Question
The graphs of the equations 3x-20y-2=0 and 11x-5y +61=0
intersect at P(a,b). What is the value of (a² + b² − ab)/(a² − b² + ab)?Solution
3x – 20y – 2 = 0 11x – 5y + 61 = 0 3x – 20y = 2 → (1) 11x – 5y = ̶ 61 → (2) 44x – 20y = 244→ (3) Equation (1) – (3) = (3x – 44x) = 2 + 244 = - 41x = 246 = [x = -6] Put x = - 6 in equation (1) -18 -20y = 2 = [y = -1] both equation intersect each other P(a,b) = P(x,y) a = 6, b = -1 on putting we get 31/41
I. 3x2 = 2x2 + 9x – 20
II. 3y2 = 75
I. 2x2 - 9 x + 9 = 0
II. 2y2 - 7 y + 3 = 0
I. 2x² + 11 x + 15 = 0
II. 2y² - 19 y + 44 = 0
I. 2x² - 9x + 10 = 0
II. 3y² + 11y + 6 = 0
I. 2y2 - 15y + 18 = 0
II. 2x2 + 9x - 18 = 0
I. 144x² - 163x - 65 = 0
II. 91y² - 128y -48 = 0
I. 2b2 + 31b + 99 = 0
II. 4a2 + 8a - 45 = 0
I. x2 - 9x - 52 = 0
II. y2 - 16y + 63 = 0
I. 104x² + 9x - 35 = 0
II. 72y² - 85y + 25 = 0
(i) 2x² + 14x - 16 = 0
(ii) y² – y – 12 = 0