Conditions for solvability

For infinitely many solutions, both equations must represent the same straight line. Therefore, the ratios of corresponding coefficients must be equal: 4/12 = 9/27 = 1/3. The coefficient of y must follow the same ratio, so (m − 5)/6 = 1/3. Multiplying by 6 gives m − 5 = 2, and hence m = 7. Thus option C is correct. The other options make the y-coefficient ratio different from 1/3, so the three ratios are not equal. In that case the equations do not describe coincident lines and cannot possess infinitely many common solutions.

A pair of linear equations has a unique solution when the ratios of the coefficients of x and y are unequal. Here, the condition is \(\frac{p+4}{5}\neq\frac{-9}{-15}\). Since \(\frac{-9}{-15}=\frac{3}{5}\), we need \(\frac{p+4}{5}\neq\frac{3}{5}\), which gives \(p\neq -1\). At \(p=-1\), the x- and y-coefficient ratios become equal, so the solution cannot be unique. Exam tip: for a unique solution, first compare the ratios of the coefficients of x and y.

A pair of linear equations has no solution when the ratios of the coefficients of x and y are equal, but this common ratio is different from the ratio of the constant terms. Here, the x-coefficient ratio is (1/2)/2 = 1/4. To make the lines parallel, the y-coefficient ratio must also be q/8 = 1/4, which gives q = 2. The constant ratio is 3/5, not 1/4, so the equations are inconsistent and have no solution. Therefore option B is correct. Values q = 1, 4, or 8 do not make the two coefficient ratios equal, giving a unique intersection instead.

For infinitely many solutions, the two linear equations must represent the same line; hence \(\frac{r}{8}=\frac{\frac{3}{2}}{6}=\frac{6}{24}\). The last two ratios are \(\frac{1}{4}\), so \(\frac{r}{8}=\frac{1}{4}\) gives \(r=2\). If \(r=4\), then \(\frac{r}{8}=\frac{1}{2}\), which is not equal to the other ratios. Exam tip: for infinitely many solutions, verify \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

For a unique solution, (\frac{\alpha}{9}\neq\frac{4}{12}) must hold. Thus (\alpha\neq3) is correct.

Equating coefficient ratios gives (\beta=3). The constant ratio (\frac{5}{9}) is different, so there is no solution.

The coefficient ratio is (\frac{1}{4}). For infinitely many solutions, (\frac{\gamma}{36}=\frac{1}{4}), so (\gamma=9).

A pair of linear equations has a unique solution when its coefficient determinant is non-zero. Here, \(D=(s+2)(s-1)-3\times5=s^2+s-17\). Therefore, the required condition is \(s^2+s-17\ne 0\). In option A, the determinant becomes zero, so a unique solution is not possible. Exam tip: use \(a_1b_2-a_2b_1\ne0\) to test for a unique solution.

The first two ratios are equal. For inconsistency, the constant ratio must be different so (t\ne192).

For two linear equations to have infinitely many solutions, the condition is \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). Here, \(\frac{6}{18}=\frac{42}{126}=\frac{1}{3}\). Therefore, \(\frac{a}{33}=\frac{1}{3}\), giving \(a=11\). Hence, option C is correct. Exam tip: For infinitely many solutions, the ratios of the corresponding coefficients and constant terms must all be equal; checking only one ratio is not sufficient.

For two linear equations to have no solution, the ratios of the coefficients of x and y must be equal, while the ratio of the constant terms must be different. Thus, p/14 = 10/35 = 2/7, giving p = 4. Also, 50/122 = 25/61, which is not equal to 2/7, so the equations are inconsistent. Exam tip: Remember the condition a₁/a₂ = b₁/b₂ ≠ c₁/c₂ for no solution.

Two linear equations have a unique solution when the ratios of the coefficients of the variables are unequal. Here, the required condition is \(\frac{11}{5}\ne\frac{k}{4}\). On simplifying, this gives \(44\ne5k\), or \(k\ne\frac{44}{5}\). Therefore, option B is correct. Exam tip: for a unique solution, use \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). In option A, the two ratios become equal, so it does not represent a unique solution.

Here, \(a_1/a_2=12/36=1/3\) and \(b_1/b_2=(-7)/(-21)=1/3\), but \(c_1/c_2=29/91\), which is not equal to \(1/3\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the two lines are parallel and distinct. Therefore, the pair has no solution. Exam tip: First compare the ratios of the coefficients of x and y, and then compare them with the ratio of the constant terms.

For the pair a1x + b1y = c1 and a2x + b2y = c2, the relationship between a1/a2 and b1/b2 determines whether the lines have the same slope. Here, 7/13 is not equal to 19/35: cross-multiplication gives 7 × 35 = 245, whereas 19 × 13 = 247. Thus the first two ratios are unequal. The lines have different slopes and must intersect at exactly one point, so the pair has a unique solution. Option C is correct. Infinite solutions require all three ratios to be equal, while no solution requires the first two ratios equal but the constant ratio different.

Here, \(a_1/a_2=10/20=1/2\) and \(b_1/b_2=3/6=1/2\), but \(c_1/c_2=470/955\neq1/2\). Thus, the two lines have the same slope but different intercepts, so they are parallel and have no point of intersection. Therefore, the system is inconsistent. A common mistake is to conclude that equal first two ratios imply infinitely many solutions; the third ratio must also be equal for coincident lines. Exam tip: if \(a_1/a_2=b_1/b_2\neq c_1/c_2\), the system is inconsistent.

Multiply the first equation by 2: 2(21x+8y=97) gives 42x+16y=194, which is exactly the second equation. Hence the two equations are dependent and represent the same straight line. Every point on this common line satisfies both equations, so the pair has infinitely many solutions. Option A and option D describe distinct parallel lines, while option C would require non-proportional coefficients.

A pair of linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\) has a unique solution when \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\). Here, \(\frac{k}{6}\neq\frac{5}{15}=\frac{1}{3}\), which gives \(k\neq2\). At \(k=2\), the ratios of the coefficients on the left-hand sides are equal, so the solution cannot be unique. Exam tip: for a unique solution, compare the ratios of the coefficients of \(x\) and \(y\).

Equating coefficient ratios gives (m=6). The constant ratio is different, so there is no solution.

For infinitely many solutions, all three ratios must be equal. From (\frac{p-2}{9}=\frac{7}{21}), (p=5).

For infinitely many solutions, the two linear equations must represent the same line; hence \(\frac{8}{20}=\frac{a}{10}=\frac{12}{30}\). Here, \(\frac{8}{20}=\frac{12}{30}=\frac{2}{5}\). Therefore, \(\frac{a}{10}=\frac{2}{5}\), which gives \(a=4\). For \(a=3,5\), or \(6\), the ratios of the corresponding coefficients are not equal. Exam tip: for infinitely many solutions, check that \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).