Solve the equations y 2 4 x 2 - 108 = x x 3 - 9 y 3 , 2 x 2 + 9 x y + y 2 = 108 .

Given y 2 4 x 2 - 108 = x x 3 - 9 y 3 . . . i 2 x 2 + 9 x y + y 2 = 108 . . . i i Rewriting i with the help of i i , we have y 2 4 x 2 - 2 x 2 - 9 x y - y 2 = x x 3 - 9 y 3 ⇒ 4 x 2 y 2 - 2 x 2 y 2 - 9 x y 3 - y 4 = x 4 - 9 x y 3 ⇒ x 4 - 2 x 2 y 2 + y 4 = 0 ⇒ x 2 - y 2 2 = 0 ∴ x 2 - y 2 = 0 i.e. x = ± y Putting x = y in i i , we have 2 x 2 + 9 x 2 + x 2 = 108 ⇒ 12 x 2 = 108 ⇒ x 2 = 9 ∴ x = ± 3 and y = ± 3 Now, putting y = - x in i i , we have 2 x 2 - 9 x 2 + x 2 = 108 ⇒ 6 x 2 = - 108 ⇒ x 2 = - 18 ∴ x = ± - 18 and y = ∓ - 18 Hence, solution of the given equations are x = ± 3 , ± - 18 (imaginary) y = ± 3 , ∓ - 18 .

Solve the equations x - a a 2 + y - b b 2 = 1 x - b - 1 y - a - 1 a - b = 0 .

We have x - a a 2 + y - b b 2 = 0 . . . ( i ) 1 x - b - 1 y - a = 1 a - b . . . ( ii ) From ( i ) , we have x - a = a 2 ( b - y ) b 2 ⇒ x = a + a 2 ( b - y ) b 2 Subtracting b from both sides, we have x - b = a + a 2 ( b - y ) b 2 - b = a b 2 + a 2 b - a 2 y - b 3 b 2 . . . ( iii ) Therefore, from ( ii ) after substituting this value of ( x - b ) , we get b 2 a b 2 + a 2 b - a 2 y - b 3 - 1 y - a = 1 a - b ⇒ 1 y - a = b 2 a b 2 + a 2 b - a 2 y - b 3 - 1 a - b ⇒ 1 y - a = a b 2 - b 3 - a b 2 - a 2 b + a 2 y + b 3 ( a - b ) a b 2 + a 2 b - a 2 y - b 3 ⇒ 1 y - a = a 2 y - a 2 b ( a - b ) a b 2 + a 2 b - a 2 y - b 3 ⇒ ( y - a ) a 2 y - a 2 b = ( a - b ) a b 2 + a 2 b - a 2 y - b 3 ⇒ a 2 y 2 - a 2 b y - a 3 y + a 3 b = a 2 b 2 + a 3 b + a 3 y - a b 3 - a b 3 - a 2 b 2 + a 2 b y + b 4 ⇒ a 2 y 2 + y - 2 a 2 b + 2 a b 3 - b 4 = 0 ⇒ a 2 y 2 - 2 a 2 b y + 2 a b 3 - b 4 = 0 ∴ y = 2 a 2 b ± 4 a 2 b 2 - 8 a 2 b 4 + 4 a 2 b 4 2 a 2 = 2 a 2 b ± 2 a b a 2 - 2 a b + b 2 2 a 2 = a b ± b a - b a ⇒ y = b 2 a - b a or b 2 a Putting these values of y in ( iii ) , we get x = a 2 b when y = b 2 a x = a ( 2 b - a ) b when y = b ( 2 a - b ) a .

Solve the equations 2 a x y - y x + 4 a 2 = 4 x 2 + x y 2 a - y 2 a 2 = 1 .

Given, 2 a x y - y x + 4 a 2 = 1 . . . ( i ) 4 x 2 + x y 2 a - y 2 a 2 = 1 . . . ( ii ) From ( i ) , we have 2 a x 2 - y 2 + 4 a 2 x y = x y or 2 a x 2 - y 2 = x y 1 - 4 a 2 . . . ( iii ) And from ( ii ) , we have 8 a 2 x 2 - 2 y 2 - 2 a 2 + a x y = 0 . . . ( iv ) ( ∵ this is a homogenous equation) Putting y = m x in ( iii ) and ( iv ) , we get 2 a x 2 - m 2 x 2 = x 2 m 1 - 4 a 2 ⇒ 2 a 1 - m 2 = m 1 - 4 a 2 . . . ( v ) and 8 a 2 x 2 - 2 m 2 x 2 + a x 2 m = 2 a 2 . . . ( vi ) [from equation ( iv ) ] From ( v ) , we have 2 a - 2 a m 2 = m - 4 a 2 m ⇒ 2 a m 2 + m 1 - 4 a 2 - 2 a = 0 ⇒ 2 a m 2 + 1 - 4 a 2 m - 2 a = 0 ⇒ m = 4 a 2 - 1 ± 1 - 4 a 2 2 + 16 a 2 4 a ⇒ m = 4 a 2 - 1 ± 16 a 4 + 1 - 8 a 2 + 16 a 2 4 a ⇒ m = 4 a 2 - 1 ± 4 a 2 + 1 2 4 a ⇒ m = 4 a 2 - 1 ± 4 a 2 + 1 4 a ∴ m = 2 a , - 1 2 a Putting this value of m in ( vi ) , we get 8 a 2 x 2 - 8 a 2 x 2 + 2 a 2 x 2 = 2 a 2 or x = ± 1 ∴ From y = m x , y = ± 2 a . And, putting y = - 1 2 a in vi , we get 8 a 2 x 2 - 2 m 2 x 2 + a x 2 m = 2 a 2 ⇒ 8 a 2 x 2 - 2 - 1 2 a 2 x 2 + a x 2 - 1 2 a = 2 a 2 ⇒ 8 a 2 x 2 - 1 2 a 2 x 2 - x 2 2 = 2 a 2 ⇒ 16 a 4 x 2 - x 2 - a 2 x 2 = 4 a 4 ⇒ x 2 16 a 4 - 1 - a 2 = 4 a 4 ⇒ x = ± 4 a 4 16 a 4 - 1 - a 2 ⇒ y = ∓ 1 2 a 4 a 4 16 a 4 - 1 - a 2

E M B I B E

Higher Algebra>Chapter 10 - Miscellaneous Equations>EXAMPLES. X.b.

H S Hall and S R Knight Solutions for Chapter: Miscellaneous Equations, Exercise 2: EXAMPLES. X.b.

Author:H S Hall & S R Knight

H S Hall Mathematics Solutions for Exercise - H S Hall and S R Knight Solutions for Chapter: Miscellaneous Equations, Exercise 2: EXAMPLES. X.b.

Attempt the free practice questions on Chapter 10: Miscellaneous Equations, Exercise 2: EXAMPLES. X.b. with hints and solutions to strengthen your understanding. Higher Algebra solutions are prepared by Experienced Embibe Experts.