Solve for x and y . 5 x + 4 y = 9 , 7 x - 2 y = 5 , y ≠ 0

Given system of equation are, 5 x + 4 y = 9 . . . . ( i ) 7 x - 2 y = 5 . . . . ( ii ) Multiplying equation ( ii ) by 2 , we get 14 x - 4 y = 10 . . . . ( iii ) Adding ( i ) and ( iii ) , we get, 19 x = 19 x = 1 Now, substituting x value in equation ( i ) , we get 5 ( 1 ) + 4 y = 9 4 y = 4 y = 1 Therefore, x = 1 & y = 1 are the required solution for the given system of equation.

Solve for x and y . 5 x + y - 2 x - y = - 1 15 x + y + 7 x - y = 10 , x + y ≠ 0 , x - y ≠ 0

Given system of equation are, 5 x + y - 2 x - y = - 1 15 x + y + 7 x - y = 10 Let, u = 1 x + y & v = 1 x - y Therefore, we can write above system of equations as, 5 u - 2 v = - 1 . . . . ( i ) 15 u + 7 v = 10 . . . . ( ii ) Multiplying equation ( i ) by 3 , we get 15 u - 6 v = - 3 . . . . ( iii ) Subtracting equation ( ii ) from ( iii ) , we get, - 13 v = - 13 ⇒ v = 1 Now, substituting v value in equation ( i ) , we get 5 u - 2 ( 1 ) = - 1 ⇒ 5 u = 2 - 1 ⇒ u = 1 5 Now, we have, u = 1 x + y & v = 1 x - y Substituting u & v values, we get 1 5 = 1 x + y & 1 = 1 x - y ⇒ x + y = 5 . . . . ( iv ) x - y = 1 . . . . ( v ) Adding ( iv ) & ( v ) , we get 2 x = 6 ⇒ x = 3 Substituting x value in ( iv ) , we get 3 + y = 5 ⇒ y = 2 Therefore, x = 3 & y = 2 are the required solution for the given system of equation.

Solve for x and y . 2 3 x + 2 y + 3 3 x - 2 y = 17 5 5 3 x + 2 y + 1 3 x - 2 y = 2 , 3 x + 2 y ≠ 0 , 3 x - 2 y ≠ 0

Given system of equation are, 2 3 x + 2 y + 3 3 x - 2 y = 17 5 5 3 x + 2 y + 1 3 x - 2 y = 2 Let, u = 1 3 x + 2 y & v = 1 3 x - 2 y Therefore, we can write above system of equations as, 2 u + 3 v = 17 5 . . . . ( i ) 5 u + v = 2 . . . . ( ii ) Multiplying equation ( ii ) by 3 , we get 15 u + 3 v = 6 . . . . ( iii ) Subtracting equation ( iii ) from ( i ) , we get, - 13 u = - 13 5 u = 1 5 Now, substituting u value in equation ( i ) , we get 2 ( 1 5 ) + 3 v = 17 5 3 v = 17 - 2 5 = 15 5 = 3 v = 1 Now, we have u = 1 3 x + 2 y & v = 1 3 x - 2 y ⇒ 1 5 = 1 3 x + 2 y & 1 = 1 3 x - 2 y ⇒ 3 x + 2 y = 5 . . . . . ( iv ) 3 x - 2 y = 1 . . . . . ( v ) Adding ( iv ) & ( v ) , we get 6 x = 6 x = 1 Substituting x value in ( iv ) , we get 3 ( 1 ) + 2 y = 5 2 y = 2 y = 1 Therefore, x = 1 & y = 1 are the required solution for the given system of equation.

