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II. 20y2 – 41y + 20 = 01). x > y2). x ≥ y3). x< y4). x ≤ y |
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Answer» We will SOLVE both the equations separately. Equation I: 12x2 – x – 1 = 0 ⇒ 12x2 – 4X + 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 ⇒ 20y2 – 25Y – 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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