1.

II. 35y2 – 12y + 1 = 01). if x < y2). if x > y3). if x ≤ y4). if x ≥ y

Answer»

I. 10x2 – 7x + 1 = 0

⇒ 10x25X – 2x + 1 = 0

⇒ 5x(2x – 1) – 1(2x – 1) = 0

⇒ (5x – 1)(2x – 1) = 0

Then, x = (1/5) or x = (1/2)

II. 35y2 – 12y + 1 = 0

⇒ 35y2 – 7y – 5y + 1 = 0

⇒ 7y(5y – 1) – 1(5y – 1) = 0

⇒ (7y – 1)(5y – 1) = 0

Then, y = (1/7) or y = (1/5)

So, when x = (1/5), x = y for y = (1/5) and x > y for y = (1/7)

And when x = (1/2), x > y for y = (1/5) and x > y for y = (1/7)

∴ x ≥ y


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