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 .

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

The given system of equations is: b a x + a b y = a 2 + b 2 . . . . . . . . . . . 1 x + y = 2 a b . . . . . . . . . . . . . . . . . . . . . . 2 Equation 1 can be rewritten as: b 2 x + a 2 y = a 3 b + b 3 a . . . . . . 3 Multiplying equation 2 by a 2 , we get, a 2 x + a 2 y = 2 a 3 b . . . . . . . . 4 Subtracting 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 ⇒ x b 2 - a 2 = a b b 2 - a 2 ⇒ x = a b Substituting this value in equation 2 , we get, a b + y = 2 a b ⇒ y = 2 a b - a b = a b Hence, the required solutions are x = a b , y = a b .

Solve: x + y - 8 2 = x + 2 y - 14 3 = 3 x + y - 12 11 .

Given, x + y - 8 2 = x + 2 y - 14 3 = 3 x + y - 12 11 Consider the equations in pairs as shown: x + y - 8 2 = 3 x + y - 12 11 ⇒ 11 x + y - 8 = 2 3 x + y - 12 ⇒ 11 x + 11 y - 88 = 6 x + 2 y - 24 ⇒ 5 x + 9 y = 64 . . . . . . . . . . . . 1 Now consider, x + y - 8 2 = x + 2 y - 14 3 ⇒ 3 x + y - 8 = 2 x + 2 y - 14 ⇒ 3 x + 3 y - 24 = 2 x + 4 y - 28 ⇒ x - y = - 4 ⇒ 9 x - 9 y = - 36 . . . . . . . . . . . . . . . . . . 2 Adding equations 1 and 2 , we get, 5 x + 9 y = 64 9 x - 9 y = - 36 14 x + 0 = 28 ⇒ x = 28 14 = 2 Substituting this value in equation 1 , we get, 5 × 2 + 9 y = 64 ⇒ 10 + 9 y = 64 ⇒ 9 y = 64 - 10 = 54 ∴ y = 54 9 = 6 Hence, the solutions of the given system of equations is x = 2 , y = 6 .

E M B I B E

Modern's abc+ of Mathematics for Class 10>Chapter 3 - Pair of Linear Equations in Two Variables>EXERCISE

J P Mohindru and Bharat Mohindru Solutions for Chapter: Pair of Linear Equations in Two Variables, Exercise 3: EXERCISE

Author:J P Mohindru & Bharat Mohindru

J P Mohindru Mathematics Solutions for Exercise - J P Mohindru and Bharat Mohindru Solutions for Chapter: Pair of Linear Equations in Two Variables, Exercise 3: EXERCISE

Attempt the free practice questions on Chapter 3: Pair of Linear Equations in Two Variables, Exercise 3: EXERCISE with hints and solutions to strengthen your understanding. Modern's abc+ of Mathematics for Class 10 solutions are prepared by Experienced Embibe Experts.