10/x+y +2/x-y , 15/x+y - 5/x-y solve the pair of equation by them to a pair of linear equation

10/x+y +2/x-y , 15/x+y - 5/x-y solve the pair of equation by them to a

1.	10/x+y +2/x-y , 15/x+y - 5/x-y solve the pair of equation by them to a pair of linear equation
Answer» Given equations are ,\xa010/( x + y ) + 2/( x - y ) = 4 ---( 1 )\xa015/( x + y ) - 5/( x - y ) = -2 --( 2 )\xa0Let ,\xa01/(x+y) = a , 1/(x-y) = b\xa010a + 2b = 4\xa0Divide each term with 2 , we get\xa05a + b = 2\xa0=> b = 2 - 5a ----( 3 )\xa015a - 5b = -2 -----( 4 )\xa0Substitute b = 2 - 5a in equation\xa0( 4 ) , we get\xa015a - 5( 2 - 5a ) = -2\xa0=> 15a - 10 + 25a = -2\xa0=> 40a = -2 + 10\xa0=> 40a = 8\xa0=> a = 8/40\xa0=> a = 1/5\xa0Put a = 1/5 in equation ( 3 ) , we\xa0get\xa0b = 2 - 5 × 1/5\xa0=> b = 2 - 1\xa0b = 1\xa0Therefore ,\xa01/( x + y ) = 1/5 => x + y = 5 --( 5 )\xa01/(x-y) = 1/1 => x - y = 1 ---( 6 )\xa0Add equations ( 5 ) and ( 6 ) ,\xa0We get\xa02x = 6\xa0=> x = 6/2 = 3 ,\xa0Put x = 3 in equation ( 5 ) , we get\xa03 + y = 5\xa0=> y = 5 - 3 = 2\xa0Therefore ,\xa0x = 3 , y = 2