1you can use gadha- ghora method of substitutinng
the options & check if they satisfy both!!: ;)
8 Answers
1
19yup..i did just that ! there was no time to be a good mathematician at that stage ! [3]
11So did I....Deobotosh.
But I wanna see if there's any general way without consulting the options....I cud not find any in the exam hall...
U can call this my doubt Nishant bhaiya.
112y-x(x+y)=1
=>x+y=2x-y
puting this value in another equation( 2x-y)x-y=2(x-y)2=2
=>(x-y)2=1
=>x-y=1
=>x=y+1
now puting the value of x in 1st eqn 2-1[y+1 + y]=1
=>2-1(2y+1)=1
=>y=1/2
=>x=3/2
now for x-y=-1
=>x=y-1
now puting the value of x in 1st eqn 21[y-1 + y]=1
=>21(2y-1)=1
=>y=3/4
=>x=-1/4
11There's another ans.....
I guess that comes when (x-y)=-1.
DAMN I MISSED IT - IDIOTIC OF ME!!!
11arre bhaiya, my above two solns satisfy both the equation .so i think it should be the answer.