The system of equations $4x + 14y = 18$ and $6x + 21y = 33$ has

Consider the system of equations $4x + 14y = 18$ and $6x + 21y = 33$.

For the first equation $4x + 14y = 18$, dividing by 2:

$2x + 7y = 9$

For the second equation $6x + 21y = 33$, dividing by 3:

$2x + 7y = 11$

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The simplified equations are:

$2x + 7y = 9$

$2x + 7y = 11$

The left sides are identical, but the right sides are different: $9 \neq 11$.

The same expression $2x + 7y$ cannot equal both 9 and 11 simultaneously. These represent parallel lines that never intersect, so no point $(x, y)$ satisfies both equations.

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Testing $x = 1, y = 1$ in the original equations:

$4(1) + 14(1) = 18$ (satisfies first equation)

$6(1) + 21(1) = 27 \neq 33$ (does not satisfy second equation)

Similarly, $x = 2, y = 1$ does not satisfy both equations.

For infinite solutions, both sides would need to be identical after simplification, which is not the case here.

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Therefore, the system has no solution (Option 4).