Directions (Q. 1-5): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer

1) if x > y

2) if x ? y

3) if x < y

4) if x ? y

5) if x = y or relationship between x and y cannot be established.

1. I. 676x2 – 1 = 0 II. II. 1/3√13824

2. I. 8x + 13y = 62 II. 13x – 17y + 128 = 0

3. I. 7x2 + 2x = 120 II. y2 + 11y + 30 = 0

4. I. x2 = 7x II. (y + 7)2 = 0

5. I. 2x2 + 5x – 33 = 0 II. y2 – y – 6 = 0

Answers

1. 3; I. 676x2 – 1 = 0

or, x2 1/676

x= 1/26

II. 3

y 1/cuberoot 13824=

y = 1/24

ie, x < y

2. 3 On solving these two equations, we get x = –2, y = 6

ie, x < y

3. 1; I. 7x2 – 28x + 30x – 120 = 0

or, 7x(x – 4) +30(x – 4) = 0

or, (x – 4) (7x + 30) = 0

x = 4, 30/7

II. y2 + 6y + 5y + 30 = 0

or, y(y + 6) + 5(y + 6) = 0

or, (y + 5) (y + 6) = 0

y = –5, –6

ie, x > y

4. 1; I. x2 = 7x

or, x2 – 7x = 0

or, x(x – 7) = 0

x = 0, 7

II. (y + 7)2 = 0

or, (y + 7) = 0

y = –7

ie, x > y

5. 5; I. 2x2 – 6x + 11x – 33 = 0

or, 2x(x – 3) +11(x – 3) = 0

or, (2x + 11) (x – 3) = 0

x = 3, 11/2

II. y2 – 3y + 2y – 6 = 0

or, y(y – 3) +2(y – 3) = 0

or, (y + 2) (y – 3) = 0

y = –2, 3

1) if x > y

2) if x ? y

3) if x < y

4) if x ? y

5) if x = y or relationship between x and y cannot be established.

1. I. 676x2 – 1 = 0 II. II. 1/3√13824

2. I. 8x + 13y = 62 II. 13x – 17y + 128 = 0

3. I. 7x2 + 2x = 120 II. y2 + 11y + 30 = 0

4. I. x2 = 7x II. (y + 7)2 = 0

5. I. 2x2 + 5x – 33 = 0 II. y2 – y – 6 = 0

Answers

1. 3; I. 676x2 – 1 = 0

or, x2 1/676

x= 1/26

II. 3

y 1/cuberoot 13824=

y = 1/24

ie, x < y

2. 3 On solving these two equations, we get x = –2, y = 6

ie, x < y

3. 1; I. 7x2 – 28x + 30x – 120 = 0

or, 7x(x – 4) +30(x – 4) = 0

or, (x – 4) (7x + 30) = 0

x = 4, 30/7

II. y2 + 6y + 5y + 30 = 0

or, y(y + 6) + 5(y + 6) = 0

or, (y + 5) (y + 6) = 0

y = –5, –6

ie, x > y

4. 1; I. x2 = 7x

or, x2 – 7x = 0

or, x(x – 7) = 0

x = 0, 7

II. (y + 7)2 = 0

or, (y + 7) = 0

y = –7

ie, x > y

5. 5; I. 2x2 – 6x + 11x – 33 = 0

or, 2x(x – 3) +11(x – 3) = 0

or, (2x + 11) (x – 3) = 0

x = 3, 11/2

II. y2 – 3y + 2y – 6 = 0

or, y(y – 3) +2(y – 3) = 0

or, (y + 2) (y – 3) = 0

y = –2, 3