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

The correct answer is Option (4) → no solution

Two equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\) have no solution when their lines are parallel, that is, when \(\displaystyle\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\).

• \(\displaystyle\frac{a_1}{a_2}=\frac{4}{6}=\frac{2}{3}\)

• \(\displaystyle\frac{b_1}{b_2}=\frac{14}{21}=\frac{2}{3}\)

• \(\displaystyle\frac{c_1}{c_2}=\frac{18}{33}=\frac{6}{11}\)

The first two ratios are equal but the third is not, so the lines are parallel and never meet.

Why the other options are incorrect

• Option (1) infinite solutions — Incorrect. That needs all three ratios equal, but \(\frac{6}{11}\ne\frac{2}{3}\).

• Option (2) a unique solution \(x = 1, y = 1\) — Incorrect. It fits the first equation (\(4+14=18\)) but not the second (\(6+21=27\), not 33).

• Option (3) a unique solution \(x = 2, y = 1\) — Incorrect. It fits neither equation: \(8+14=22\), not 18.