Conditions for solvability

The second equation is exactly twice the first: multiplying 14x+9y=61 by 2 gives 28x+18y=122. Hence both equations represent the same line and the pair has infinitely many solutions. Option A is incorrect because distinct parallel lines require the ratio of the constants to differ from the ratios of the corresponding coefficients. Option C represents a unique solution, which is not possible here. Exam tip: if a₁/a₂ = b₁/b₂ = c₁/c₂, the two lines are coincident.

Comparing the corresponding coefficients gives 18/3 = 6 and (-12)/(-2) = 6, but 72/13 is not equal to 6. Thus, a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so the pair is inconsistent and has no solution. Equivalently, multiplying the second equation by 6 gives 18x − 12y = 78, which contradicts the first equation, 18x − 12y = 72. Exam tip: When a₁/a₂ = b₁/b₂ but c₁/c₂ is different, the lines are parallel and distinct, so the number of solutions is zero.

The coefficient ratios are unequal: \(\frac{19}{9} \ne \frac{8}{4}\). Equivalently, the determinant is \(19 \times 4 - 8 \times 9 = 4 \ne 0\). Therefore, the two lines intersect at exactly one point, so the equations have one unique solution. Option B would apply if both equations represented the same line, while option A describes parallel lines. Exam tip: when \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\), a pair of linear equations has a unique solution.

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, \(\frac{10}{20}=\frac{44}{88}=\frac{1}{2}\), so \(\frac{3}{a}=\frac{1}{2}\), which gives \(a=6\). The nearby distractor 5 is incorrect because \(\frac{3}{5}\neq\frac{1}{2}\). Exam tip: for infinitely many solutions, verify that all three coefficient-to-constant ratios are equal.

For the two equations, \(\frac{a_1}{a_2}=\frac{11}{22}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{6}{12}=\frac{1}{2}\), but \(\frac{c_1}{c_2}=\frac{25}{53}\neq\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\), which represents an inconsistent pair of equations and two distinct parallel lines. Therefore, there is no solution. Option C would be correct only if all three ratios were equal. Exam tip: compare the ratios in the order \(a_1/a_2\), \(b_1/b_2\), and then \(c_1/c_2\).

For the pair 7x + 10y = 39 and 5x + 8y = 31, compare the ratios of the corresponding coefficients. The x-coefficient ratio is 7/5 and the y-coefficient ratio is 10/8 = 5/4. These are not equal: cross multiplication gives 7 × 8 = 56, while 10 × 5 = 50. Therefore the two lines have different slopes and must intersect at exactly one point. The pair is consequently consistent and independent, with one unique solution. Comparing the constant ratio is unnecessary once the first two ratios differ. No solution would require equal x- and y-coefficient ratios but a different constant ratio; infinitely many solutions would require all three ratios to be equal. Thus option C states the correct conclusion.

For two linear equations to have infinitely many solutions, the condition is 5/15 = 2/a = 18/54. Since 5/15 and 18/54 are both equal to 1/3, 2/a = 1/3 gives a = 6. Therefore, option C is correct. Exam tip: all three corresponding coefficient and constant ratios must be equal; checking only two ratios is not sufficient.

For two linear equations to have no solution, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\) must hold. Here, \(\frac{8}{20}=\frac{2}{5}\), so \(\frac{p}{10}=\frac{2}{5}\) gives \(p=4\). Also, \(\frac{24}{61}\neq\frac{2}{5}\), so the two lines are parallel and distinct. Exam tip: For ‘no solution’, first equate the ratios of the variable coefficients and then verify that the constants’ ratio is different.

The first two ratios are equal. For inconsistency, the constant ratio must be different so (m \ne 50).

The first equation is 9x − 2y = 31. Multiplying it by 3 produces 27x − 6y = 93, exactly the second equation. Therefore, the pair represents two coincident lines, not two intersecting or distinct parallel lines. The coefficient ratios also confirm this: 27/9 = (−6)/(−2) = 93/31 = 3. Every point satisfying the first equation automatically satisfies the second, so infinitely many solutions exist.

For the two equations, 6/12 = 11/22 = 1/2, but 29/63 is not equal to 1/2. Thus, the ratios of the coefficients of x and y are equal, while the ratio of the constant terms is different. This represents two distinct parallel lines, so they have no common point and the pair has no solution. Exam tip: if a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair of linear equations has no solution.

Here (8/15 \ne 13/24) so the lines intersect at one point. If the first two ratios are different, one unique solution is obtained.

A pair of linear equations has infinitely many solutions when the ratios of corresponding coefficients and constants are equal: k/15 = 12/20 = 36/60. Since both 12/20 and 36/60 equal 3/5, we get k/15 = 3/5, so k = 9. Therefore, option C is correct. Exam tip: For infinitely many solutions, verify a₁/a₂ = b₁/b₂ = c₁/c₂.

For two linear equations to have infinitely many solutions, the ratios of their corresponding coefficients and constant terms must be equal. Here, the second equation must be twice the first, so 18 = 2q. Therefore, q = 9. Exam tip: For infinitely many solutions, check \(a_1/a_2=b_1/b_2=c_1/c_2\).

For two linear equations to have a unique solution, the ratios of the corresponding coefficients must be unequal: \(\frac{10}{2}\ne\frac{p}{7}\). Thus, \(5\ne\frac{p}{7}\), which gives \(p\ne35\). If \(p=35\), the ratios of the coefficients of \(x\) and \(y\) become equal; since \(\frac{50}{19}\ne\frac{10}{2}\), the equations then have no solution rather than a unique solution. Exam tip: for a unique solution, check \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\).

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, \(\frac{9}{27}=\frac{63}{189}=\frac{1}{3}\), so \(\frac{a}{24}=\frac{1}{3}\), giving \(a=8\). Therefore, option C is correct. Exam tip: For infinitely many solutions, verify equality of all three ratios, not just two of them.

\(a_1b_2\ne a_2b_1\) makes the lines non-parallel, so they meet once. In option B, proportional coefficients with unequal constants give no solution. Exam tip: compare cross-products.

When the ratios of the coefficients of \(x\) and \(y\) are equal, the two lines have the same slope. To decide whether they are the same line or distinct parallel lines, compare the ratio of the constants too. Equal ratios throughout mean coincident lines; a different constant ratio means distinct parallel lines.

Here, \(20/2=10\) and \(30/3=10\), but \(90/11\ne10\). Thus the coefficients of the variables are proportional, while the constants are not. The equations represent two separate parallel lines. They have no common point, so the graph is of distinct parallel lines, as stated in option C.

The first equation is five times the second: \(25x+10y=5(5x+2y)=5\times19=95\). Thus, both equations represent the same equation, so their graphs coincide as one line. Exam tip: when \(a_1/a_2=b_1/b_2=c_1/c_2\), the pair has infinitely many solutions and the two lines are coincident, not merely distinct parallel lines.

The coefficient ratios are unequal: \\(\frac{4}{9} \ne \frac{15}{31}\\). Equivalently, the determinant is \\(4\times31-9\times15=-11\ne0\\), so the two lines intersect at exactly one point and the pair has a unique solution. Coincident or distinct parallel lines require the relevant coefficient ratios to be equal. Exam tip: if \\(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\\), the lines intersect at one point.