Conditions for solvability

Dividing every term of the second equation by 2 gives \(8x+3y=17\), which is exactly the first equation. Hence, both equations represent the same line and have infinitely many solutions. Option C is incorrect because distinct parallel lines do not have all three corresponding ratios equal. Exam tip: if \(a_1/a_2=b_1/b_2=c_1/c_2\), the two lines are coincident and the system has infinitely many solutions.

The ratios of the coefficients are equal: 9/3 = (-6)/(-2) = 3, but the ratio of the constant terms is 12/7, which is not 3. Hence, the two lines are parallel and distinct, so the pair has no solution. If all three ratios were equal, the lines would be coincident and would have infinitely many solutions. Exam tip: when a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair has no solution.

For a pair of linear equations, the condition \(a_1/a_2 \ne b_1/b_2\) indicates a unique solution. Here, \(13/6 \ne 2/1\), so the two lines intersect at exactly one point. Therefore, the correct answer is one unique solution. Infinitely many solutions require the corresponding coefficient and constant ratios to be equal, while exactly two solutions are not possible for a pair of linear equations. Exam tip: first compare \(a_1/a_2\) and \(b_1/b_2\).

Here,
(4/12)=((-3)/(-9))=1/3, but (11/35) ≠ 1/3. Thus, the two lines have the same slope but are not the same line; they are parallel and distinct. Therefore, the system is inconsistent and has no solution. Exam tip: If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, a pair of linear equations has no solution. Exactly two solutions are impossible because two lines can intersect at most once or coincide completely.

For two linear equations to have infinitely many solutions, the ratios a₁/a₂, b₁/b₂ and c₁/c₂ must be equal. Here, 2/8 = 14/56 = 1/4, so m/20 must also equal 1/4. Therefore, m = 5, making option A correct. For example, if m = 6, then m/20 = 3/10, which is not equal to 1/4. Exam tip: For infinitely many solutions, always check that all three coefficient and constant ratios are equal.

Here (7/3 \ne 4/2), so the lines will intersect at one point. If the first two ratios differ, one unique solution is obtained.

The second equation must be (2) times the first, so (k=36). For infinitely many solutions, keep all three ratios equal.

A pair of linear equations has infinitely many solutions when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). Here, \(\frac{3}{9}=\frac{21}{63}=\frac{1}{3}\). Thus, \(\frac{a}{12}=\frac{1}{3}\), giving \(a=4\). Therefore, option C is correct. Exam tip: for infinitely many solutions, all three corresponding coefficient and constant ratios must be equal; equality of only two ratios is not sufficient.

A pair of linear equations has no solution when \(a_1/a_2=b_1/b_2\), but this common ratio is not equal to \(c_1/c_2\). Here, \(7/14=1/2\), whereas \(28/60=7/15\), so the constant ratio is different. Thus, \(k/8=1/2\), giving \(k=4\). Exam tip: For inconsistent equations, the coefficient ratios must be equal while the constant ratio must be different.

The first two ratios are equal, so for inconsistency the constant ratio must differ. Hence, (m \ne 32).

The second equation is (2) times the first, so both lines are coincident. Coincident lines have infinitely many solutions.

Multiplying the left-hand side of the first equation by 2 gives \(12x-10y\), but the right-hand side would be \(2\times14=28\), not 31. Thus, the coefficient ratios are equal: \(\frac{6}{12}=\frac{-5}{-10}=\frac{1}{2}\), whereas the ratio of the constants is \(\frac{14}{31}\), which is different. Therefore, the two lines are distinct and parallel, so the pair has no solution. Exam tip: If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the pair of linear equations has no solution.

Here (4/8 \ne 9/17), so the lines intersect at one point. Different first two ratios give a unique solution.

When all three ratios are equal, both equations form the same line. Therefore, the pair is consistent and dependent.

Equal first two ratios give the same slope and a different third ratio keeps the lines separate. Hence, they are distinct parallel.

For two linear equations to have infinitely many solutions, the ratios of the corresponding coefficients and constants must be equal: \(\frac{p}{12}=\frac{3}{6}=\frac{15}{30}=\frac{1}{2}\). Thus, \(\frac{p}{12}=\frac{1}{2}\), giving \(p=6\). Therefore, option C is correct. Exam tip: For infinitely many solutions, check whether \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

For two linear equations to have a unique solution, the ratios of the corresponding coefficients must be unequal. Thus, \(\frac{3}{7}\ne\frac{q}{14}\). On cross-multiplying, this gives \(42\ne 7q\), or \(q\ne 6\). If \(q=6\), the ratios of the coefficients of \(x\) and \(y\) become equal, so the system does not have a unique solution. Exam tip: For a unique solution, remember \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\).

The first equation must be (5) times the second, so (k=20). A common multiplier makes both lines coincident.

For two linear equations to have no solution, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\) must hold. Here, \(\frac{8}{2}=4\) and \(\frac{40}{13}\ne4\). Therefore, \(\frac{a}{6}=4\), giving \(a=24\). Hence, option D is correct. Exam tip: equal ratios of the variable coefficients but an unequal ratio of the constants indicate no solution.

For the two equations, \\(a_1/a_2=10/2=5\\) and \\(b_1/b_2=15/3=5\\), but \\(c_1/c_2=35/9\\). Thus, \\(a_1/a_2=b_1/b_2\\ne c_1/c_2\\), so the two lines are distinct and parallel. They neither intersect nor coincide. Exam tip: the condition \\(a_1/a_2=b_1/b_2\\ne c_1/c_2\\) represents distinct parallel lines.