Conditions for solvability

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 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).

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.

Check the three ratios: \(a_1/a_2=8/24=1/3\), \(b_1/b_2=13/39=1/3\), and \(c_1/c_2=(-59)/(-177)=1/3\). Thus, \(a_1/a_2=b_1/b_2=c_1/c_2\); the two lines are coincident and the pair has infinitely many solutions. Options A and D incorrectly state that one ratio is unequal. Exam tip: if all three ratios are equal, there are infinitely many solutions; if only the first two are equal and the third differs, there is no solution.

Here, \(a_1=12, b_1=-5, c_1=37\) and \(a_2=24, b_2=-10, c_2=79\). Thus, \(\frac{a_1}{a_2}=\frac{12}{24}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}\), whereas \(\frac{c_1}{c_2}=\frac{37}{79}\ne\frac{1}{2}\). Therefore, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the two lines are parallel and distinct, giving no solution. Exam tip: When the first two coefficient ratios are equal but the constant-term ratio is different, the pair has no solution.

Here (13/26 \ne 6/13) so the lines intersect at one point. In this case, one unique solution is obtained.

For two linear equations to have no solution, the condition is \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). Here, \(\frac{16}{42}=\frac{8}{21}\), so \(\frac{c}{21}=\frac{8}{21}\), giving \(c=8\). Also, \(\frac{64}{130}=\frac{32}{65}\), which is not equal to \(\frac{8}{21}\); therefore, the equations are inconsistent. The other values make the coefficients' ratios unequal and lead to a unique solution. Exam tip: equate the ratios of the variable coefficients first, then check that the constant-term ratio is different.

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, \(\frac{6}{18}=\frac{54}{162}=\frac{1}{3}\), so \(\frac{d}{30}=\frac{1}{3}\), giving \(d=10\). If d were 9 or 11, the coefficient ratios would not remain \(1:3\). Exam tip: first compare the ratios of the known coefficients and constants, then use that ratio to find the unknown coefficient.

Two linear equations have a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, the condition is \(\frac{15}{7}\ne\frac{p}{2}\), which gives \(30\ne7p\), or \(p\ne\frac{30}{7}\). Hence, option B is correct. In option A, the two coefficient ratios become equal, so the system does not have a unique solution. Exam tip: For a unique solution, check that the ratios of corresponding coefficients are unequal.

For two linear equations to have infinitely many solutions, the ratios of their corresponding coefficients and constants must be equal. Here, \(27/9=3\) and \(15/5=3\), so the second equation must be 3 times the first one. Hence, \(n=3\times47=141\), making option C correct. Exam tip: For infinitely many solutions, check \(a_1/a_2=b_1/b_2=c_1/c_2\).

When \(a_1/a_2=b_1/b_2\), the two lines have the same slope. Since \(c_1/c_2\) is different, they cannot be the same line; hence they are distinct parallel lines with no solution. Exam tip: equal first two ratios and an unequal third ratio indicate an inconsistent pair.

Every term in the second equation is three times the corresponding term in the first: 27=3×9, 42=3×14, and 213=3×71. Thus, both equations represent the same line, so every point on that line satisfies both equations. Therefore, the pair has infinitely many solutions. A no-solution or distinct-parallel-lines conclusion would require the coefficients of x and y to be proportional but the constants not to be proportional; here all three ratios are equal. Exam tip: If a₁/a₂ = b₁/b₂ = c₁/c₂, the pair has infinitely many solutions.