Conditions for solvability

Here, \(a_1=10, b_1=-7, c_1=31\) and \(a_2=20, b_2=-14, c_2=65\). We get \(\frac{a_1}{a_2}=\frac{10}{20}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-7}{-14}=\frac{1}{2}\), but \(\frac{c_1}{c_2}=\frac{31}{65}\ne\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the pair is inconsistent and has no solution because the corresponding lines are parallel. Exam tip: Compare the ratios of the coefficients of both variables first, and then compare them with the ratio of the constants.

Here (11/22 \ne 4/9), so the lines intersect at one point. In this case, one unique solution is obtained.

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, \(\frac{c}{15}=\frac{14}{35}=\frac{2}{5}\), giving \(c=6\). Also, \(\frac{56}{121}\neq\frac{2}{5}\), so the two lines are distinct and parallel. Hence, option (6) is correct. Exam tip: For no solution, use \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\).

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, \(\frac{5}{20}=\frac{40}{160}=\frac{1}{4}\). Thus, \(\frac{d}{28}=\frac{1}{4}\), which gives \(d=7\). Therefore, option C is correct. Remember that the ratios of both variable coefficients and the constants must be equal; matching only one ratio is insufficient.

Coincident lines represent the same line, so every point on that line satisfies both equations and gives infinitely many solutions. Hence all three coefficient ratios must be equal. In option B, the lines are parallel but distinct. Exam tip: for infinitely many solutions, check equality of all three ratios.

For two linear equations to have infinitely many solutions, one equation must be an exact multiple of the other. Here, 24=3×8 and 15=3×5, so the right-hand side must also be three times 43. Hence, n=3×43=129, making option C correct. Exam tip: For infinitely many solutions, check that the ratios of all corresponding coefficients and constants are equal.

The equations have the form 12x − 6y = 42 and 2x − y = t. Their x-coefficient ratio is 12/2 = 6, and their y-coefficient ratio is (−6)/(−1) = 6. Thus the corresponding lines have equal slopes. To be inconsistent, they must be different parallel lines, so the constant ratio must not equal the same value: 42/t ≠ 6. Solving the equality 42/t = 6 gives 42 = 6t and therefore t = 7. This value would make the equations dependent, because multiplying 2x − y = 7 by 6 gives 12x − 6y = 42, producing infinitely many solutions. Every value t other than 7 makes the lines distinct and parallel, with no common solution. Hence option B, t ≠ 7, is correct.

Every term of the second equation is three times the corresponding term of the first: 21x=3(7x), 36y=3(12y), and 159=3(53). Hence, both equations represent the same line, so they have infinitely many common points. Option B is incorrect because no solution occurs when the lines are distinct and parallel. Also, the non-zero constants show that the common line does not pass through the origin. Exam tip: If a₁/a₂ = b₁/b₂ = c₁/c₂, the pair has infinitely many solutions.

The ratios of the corresponding coefficients are 15/3 = 5 and 20/4 = 5, whereas the ratio of the constant terms is 55/12, which is not 5. Thus, a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so the lines are distinct and parallel and the pair has no solution. Option A is incorrect because all three ratios are not equal. Exam tip: When a₁/a₂ = b₁/b₂ ≠ c₁/c₂, a pair of linear equations has no solution.

A pair of linear equations has one unique solution when the ratios of the coefficients of x and y are unequal. Geometrically, this means the corresponding lines have different slopes, so they cross at one point. If both variable ratios were equal, the constant ratio would then distinguish coincident lines from distinct parallel lines.

Here, \\(13/6\\) is approximately \\(2.167\\), while \\(8/4=2\\). Therefore \\(13/6\\ne8/4\\). The lines have different slopes and must intersect exactly once. The constant terms do not need further comparison for this classification. Hence the system has one unique solution, and option C is correct.

