Subtracting (1) from (2) .....
=x2-y2+y-x=4
=(x-y)(x+y-1)=4
Now this can either be ......
=(x-y)(x+y-1)=2*2
or, (x-y)(x+y-1)=4*1 or, (x-y)(x+y-1)=1*4
By solving .....
x,y=(2.5,0.5), (3,-1), (3,2)
Subtracting (1) from (2) .....
=x2-y2+y-x=4
=(x-y)(x+y-1)=4
Now this can either be ......
=(x-y)(x+y-1)=2*2
or, (x-y)(x+y-1)=4*1 or, (x-y)(x+y-1)=1*4
By solving .....
x,y=(2.5,0.5), (3,-1), (3,2)
@arka .. that is definitely NOT the case.. u can break 4 into infinitely many ways...
add both equations. we get
(x+y)2 + (x+y) = 30
taking x+y = t, we get
t2 +t -30 = 0
=> t = 5, -6
=> x+y = 5, -6
CASE I, x+y = 5
subtract both equations,
(x+y)(x-y) - (x-y) = 4
=> 4(x-y) = 4
=> x-y = 1 and x+y = 5
solving we get x = 3, y = 2
CASE II, x+y = -6
subtracting both equations,
(x+y)(x-y) - (x-y) = 4
=> -7(x-y) = 4
=> x-y = -4/7 and x+y = -6
=> x = -23/7 and y = -19/7
just hope there arent any calculation errors
Yes u r absolutely right.
Sorry ........ I just considered the equation to be the product of 2 integers.