Question

find m ∈ r such that in the solution of the system mx + (m + 2)y = 1, x + my = m is x > y.​

Answer 1

Given : solution of the system mx + (m + 2)y = 1, x + my = m is x > y.​

To Find :  m ∈ r

Solution:

mx + (m + 2)y = 1

 x + my = m  

Assume infinite solution :

m/1  = (m + 2)/m   = 1/m

from 1 and 3   m = ± 1

from 2  and 3  =>   m + 2 = 1 => m = - 1

from 1 and 2  => m² = m + 2   =>  (m² - m - 2 = 0  (m - 2)(m + 1) = 0

=> m = 2 , m = - 1

m =2  no solutions

Hence m = - 1  gives infinte solutions

-x  + y  = 1    => y = 1 + x

=> x = y - 1

Hence x < y  

so this is not possible

mx + (m + 2)y = 1

 mx + m²y = m²  

(m + 2)y  - m²y = 1  - m²

=> y ( m² - m - 2)  = m² - 1

=> y  =   (m - 1)/(m - 2)

x + my = m  

=>x = m (1  - y)

=> x  = m  (  1   -    (m - 1)/(m - 2) )

=> x = m ( - 1/(m - 2)

=> x  = -m/(m - 2)  

for m > 2  x is -ve while y is + ve  hence no feasible solution x > y

for m ∈ [1 , 2)   x  > y   as  y is non positive and x is +ve  => x > y

now for m < 1  

y is + ve    and  x  is + ve    

-m/(m - 2) >  (m - 1)/(m - 2)

as m - 2 is -ve

=>  - m < m - 1

=> 2m > 1

=> m > 1/2

Hence  m ∈ ( 1/2 , 2)  

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