The ratios of the first two coefficients are equal: \(\frac{20}{5}=\frac{28}{7}=4\). However, the ratio of the constants, \(\frac{84}{23}\), is not equal to 4. Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the lines are parallel and have no common solution. Hence, the pair is inconsistent. Exam tip: Compare all three ratios to identify the consistency of a pair quickly.

For the two equations, \(\frac{a_1}{a_2}=\frac{14}{7}=2\) and \(\frac{b_1}{b_2}=\frac{5}{3}\). Since \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), the two lines intersect at exactly one point, so the pair has a unique solution. Therefore, it is consistent and independent. In a consistent dependent pair, all three corresponding ratios are equal. Exam tip: first compare the ratios of the coefficients of \(x\) and \(y\).

The second equation is (3) times the first. Therefore, both conditions give the same information and have infinitely many solutions.

Here, \(\frac{a_1}{a_2}=\frac{7}{14}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{5}{10}=\frac{1}{2}\), but \(\frac{c_1}{c_2}=\frac{260}{535}=\frac{52}{107}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), which is the condition for an inconsistent system. The two lines have the same slope but different intercepts, so they have no common solution. Exam tip: if all three ratios are equal, the system is dependent; here, the constant-term ratio is different.

For the given equations, \(a_1=5, b_1=2\) and \(a_2=2, b_2=-3\). Since \(\frac{a_1}{a_2}=\frac{5}{2}\) is not equal to \(\frac{b_1}{b_2}=\frac{2}{-3}\), the two lines intersect at exactly one point. Hence, the pair has one unique solution. No solution or infinitely many solutions can occur only when these two ratios are equal. Exam tip: compare \(\frac{a_1}{a_2}\) and \(\frac{b_1}{b_2}\) first.

The three ratios are 7/21=1/3, 10/30=1/3, and 46/138=1/3. Thus, a₁/a₂=b₁/b₂=c₁/c₂. Therefore, the two linear equations represent the same line and the pair has infinitely many solutions. Exam tip: when all three ratios are equal, the equations are coincident.

Coincident lines represent the same line, so all corresponding coefficients must be in the same ratio: \(a_1/a_2=b_1/b_2=c_1/c_2\). Hence, there are infinitely many solutions. Exam tip: if only the first two ratios are equal and the third differs, the pair has no solution.

Here, \(\frac{a_1}{a_2}=\frac{8}{16}=\frac{1}{2}\), whereas \(\frac{b_1}{b_2}=\frac{9}{17}\). Since these two ratios are unequal, the two lines intersect at exactly one point, so the pair has a unique solution. Option A describes the case of parallel distinct lines, which does not apply here. Exam tip: first compare \(\frac{a_1}{a_2}\) and \(\frac{b_1}{b_2}\); if they are unequal, the pair has a unique solution.

For a pair of consistent and dependent linear equations, the ratios of the corresponding coefficients and constant terms must be equal. Here, \(18/6=33/11=3\), so \(s/53=3\), giving \(s=159\). If \(s\) were 158 or 160, the coefficients would still be in the ratio 3, but the constant terms would not be; the equations would then be inconsistent rather than dependent. Exam tip: In a dependent pair, one equation is an exact multiple of the other.

The first two ratios are equal, so the constant ratio must be different for inconsistency. Hence, (r \ne 62).

Infinitely many solutions occur when both equations represent the same line, so every point on it is common. Exam tip: check \(a_1/a_2=b_1/b_2=c_1/c_2\). Distinct parallel lines have no common 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, \(\frac{9}{3}=\frac{81}{27}=3\), so \(\frac{k}{8}=3\), giving \(k=24\). Therefore, option C is correct. Exam tip: For infinitely many solutions, verify equality of all three ratios, not just two of them.

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, l/14 = 13/26 = 1/2, giving l = 7. Also, 52/109 is not equal to 1/2, so the two lines are parallel and inconsistent. Exam tip: compare the coefficient ratios first, then verify that the constant-term ratio is different.

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.