B

1. x-y = y

x = 2/3 or 3/8

y = x/2 or x/3

lets substituate the value of y in x

for y=x/2

x-(x/2) = x/2 = y ....true

for y=x/3

x-(x/3) = 2x/3 not equal to y...false.

so now we know for the first equation, y=x/2, since x can be 2/3 or 3/8, we cannot get a single value for y

2. x+y = 1

x = 2/3 or 3/8

y = x/2 or x/3

lets substituate the value of y in x

for y=x/2

x+(x/2) = 1

3x/2 = 1 => x = 2/3(it one of the given values of x)

for x=2/3 , subsitute x for y => y = (2/3)/2 = 1/3

for y=x/3

x+(x/3) = 1

4x/3 = 1 => x = 3/4(is not the given value of x...so invalid case)

From 2, we got x=2/3 and y = 1/3.

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