Question

mx+ny= m^2- n^2 and ny+mx=0 find the value of x+y

Answer 1

Heya User,

--> mx + ny = m² - n² = 0
=> m² - n² = 0
=> ( m + n )( m - n ) = 0
=> m = ± n

=> mx + ny = 0
=> ±nx + ny = 0
=> n ( x + y ) = 0  ||  n ( y - x ) = 0

--> x + y = 0  ---> [ m ± n not equal to zero ]
--> x = y = ( m - n ) --> [ m = ± n not equal to zero ]

Also, we have infinite possible values for x, y if m = ± n = 0 |||

Answer 2

Hi ,

mx + ny = m² - n² ----( 1 )

nx + my = 0 ---( 2 )

do [ n × ( 1 ) , m × ( 2 ) ]

mnx + n² y = n( m² - n² ) ----( 3 )

mnx + m² y = 0 ----( 4 )

subtract (4 ) from ( 3 ) , we get

y ( n² - m² ) = n ( m² - n² )

y = [ - n ( n² - m² ) ] / ( n² - m² )

y = -n ----( 5 )

put y value in equation ( 2 ) , we get

nx + m ( -n ) = 0

x - m = 0

x = m ---( 6 )

Therefore ,

add ( 6 ) and ( 5 ) ,

x + y = m - n

I hope this helps you.

:)