1). x< y2). x> y3). x = y OR the relationship cannot be determined4).

1.	1). x< y2). x> y3). x = y OR the relationship cannot be determined4). x ≥ y
Answer» I. 2x² – 7x + 5 = 0 ⇒ 2x² – 5X – 2x + 5 = 0 ⇒ x(2x – 5) – 1(2x – 5) = 0 ⇒ (2x – 5)(x – 1) = 0 Then, x = + 5/2 or x = + 1 II. 2Y² – 3Y + 1 = 0 ⇒ 2y² – 2y – y + 1 = 0 ⇒ 2y(y – 1) – 1(y – 1) = 0 → (y – 1)(2y – 1) = 0 Then, y = + 1 or y = + ½ So, when x = + 5/2, x > y for y = + 1 and x > y for y = + ½ And when x = + 1, x = y for y = + 1 and x > y for y = + ½ ∴ So, we can clearly observe that x ≥ y.

1). x< y2). x> y3). x = y OR the relationship cannot be determined4). x ≥ y

Answer»

I. 2x² – 7x + 5 = 0

⇒ 2x² – 5X – 2x + 5 = 0

⇒ x(2x – 5) – 1(2x – 5) = 0

⇒ (2x – 5)(x – 1) = 0

Then, x = + 5/2 or x = + 1

II. 2Y² – 3Y + 1 = 0

⇒ 2y² – 2y – y + 1 = 0

⇒ 2y(y – 1) – 1(y – 1) = 0

→ (y – 1)(2y – 1) = 0

Then, y = + 1 or y = + ½

So, when x = + 5/2, x > y for y = + 1 and x > y for y = + ½

And when x = + 1, x = y for y = + 1 and x > y for y = + ½

∴ So, we can clearly observe that x ≥ y.