1.

II. 12y2 – 22y + 8 = 01). If x > y2). If x ≥ y3). If x< y4). If x ≤ y

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:

12y222Y + 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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