182 + 10 × 12 - ? =

I. (y – 5)² – 9 = 0

II. x² – 3x + 2 = 0

In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y.

I. x

I. x² - 33x + 270 = 0

II. y² - 41y + 414 = 0

Find the remainder when x⁵ − 3x⁴ + 4x³ − 6x² + 8x − 3 is divided by (x − 2).

I. x² – 18x + 81 = 0

II. y² – 3y - 28 = 0

I. x² + 24x + 143 = 0

II. y² + 12y + 35 = 0

Solve the quadratic equations and determine the relation between x and y:

Equation 1: 4x² - 12x + 9 = 0

Equation 2: 2y² + 8y + 6 = 0

I. 12 x ² - 3 x – 15 = 0

II. 2 y² + 12

I. 6x² + 19x + 10 = 0

II. y² + 10y + 25 = 0

I. 49y² + 35y + 6 = 0

II. 12x² + 17 x + 6 = 0