Solve the following system of equations: x a + y b = 2 a x - b y = a 2 - b 2

The given system of equations is: x a + y b = 2 . . . . . . . . . . . . . 1 a x - b y = a 2 - b 2 . . . . . . 2 Both the equations can be rewritten as: a y + b x = 2 a b . . . . . . . 3 a x - b y = a 2 - b 2 . . . . . . 4 Multiply the equations 1 by a and 2 by b , then subtracting equation ( 4 ) from equation ( 3 ) , we get, a b x + a 2 y = 2 a 2 b - a b x + b 2 y = - b a 2 + b 3 0 + y ( a 2 + b 2 ) = 2 a 2 b - b a 2 + b 3 ⇒ y a 2 + b 2 = 2 a 2 b - b a 2 + b 3 ⇒ y a 2 + b 2 = a 2 b + b 3 ⇒ y a 2 + b 2 = b a 2 + b 2 ⇒ y = b Putting this value in equation 3 , we get, a y + b x = 2 a b ⇒ b x = 2 a b - a b ⇒ b x = a b ⇒ x = a Hence, the required solutions are x = a , y = b .

Solve the following system of equations: x a - y b = 0 a x + b y = a 2 + b 2

The given system of equations is: x a - y b = 0 . . . . . . . . . . . . . 1 Multiply equation 1 by a b b x - a y = 0 . . . . . . . 2 a x + b y = a 2 + b 2 . . . . . . 3 Multiply the equations 2 by a and 3 by b , then subtracting equation, we get, a b x - a 2 y = 0 a b x + b 2 y = b a 2 + b 3 - - - 0 - y ( a 2 + b 2 ) = - b a 2 - b 3 ⇒ - y a 2 + b 2 = - b a 2 + b 2 ⇒ y = b Putting this value in equation 2 , we get, b x - a b = 0 ⇒ b x = a b ⇒ x = a Hence, the required solutions are x = a , y = b .

Solve the following system of equations: b x a + a y b = a 2 + b 2 x + y = 2 a b

Given system of linear equations are, b x a + a y b = a 2 + b 2 . . . . . . . . . 1 x + y = 2 a b . . . . . . . . . . . . 2 Rewriting the equation 1 in its standard form by multiplying by a b and the equation 2 by a 2 , we get, b 2 x + a 2 y = a 3 b + b 3 a . . . . . ( 3 ) a 2 x + a 2 y = 2 a 3 b . . . . . ( 4 ) Subtract equation ( 4 ) from equation ( 3 ) , we get, b 2 x + a 2 y = a 3 b + b 3 a a 2 x + a 2 y = 2 a 3 b - - - b 2 x - a 2 x + 0 = b 3 a - a 3 b ⇒ b 2 - a 2 x = a b b 2 - a 2 ⇒ x = a b b 2 - a 2 b 2 - a 2 ⇒ x = a b Putting this value in equation 2 , we get, a b + y = 2 a b ⇒ y = 2 a b - a b ⇒ y = a b Hence, the required solution of the given equations is x = y = a b .

Solve the following system of equations: b x a - a y b + a + b = 0 b x - a y + 2 a b = 0

The given system of equations is: b x a - a y b + a + b = 0 . . . . . . . . 1 b x - a y + 2 a b = 0 . . . . . . . . . . 2 Dividing equation 2 by b , we get, x - a y b + 2 a = 0 . . . . . . . . 3 Now, subtracting equation 3 from 1 , we get, b x a - x + a + b - 2 a = 0 ⇒ x b a - 1 + b - a = 0 ⇒ x a b - a + b - a = 0 ⇒ b - a x a + 1 = 0 ⇒ x a + 1 = 0 ∴ x a = - 1 ⇒ x = - a Substituting value of x in equation 3 , we get, - a - a y b + 2 a = 0 ⇒ a = a y b ⇒ y = b Hence, the required solutions are x = - a , y = b .

Solve the following system of equations: x a + y b = a + b x a 2 + y b 2 = 2

The given system of equations is: x a + y b = a + b . . . . . . 1 x a 2 + y b 2 = 2 . . . . . . . . . . 2 Equation 1 can be rewritten as, x = a 2 + a b - a y b . . . . . 3 Putting this value in equation 2 , we get 1 + b a - y a b + y b 2 = 2 ⇒ 1 - b a = y b 2 1 - b a ⇒ 1 = y b 2 ∴ y = b 2 Putting this value of y in equation 3 , we get x = a 2 + a b - a × b 2 b ⇒ x = a 2 + a b - a b ∴ x = a 2 Hence, the required solutions are x = a 2 , y = b 2 .

Solve the following system of equations: 2 x a + y b = 2 ; x a - y b = 4

The given system of equations is: 2 x a + y b = 2 . . . . . . . . . . . 1 x a - y b = 4 . . . . . . . . . . . . . . . . 2 Adding equations 1 and 2 , we get, 2 x a + y b = 2 x a - y b = 4 3 x a + 0 = 6 ⇒ x = 6 a 3 = 2 a Substituting this value in equation 2 , we get, 2 a a - y b = 4 ⇒ 2 - y b = 4 ⇒ - y b = 4 - 2 ⇒ - y = 2 b ⇒ y = - 2 b Hence, the required solutions are x = 2 a , y = - 2 b .