Conditions for solvability

For two linear equations, the ratios of corresponding coefficients determine how their lines are related. If \(a_1/a_2\) is different from \(b_1/b_2\), the lines have different slopes and therefore intersect at exactly one point. The pair is then consistent and independent, with one unique solution. Equal first two ratios alone would not be enough to conclude infinitely many solutions; the constant-term ratio would also need to agree.

Here, \(19/9\) is not equal to \(12/6\), because \(12/6=2\), whereas \(19/9\) is about \(2.11\). Thus the coefficient ratios are unequal, so the two lines meet once and the equations have a unique solution. Choice C correctly states this conclusion. Choice A is not valid because its displayed equality is false, and choice D also uses an equality that does not hold.

When one equation in a pair is a non-zero multiple of the other, both equations describe the same line. The pair is then called consistent and dependent. It has infinitely many solutions, because every point on that common line satisfies both equations. A quick test is to multiply the coefficients and the constant term of one equation by the same number and check whether the other equation results exactly.

Multiplying \(2x+3y=14\) by \(15\) gives \(30x+45y=210\), which is precisely the first equation. Thus the two equations are not different lines; they coincide completely. Therefore the pair is consistent and dependent, and choice A is correct. It would be incorrect to call it independent, since independent lines intersect at only one point, while these equations have infinitely many common solutions.

Here, a₁/a₂ = 26/2 = 13 and b₁/b₂ = 39/3 = 13, but c₁/c₂ = 117/10, which is not equal to 13. Thus, a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so the two lines are distinct and parallel. The pair has no solution and is therefore inconsistent. Option B would be correct only if all three ratios were equal. Exam tip: first compare a₁/a₂ and b₁/b₂; if they are equal but c₁/c₂ is different, the pair is inconsistent.

Here, \\(\frac{18}{9}=2\\), whereas \\(\frac{7}{4}\\) is different. Since the ratios of the coefficients of x and y are unequal, the two lines intersect at exactly one point, giving a unique solution. Therefore, the pair is consistent and independent. A consistent dependent pair would require the corresponding coefficient ratios to be equal. Exam tip: if \\(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\\), a pair of linear equations has a unique solution.

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

Here, \(a_1/a_2=9/18=1/2\) and \(b_1/b_2=4/8=1/2\), but \(c_1/c_2=380/775\neq1/2\). Thus, \(a_1/a_2=b_1/b_2\neq c_1/c_2\), which is the condition for an inconsistent pair of linear equations; the corresponding distinct parallel lines do not intersect, so there is no solution. For option B, all three ratios would have to be equal. Exam tip: Compare the ratios of the coefficients of both variables first, and then compare the ratio of the constant terms.

For the coefficients, \\(\frac{7}{4} \ne \frac{5}{-3}\\). Equivalently, the determinant of the coefficient matrix is \\(7(-3)-5(4)=-41\\), which is non-zero. Therefore, the two lines intersect at exactly one point, so the pair has one unique solution. In fact, \\(x=\frac{229}{41}, y=\frac{155}{41}\\). Exam tip: If \\(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\\), a pair of linear equations has a unique solution.

The three ratios are \(11/33\), \(18/54\), and \(86/258\). Each ratio equals \(1/3\), so \(11/33=18/54=86/258\). Hence, all three ratios are equal, and the two equations represent the same line; therefore, they are coincident equations. Exam tip: If \(a_1/a_2=b_1/b_2=c_1/c_2\), the pair of linear equations has infinitely many solutions.

Here, \(a_1=18, b_1=27, c_1=126\) and \(a_2=2, b_2=3, c_2=16\). Thus, \(a_1/a_2=18/2=9\) and \(b_1/b_2=27/3=9\), whereas \(c_1/c_2=126/16=63/8\), which is not 9. Therefore, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the pair of linear equations is inconsistent and has no solution. Exam tip: equal ratios of the coefficients but an unequal ratio of the constant terms indicate no solution.

For two linear equations, the condition a₁/a₂ ≠ b₁/b₂ indicates a unique solution. Here, 12/24 = 1/2 and 13/25 = 0.52, so the two ratios are unequal. Solving the equations gives y = 3 and x = 4/3. Therefore, B is correct; A and C incorrectly assume equal ratios, while D reverses the values of x and y. Exam tip: Compare a₁/a₂ and b₁/b₂ first to determine the number of solutions.

For a pair of consistent and dependent linear equations, the corresponding coefficients and constant terms must have the same ratio. Here, 27=3×9 and 48=3×16, so the second equation must be three times the first. Therefore, s=3×77=231, making option C correct. Exam tip: In dependent equations, every corresponding term of one equation is obtained by multiplying the other equation by the same non-zero constant.

The first two ratios are equal. For inconsistency, the constant ratio must be different so (r\ne98).

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, \(\frac{10}{20}=\frac{1}{2}\), so \(\frac{9}{a}=\frac{1}{2}\), which gives \(a=18\). Also, \(\frac{38}{91}\ne\frac{1}{2}\), so the two lines are parallel and inconsistent. Therefore, the correct answer is 18. Exam tip: For ‘no solution,’ first check that the ratios of the coefficients of both variables are equal but the ratio of the constant terms is different.

Two linear equations have infinitely many solutions when a₁/a₂ = b₁/b₂ = c₁/c₂. Here, 12/3 = 132/33 = 4. Therefore, k/10 = 4, giving k = 40. Hence, option C is correct. Exam tip: In such questions, first compare the ratios of the known coefficients and constants, then determine the unknown coefficient.

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, \(\frac{l}{20}=\frac{17}{34}=\frac{1}{2}\), so \(l=10\). Also, \(\frac{68}{139}\ne\frac{1}{2}\), confirming that the lines are parallel and inconsistent. Exam tip: First equate the ratios of the coefficients to find the parameter, and then verify that the ratio of constants is different.

The second equation is exactly twice the first: multiplying \(18x+13y=89\) by 2 gives \(36x+26y=178\). Hence, both equations represent the same line and the pair has infinitely many solutions. Option A is incorrect because distinct parallel lines have equal ratios of the coefficients of \(x\) and \(y\), but a different ratio for the constant terms. Exam tip: compare \(a_1/a_2\), \(b_1/b_2\), and \(c_1/c_2\); if all three ratios are equal, the lines are coincident.

For the two equations, a₁/a₂ = 24/4 = 6 and b₁/b₂ = (-18)/(-3) = 6, but c₁/c₂ = 102/18 = 17/3, which is not 6. Thus, a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so the lines are parallel and distinct and the pair has no solution. Infinitely many solutions would occur only when all three corresponding ratios are equal. Exam tip: compare the ratios of the coefficients first, then compare them with the ratio of the constants.

Here, \\(\frac{29}{14} \ne \frac{12}{6}\\). Therefore, the two lines intersect at exactly one point, so the pair of equations has a unique solution. In fact, multiplying the second equation by 2 and subtracting it from the first gives \(x=3\), followed by \(y=\frac{7}{6}\). Thus, the solution is unique but not an integer, so option D is incorrect. Exam tip: If \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\), a pair of linear equations always has a unique solution.

For a unique solution, coefficient ratios must not be equal. Here (k\neq2); check ratios directly in exams.

For no solution, coefficient ratios are equal but the constant ratio is different. At (k=4), this condition holds.