If 3x + 5y + 7z = 49 and 9x + 8y + 21z = 126, then what is the value of y?

By eliminating variable 'z' as there are three unknowns & only 2 equations.
By putting z = 0
3x + 5y = 49 . . . . . . (i)
9x + 8y = 126 . . . . . . (ii)
Multiplying by 3 in equation (i) and subtracting equation (ii)
9x + 15y = 147
9x + 8y = 126
$$\overline {\,\,\,\,\,\,\,\,\,\,\,\,7{\text{y}} = 21\,\,} $$
y = 3

Alternate solution:
3x + 5y + 7z = 49
9x + 8y + 21z = 126
Assume value x, y, z
x = 2, y = 3, z = 4
3(2) + 5(3) + 7(4) = 49
6 + 15 + 28 = 49
49 = 49 value satisfied
9(2) + 8(3) + 21(4) = 126
18 + 24 + 84 = 126
126 = 126 value satisfied
y = 3