1.

II. 20y2 – 41y + 20 = 01). x > y2). x ≥ y3). x< y4). x ≤ y

Answer»

We will SOLVE both the equations separately.

Equation I:

12x2 – x – 1 = 0

⇒ 12x24X + 3x – 1 = 0

⇒ 4x(3x – 1) + 1(3x – 1) = 0

⇒ (4x + 1) (3x – 1) = 0

⇒ x = -(1/4) = - 0.25

Or, x = 1/3 = 0.33

Equation II:

20y2 – 41y + 20 = 0

⇒ 20y225Y – 16y + 20 = 0

⇒ 5y (4y – 5) – 4(4y – 5) = 0

⇒ (5y – 4) (4y – 5) = 0

⇒ y = 4/5 = 0.8 or y = 5/4 = 1.25

Comparing the VALUES of x and y we get,

x < y


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