If the system of equations 7x + ky = 27 and kx + 7y = 19 has a unique…

Concept: For two linear equations a1x + b1y = c1 and a2x + b2y = c2, the system has a unique solution exactly when the coefficient ratios differ, that is a1/a2 ≠ b1/b2. If instead a1/a2 = b1/b2, the two lines are parallel (or coincide) and there is no single intersection point.

Applying this to the given system:

1. From 7x + ky = 27 and kx + 7y = 19, matching to the a1x + b1y = c1 form: a1 = 7, b1 = k, a2 = k, b2 = 7.

2. The uniqueness condition a1/a2 ≠ b1/b2 becomes 7/k ≠ k/7.

3. Cross-multiplying gives 7 × 7 ≠ k × k, i.e. k2 ≠ 49.

4. Solving this inequality gives k ≠ 7 and k ≠ −7.

Cross-check: take a value of k that avoids both ±7, e.g. k = 0. The equations reduce to 7x = 27 and 7y = 19, which independently fix x and y — a single unique solution, consistent with the derived condition.

So the restriction guaranteeing a unique solution is k ≠ 7 (together with k ≠ −7, though only the exclusion of 7 is offered among the given choices).