Conditions for solvability

For two lines to be coincident, the ratios of their corresponding coefficients and constants must be equal: \(\frac{3}{3}=\frac{a}{4}=\frac{24}{24}\). Thus, \(\frac{a}{4}=1\), giving \(a=4\). Therefore, option B is correct. Exam tip: coincident lines satisfy \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Coincident lines have equations that are nonzero multiples of each other. Multiplying 2x + 3y = 10 by 2 gives 4x + 6y = 20, which must match 4x + ay = 20. Therefore, a = 6, so option A is correct.

For infinitely many solutions, both equations must represent the same line. Therefore, the ratios of corresponding coefficients must be equal: \(\frac{p}{2}=\frac{-6}{-3}=\frac{18}{9}=2\). Hence \(\frac{p}{2}=2\), so \(p=4\). If \(p=2\), the ratio of the coefficients of \(x\) would not match the other ratios. Exam tip: for infinitely many solutions, check \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Dividing the first equation by 5 gives \(x+2y=4\). On putting \(k=1\), the second equation becomes \(x+2y=7\). The coefficients of \(x\) and \(y\) are the same, but the constants are different; hence the lines are parallel and have no solution. For \(k=2\) or \(k=5\), the coefficient ratios are not equal, so the lines intersect at one point. Exam tip: for no solution, check \(a_1/a_2=b_1/b_2\ne c_1/c_2\).

Two linear equations have infinitely many solutions when the ratios of their corresponding coefficients are equal. Multiplying \(3x-4y=12\) by 2 gives \(6x-8y=24\). Therefore, the first equation matches it only when \(p=6\). For instance, if \(p=4\), the ratio of the coefficients of \(x\) is not equal to the other ratios. Exam tip: for infinitely many solutions, check \(a_1/a_2=b_1/b_2=c_1/c_2\).

For infinitely many solutions, both linear equations must represent the same line. Hence the ratios of corresponding coefficients must be equal: \(\frac{a}{2}=\frac{6}{3}=\frac{14}{7}=2\). Thus, \(\frac{a}{2}=2\) gives \(a=4\). If \(a=2\), the ratio of the coefficients of \(x\) does not match the other ratios. Exam tip: for infinitely many solutions, check \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

For no solution, the two lines must be parallel but distinct. Thus, the ratios of the coefficients of x and y must be equal, while the ratio of the constants must be different. Here, \(\frac{12}{2}=6\), so \(\frac{6}{k}=6\) gives \(k=1\). The equations then become \(x+2y=5\) and \(x+2y=8\), which represent distinct parallel lines. For \(k=2\) or other listed values, the coefficient ratios are not equal. Exam tip: first equate the coefficient ratios, then check that the constant ratio differs.

A pair of linear equations has no solution when the equations represent distinct parallel lines. Simplify the first equation by dividing every term by 6: 12x + 18y = 54 becomes 2x + 3y = 9. The second equation is 2x + 3y = c, which has exactly the same left-hand side. If c = 9, the equations are identical and therefore have infinitely many solutions, not no solution. If c differs from 9, the same expression 2x + 3y would have to equal two different constants, 9 and c, which is impossible. Geometrically, the lines are parallel but distinct. Among the choices, only c = 10 is different from 9 and therefore produces no solution. Option C is correct.

For two linear equations to have infinitely many solutions, the ratios of corresponding coefficients must be equal: a₁/a₂ = b₁/b₂ = c₁/c₂. Here, 3/9 = 12/36 = 1/3. Therefore, k/15 = 1/3, giving k = 5. Hence, option C is correct. Exam tip: Find the common ratio using the coefficients of x and the constants, then use it to determine the coefficient of y.

Two linear equations have no solution when \,\(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). Here, \,\(\frac{a}{6}=\frac{4}{8}=\frac{1}{2}\), so \,\(a=3\). Also, \,\(\frac{10}{25}=\frac{2}{5}\ne\frac{1}{2}\), so the two lines are distinct and parallel, giving no solution. Exam tip: first equate the ratios of the x- and y-coefficients, then check the constant-term ratio.

The first two ratios are equal, so the constant ratio must be different for inconsistency. Hence, (m \ne 34).

The coefficients of the first equation are proportional to the corresponding coefficients of the second equation: \\(\frac{7}{14}=\frac{4}{8}=\frac{1}{2}\\). However, the ratio of the constants is \\(\frac{19}{41}\\), which is not equal to \\(\frac{1}{2}\\). Thus, \\(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\\), so the two lines are distinct parallel lines and do not intersect. Hence, there is no solution. Exam tip: For infinitely many solutions, all three ratios must be equal.

