Conditions for solvability

For a pair of linear equations to be consistent and dependent, the corresponding coefficients and constant terms must be in the same ratio. Here, the coefficients in the second equation are three times those in the first: 21=3×7 and 39=3×13. Therefore, its constant term must also be three times 67, so s=3×67=201. Hence, option C is correct. Exam tip: First identify the common multiplier between the corresponding x- and y-coefficients, then apply it to the constant term.

The first two ratios are equal. For inconsistency, the constant ratio must be different so (r \ne 74).

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, \(\frac{8}{16}=\frac{1}{2}\). Therefore, \(\frac{7}{a}=\frac{1}{2}\), which gives \(a=14\). Also, \(\frac{34}{79}\ne\frac{1}{2}\), so the two equations represent distinct parallel lines and have no solution. Exam tip: first equate the ratios of the coefficients of the variables, then verify that the constants have a different ratio.

For two linear equations to have infinitely many solutions, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) must hold. Here, \(\frac{10}{2}=\frac{110}{22}=5\), so \(\frac{k}{9}=5\), which gives \(k=45\). Therefore, option C is correct. Exam tip: compare the constant-term ratio first, then equate the corresponding coefficient ratio to it.

For two linear equations to have no solution, the ratios of the coefficients of the variables must be equal, while the ratio of the constant terms must be different. Thus, \(\frac{l}{18}=\frac{15}{45}=\frac{1}{3}\), which gives \(l=6\). Also, \(\frac{60}{181}\neq\frac{1}{3}\), since \(180\neq181\); therefore, the two lines are distinct and parallel. If 5 or 7 were used, the coefficient ratios would not be equal, resulting in a unique solution. Exam tip: For no solution, remember \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\).

The second equation is exactly twice the first: multiplying 16x+11y=73 by 2 gives 32x+22y=146. Hence, both equations represent the same line and the pair has infinitely many common solutions. Equivalently, \(16/32=11/22=73/146=1/2\). Therefore, option B is correct; equal ratios indicate coincident lines, not distinct parallel lines. Exam tip: if \(a_1/a_2=b_1/b_2=c_1/c_2\), the two lines coincide and the pair has infinitely many solutions.

For the two equations, a₁/a₂ = 20/4 = 5 and b₁/b₂ = (-15)/(-3) = 5, but c₁/c₂ = 85/18, which is not 5. Thus, the two lines have equal slopes but different intercepts, so they are distinct parallel lines and never intersect. Therefore, the pair has no solution. Infinitely many solutions require all three ratios to be equal, while a unique solution requires the first two ratios to be unequal. Exam tip: If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair is inconsistent and has no solution.

For the two equations, the determinant of the coefficients is \(a_1b_2-a_2b_1=23\times5-11\times10=5\). Since it is non-zero, the two lines intersect at exactly one point, so the pair has one unique solution. Option B would apply when the two lines coincide, while option A would apply to distinct parallel lines. Exam tip: if \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), a pair of linear equations has a unique solution.

The first two ratios are equal. For inconsistency, the constant ratio must be different so (t \ne 93).

For two linear equations to have infinitely many solutions, the ratios of the corresponding coefficients and constants must be equal: 5/20 = a/28 = 45/180. Since 5/20 and 45/180 are both 1/4, we get a/28 = 1/4, so a = 7. Therefore, option C is correct. Exam tip: For infinitely many solutions, all three corresponding ratios must be equal; equality of only two ratios is not sufficient.

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, \(\frac{b}{16}=\frac{9}{24}=\frac{3}{8}\), giving \(b=6\). Also, \(\frac{36}{97}\ne\frac{3}{8}\), so the two lines are distinct and parallel. Exam tip: For ‘no solution’, first equate the ratios of the coefficients of the variables, then verify that the constants’ ratio is different.

Two linear equations \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\) have a unique solution when \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\). Here, we require \(\frac{13}{6} \ne \frac{p}{5}\). Thus, \(65 \ne 6p\), or \(p \ne \frac{65}{6}\). Therefore, option A is correct. In option B, the two ratios become equal, so it does not give a unique solution. Exam tip: For a unique solution, compare the ratios of the coefficients of the two variables and ensure they are unequal.

Here, \(\frac{a_1}{a_2}=\frac{14}{21}=\frac{2}{3}\) and \(\frac{b_1}{b_2}=\frac{-8}{-12}=\frac{2}{3}\), but \(\frac{c_1}{c_2}=\frac{25}{40}=\frac{5}{8}\). Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the two lines are parallel and distinct, so the pair has no solution. In an exam, apply this ratio condition directly.

Compare the ratios of the corresponding x- and y-coefficients in the equations 6x + 17y = 71 and 11x + 31y = 130. The ratios are 6/11 and 17/31. They are unequal, since 6 × 31 = 186 whereas 17 × 11 = 187. Therefore the two equations represent lines with different slopes. Two lines with different slopes intersect at one and only one point, so the pair is consistent and independent and has a unique solution. It is not necessary to compare the constant ratio after finding the first two ratios unequal. If the first two ratios were equal but the constant ratio differed, the pair would be inconsistent; if all three were equal, it would have infinitely many solutions. Hence option C is correct.

Here, \(\frac{a_1}{a_2}=\frac{9}{27}=\frac{1}{3}\) and \(\frac{b_1}{b_2}=\frac{4}{12}=\frac{1}{3}\), but \(\frac{c_1}{c_2}=\frac{360}{1090}\neq\frac{1}{3}\). Thus, the two lines are distinct and parallel, so they have no solution and the system is inconsistent. Option B would be correct only if all three ratios were equal, giving infinitely many solutions. Exam tip: When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\), the system is inconsistent.

Multiply the first equation, 18x + 5y = 79, by 2. It becomes 36x + 10y = 158, exactly the second equation. Thus both equations have the same graph and represent one coincident line. The phrase “the lines are the same” is therefore the most precise statement; such a pair has infinitely many solutions. Distinct parallel lines would have proportional x and y coefficients but a different constant ratio, while intersecting lines would not have proportional coefficients.

For two linear equations to have infinitely many solutions, the ratios of the corresponding coefficients and constants must be equal: 4/12 = a/21 = 32/96. Since 4/12 and 32/96 are both 1/3, a/21 = 1/3, giving a = 7. Therefore, option C is correct. Exam tip: all three ratios must be equal; equality of only two ratios is not sufficient for infinitely many solutions.

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, \(\frac{p}{20}=\frac{9}{30}=\frac{3}{10}\), so \(p=6\). Also, \(\frac{45}{103}\ne\frac{3}{10}\), confirming that the lines are distinct and parallel. Exam tip: first equate the ratios of the coefficients of x and y, then verify that the ratio of constants is different.

Equal ratios of \(a\) and \(b\) give the two lines the same slope. Since the \(c\)-ratios differ, the lines cannot coincide; they are distinct parallel lines with no solution. Exam tip: always compare all three ratios.

Multiply the first equation 7x − 4y = 29 by 3. This gives 21x − 12y = 87, which is precisely the second equation. Consequently, the equations are dependent and their graphs coincide. The ratio check gives 21/7 = (−12)/(−4) = 87/29 = 3, so the condition for infinitely many solutions is satisfied. Every point on the common line is a solution; therefore option C is correct, not the options for a unique or no solution.