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II. 12y2 – 22y + 8 = 01). If x > y2). If x ≥ y3). If x< y4). If x ≤ y |
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Answer» We will solve both the equations separately. Equation I: 8x2 + 6X - 5 = 0 ⇒ 8x2 + 10x – 4x – 5 = 0 ⇒ 4x (2x – 1) + 5(2x – 1) = 0 ⇒ (4x + 5) (2x – 1) = 0 ⇒ x = - 5/4 = - 1.25 or, x = ½ = 0.5 Equation II: 12y2 – 22Y + 8 = 0 ⇒ 12y2 – 16y – 6Y + 8 = 0 ⇒ 6y (2y – 1) – 8(2y – 1) = 0 ⇒ (6y – 8) (2y – 1) = 0 ⇒ y = 8/6 = 1.33 or, y = ½ = 0.5 Comparing the values of x and y, we GET, x ≤ y |
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