Solve for x and y . 5 x + 1 - 2 y - 1 = 1 2 10 x + 1 + 2 y - 1 = 5 2 , x ≠ - 1 , y ≠ 1

Given system of equation are, 5 x + 1 - 2 y - 1 = 1 2 10 x + 1 + 2 y - 1 = 5 2 Let, u = 1 x + 1 & v = 1 y - 1 Therefore, we can write above system of equations as, 5 u - 2 v = 1 2 . . . . ( i ) 10 u + 2 v = 5 2 . . . . ( ii ) Adding ( i ) & ( ii ) we get, 15 u = 6 2 = 3 u = 1 5 Now, substituting u value in equation ( i ) , we get 15 ( 1 5 ) - 2 v = 1 2 - 2 v = 1 2 - 3 - 2 v = - 5 2 v = 5 4 Now, we have, u = 1 x + 1 & v = 1 y - 1 Substituting u & v values, we get 1 5 = 1 x + 1 & 5 4 = 1 y - 1 ⇒ x + 1 = 5 & 5 y - 5 = 4 ⇒ x = 4 & y = 9 5 Therefore, x = 4 & y = 9 5 are the required solution for the given system of equation.

Solve for x and y by using method of substitution. 6 x + y - 7 x - y = 3 1 2 ( x + y ) - 1 3 ( x - y ) = 0 , x + y ≠ 0 , x - y ≠ 0

Given system of equations are, 6 x + y - 7 x - y = 3 1 2 ( x + y ) - 1 3 ( x - y ) = 0 Let, u = 1 x + y & v = 1 x - y Therefore, we can write above system of equations as, 6 u - 7 v = 3 . . . . ( i ) 1 2 u - 1 3 v = 0 . . . . ( ii ) Dividing equation ( i ) by 12 , we get 1 2 u - 7 12 v = 1 4 . . . . ( iii ) Subtracting equation ( ii ) from ( iii ) , we get, - 1 4 v = 1 4 v = - 1 Now, substituting v value in equation ( i ) , we get 6 u - 7 ( - 1 ) = 3 6 u = - 4 u = - 2 3 Now, we have, u = 1 x + y & v = 1 x - y Substituting u & v values, we get - 2 3 = 1 x + y & - 1 = 1 x - y ⇒ - 2 x - 2 y = 3 . . . . ( iv ) - x + y = 1 . . . . ( v ) Multiplying ( v ) by 2 , we get - 2 x + 2 y = 2 . . . ( vi ) Adding ( iv ) & ( vi ) , we get - 4 x = 5 x = - 5 4 Substituting x value in ( v ) , we get 5 4 + y = 1 y = 1 - 5 4 y = - 1 4 Therefore, x = - 5 4 & y = - 1 4 is the required solution for the given system of equations.

Solve for x and y . 148 x + 231 y = 527 x y 231 x + 148 y = 610 x y

Given system of equation are, 148 x + 231 y = 527 x y . . . . ( i ) 231 x + 148 y = 610 x y . . . . ( ii ) Adding ( i ) & ( ii ) , we get 379 x + 379 y = 1137 x y 379 ( 1 x + 1 y ) = 1137 x y 379 ( y + x ) = 1137 ( x + y ) = 1137 379 x + y = 3 . . . . ( iii ) Subtracting ( i ) from ( ii ) , we get 83 x - 83 y = 83 x y y - x = 1 . . . . ( iv ) Adding ( iii ) & ( iv ) , we get 2 y = 4 y = 2 Substituting y value in ( iii ) , we get x + 2 = 3 x = 1 Therefore, x = 1 & y = 2 are the required solution for the given system of equation.

E M B I B E

New Mathematics for Class X>Chapter 3 - Pair Of Linear Equations In Two Variables>EXERCISE

H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Pair Of Linear Equations In Two Variables, Exercise 4: EXERCISE

Author:H K Dass, Rama Verma & Bhagwat Swarup Sharma

H K Dass Mathematics Solutions for Exercise - H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Pair Of Linear Equations In Two Variables, Exercise 4: EXERCISE

Attempt the free practice questions on Chapter 3: Pair Of Linear Equations In Two Variables, Exercise 4: EXERCISE with hints and solutions to strengthen your understanding. New Mathematics for Class X solutions are prepared by Experienced Embibe Experts.