We have,

…(i)

…(ii)

Now, substituting and in (i) and (ii) we get:

3u + 2v = 2 …(iii)

9u – 4v = 1 …(iv)

Multiplying (iii) by 2 and adding it with (iv) we get:

6u + 9u = 4 + 1

Multiplying again (iii) by 2 and then subtracting it from (iv), we get:

6v + 4v = 6 – 1

∴ x + y = 3 …(v)

And, x – y = 2 …(vi)

Now, by adding (v) and (vi) we get:

2x = 3 + 2

Substituting the value of x in (v), we get

Hence, option B is correct