2,2/3

From II: 2y^2 -21y+54 = 0

=> 2y^2 – 12y – 9y + 54 =0

=> 2y(y-6)-9(y-6) = 0

=> (y-6)(2y-9) = 0

=> y=6 , 9/2

x=<y - I.5x^2 – 27x + 10 = 0

II.y^2 – 14y + 48 =0x=y or relation cannot be established.x=<yx>=yx<yx>yOption D

From I: 5x^2 – 27x + 10 = 0

=> 5x^2 – 25x – 2x + 10 = 0

=> 5x(x-5)-2(x-5) = 0

=>(x-5)(5x-2) = 0

=> x= 5,2/5

From II: y^2 – 14y + 48 =0

=> y^2 – 8y -6y + 48=0

=> y(y-8)-6(y-8) = 0

=> (y-8)(y-6) =0

=> y = 8,6

x<y - I.4x^2 -44x + 85 = 0

II.2y^2 +y – 28 = 0x>=yx=y or relation cannot be established.x>yx=<yx<yOption B

From I: 4x^2 -44x + 85 = 0

=> 4x^2 – 34x – 10x + 85 =0

=>2x(2x-17)-5(2x-17) = 0

=> (2x-17)(2x-5) = 0

=>x = 17/2,5/2

From II: 2y^2 +y – 28 = 0

=> 2y^2 + 8y – 7y – 28 =0

=> 2y(y+4)-7(y+4) = 0

=> (y+4)(2y-7) = 0

=> y = -4,7/2

x = y or relation cannot be established. - I.8x^2-10x+3 = 0

II.15y^2 -32y +16 = 0x<yx>=yx>yx=<yx=y or relation cannot be established.Option A

From I: 8x^2-10x+3 = 0

=> 8x^2 – 4x – 6x + 3 =0

=> 4x(2x-1)-3(2x-1) = 0

=> (4x-3)(2x-1) = 0

=>x = ¾,1/2

From II: 15y^2 -32y +16 = 0

=> 15y^2 – 12y – 20y + 16 = 0

=> 3y(5y-4)-4(5y-4) = 0

=> (3y-4)(5y-4) = 0

=> y = 4/3,4/5

x<y - I.x^2 +2x-63 = 0

II.y^2 +20y+99 = 0x>yx<yx>=yx=y or relation cannot be established.x=<yOption C

From I: x^2 +2x-63 = 0

=> x^2 -7x+9x-63 = 0

=> x(x-7)+9(x-7) = 0

=> (x-7)(x+9) = 0

=> x = 7,-9

From II: y^2 +20y+99 = 0

=> y^2 +9y+11y+99 = 0

=> y(y+9)+11(y+9) = 0

=> y = -9,-11

x>=y - I.21x^2 – 13x + 2 = 0

II. 33y^2 = 21 – 68yx=y or relation cannot be established.x>yx<yx=<yx>=yOption B

From I: 21x^2 – 13x + 2 = 0

=> 21x^2 -7x – 6x + 2 = 0

=> 7x(3x-1)-2(3x-1) = 0

=> x = 1/3,2/7

From II: 33y^2 = 21 – 68y

=> 33y^2 + 68y – 21 = 0

=> 33y^2 + 77y -9y – 21 = 0

=> 11y(3y+7)-3(3y+7) = 0

=> y = -7/3,3/11

x>y - I.16x+15y = 94

II.2x = 10 – yx>=yx>yx<yx=<yx=y or relation cannot be established.Option B

From I: x = (94-15y)/16

From II: 2*(94-15y)/16 = 10 – y

=> 188 – 30y = 160 – 16y

=> 14y = 28

=> y = 2

x = (94-15y)/16 = 4

x>y - I.9x+4y = 79

II.7x – 5y = 29x=y or relation cannot be established.x=<yx>=yx>yx<yOption D

Multiply I from 5 and II from 4, we get

x = 7

y = 4

x>y - I.12x^2 + 13x – 35 = 0

II.10y^2 = 21y – 9x>=yx=<yx>yx<yx=y or relation cannot be established.Option E

From I: 12x^2 + 13x – 35 = 0

=> 12x^2 + 28x – 15x – 35 = 0

=> 4x(3x+7)-5(3x+7) = 0

=> (3x+7)(4x-5) = 0

=> x= -7/3,5/4

From II: 10y^2 = 21y – 9

=> 10y^2 – 21y + 9 = 0

=> 10y^2 – 15y – 6y + 9 = 0

=> 5y(2y-3)-3(2y-3) = 0

=> (2y-3)(5y-3) = 0

=> y = 3/2,3/5

x =y or relation cannot be established.