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

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

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

$$2x + 7y = 9$$2x+7y=9

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

$$2x + 7y = 11$$2x+7y=11

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The simplified equations are:

$$2x + 7y = 9$$2x+7y=9

$$2x + 7y = 11$$2x+7y=11

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

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

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

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

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

Similarly, $x = 2, y = 1$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).