Conditions for solvability

For no solution, the two lines must be parallel and distinct. Algebraically, this requires a1/a2 = b1/b2 but c1/c2 must be different. From the y-coefficients, (-5)/(-15) = 1/3. Therefore, p/12 must equal 1/3, which gives p = 4. The constant ratio is 7/19, and this is not equal to 1/3, so the lines are distinct rather than coincident. Consequently, they do not intersect and the pair has no solution. Hence option A is correct; the other values do not make the x-coefficient ratio equal to the y-coefficient ratio.

For a pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to have infinitely many solutions, the coefficients must satisfy a₁/a₂ = b₁/b₂ = c₁/c₂. Comparing qx + 11y = 22 with 18x + 33y = 66 gives q/18 = 11/33 = 22/66. Both known ratios equal 1/3, so q/18 = 1/3 and q = 18/3 = 6. Therefore option C is correct. Values 4, 5, and 7 do not make the coefficient ratios equal, so they would not produce coincident lines or infinitely many common points.

Two linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\) have a unique solution when \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\). Here, \(\frac{r}{9}\neq\frac{2}{3}\) is required. Since \(\frac{2}{3}=\frac{6}{9}\), we get \(r\neq6\). If \(r=6\), the coefficients on the left-hand sides become proportional, so the solution cannot be unique. Exam tip: compare the ratios of the coefficients of \(x\) and \(y\) first.

Equating coefficient ratios, (\frac{s}{21}=\frac{4}{12}) gives (s=7). The constant ratio is not equal.

For infinitely many solutions, the two linear equations must represent the same line. Hence, the ratios of corresponding coefficients and constants must be equal: \(\frac{t}{20}=\frac{9}{15}=\frac{6}{10}\). Since \(\frac{9}{15}=\frac{6}{10}=\frac{3}{5}\), we get \(\frac{t}{20}=\frac{3}{5}\). Therefore, \(t=12\). For \(t=10\), \(\frac{t}{20}=\frac{1}{2}\), which is not equal to \(\frac{3}{5}\). Exam tip: use \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) to check for infinitely many solutions.

Under the intended interpretation that \(y\) is a parameter appearing in the coefficients, the coefficient determinant is \(D=a_1b_2-a_2b_1=(y+1)(y-2)-2\times5=y^2-y-12\). A pair has a unique solution exactly when \(D\neq0\); hence \(y^2-y-12\neq0\) is correct. In option A, \(D=0\), so the solution cannot be unique. Exam tip: test \(a_1b_2-a_2b_1\neq0\) for a unique solution.