Here (4/9 \ne 7/15), so the lines intersect at one point. If the first two ratios are different, one unique solution is obtained.

If all three ratios are equal, both equations form the same line. Therefore, the pair is consistent and dependent.

If the first two ratios are equal and the third differs, the lines are parallel and distinct. Therefore, there is no common solution.

For infinitely many solutions, (p/10=6/15=18/45) must hold. Since the common ratio is (2/5), (p=4) would be required.

For two linear equations to have infinitely many solutions, a_1/a_2=b_1/b_2=c_1/c_2 must hold. Here, a_1/a_2=7/14=1/2 and c_1/c_2=29/58=1/2 . Therefore, q/10=1/2 , giving q=5 . For instance, if q=4, the coefficient ratio becomes 4/10=2/5, so the equations are not dependent. Exam tip: For infinitely many solutions, check that all three ratios are equal.

Two linear equations have a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(a_1=6, a_2=3, b_1=k, b_2=5\), so \(\frac{6}{3}\ne\frac{k}{5}\), or \(2\ne\frac{k}{5}\). This gives \(k\ne10\). If \(k=10\), the ratios of the coefficients of \(x\) and \(y\) would be equal, so the equations would 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 corresponding coefficients and constants must be equal: \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). Here, \(\frac{5}{15}=\frac{35}{105}=\frac{1}{3}\), so \(\frac{a}{18}=\frac{1}{3}\), giving \(a=6\). Therefore, option C is correct. Exam tip: infinitely many solutions require all three corresponding ratios to be equal, not just one pair of ratios.

A pair of linear equations has no solution when the ratios of the coefficients of \(x\) and \(y\) are equal, but the ratio of the constants is different. Thus, \(\frac{b}{12}=\frac{8}{24}=\frac{1}{3}\), which gives \(b=4\). Also, \(\frac{32}{71}\ne\frac{1}{3}\), so the two lines are distinct and parallel, giving no solution. Exam tip: First equate the coefficient ratios, then check that the constant ratio is different.

For the two equations, the ratios of the coefficients of x and y are equal: 14/2 = 21/3 = 7. However, the ratio of the constant terms is 49/8, which is not equal to 7. Thus, the two lines have the same slope but different positions, so they are distinct parallel lines. Coincident lines would require all three ratios to be equal. Exam tip: If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair represents distinct parallel lines.

Every coefficient and the constant in the first equation is 5 times the corresponding term in the second: \(15x+10y=40=5(3x+2y=8)\). Hence, both equations have the same solution set, so their graphs coincide as the same line. Option B is incorrect because distinct parallel lines require the coefficients to be proportional but the constants not to be proportional. Exam tip: when one complete linear equation is a non-zero multiple of the other, the lines are coincident and the pair has infinitely many solutions.

For the two lines, \(\frac{a_1}{a_2}=\frac{2}{5}\) and \(\frac{b_1}{b_2}=\frac{11}{19}\). Since \(\frac{2}{5}\ne\frac{11}{19}\), the lines are neither parallel nor coincident; hence, they intersect at exactly one point and the pair has a unique solution. Perpendicular lines would require the product of their slopes to be \(-1\), which is not the required condition here. Exam tip: Compare \(\frac{a_1}{a_2}\) and \(\frac{b_1}{b_2}\) first to identify the number of solutions.

For the first equation, \(a_1=4, b_1=9, c_1=-31\), and for the second, \(a_2=12, b_2=27, c_2=-93\). Thus, \(\frac{a_1}{a_2}=\frac{4}{12}=\frac{1}{3}\), \(\frac{b_1}{b_2}=\frac{9}{27}=\frac{1}{3}\), and \(\frac{c_1}{c_2}=\frac{-31}{-93}=\frac{1}{3}\). Since all three ratios are equal, the two lines are coincident and the pair has infinitely many solutions. Exam tip: To identify infinitely many solutions, compare all three ratios; equality of only the first two is not sufficient.

For the first equation, \(a_1=8, b_1=-3, c_1=22\), and for the second equation, \(a_2=16, b_2=-6, c_2=47\). Thus, \(\frac{a_1}{a_2}=\frac{8}{16}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\), whereas \(\frac{c_1}{c_2}=\frac{22}{47}\), which is not equal to \(\frac{1}{2}\). Therefore, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the lines are parallel and the pair has no solution. Exam tip: when the first two coefficient ratios are equal but the constant-term ratio is different, the pair is inconsistent.