Conditions for solvability

For the two equations, \(\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-4}{-8}=\frac{1}{2}\), but \(\frac{c_1}{c_2}=\frac{17}{39}\neq\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\), so the lines are parallel and distinct and the pair has no solution. Exam tip: this ratio condition identifies inconsistent equations with zero solutions.

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

If all three ratios are equal both equations represent the same line. Such a pair is consistent and dependent.

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

For two linear equations to have infinitely many solutions, the ratios of the corresponding coefficients and constant terms must be equal: p/10 = 6/15 = 18/45. Since both 6/15 and 18/45 equal 2/5, p/10 = 2/5 gives p = 4. Therefore, option C is correct. Exam tip: For infinitely many solutions, check a₁/a₂ = b₁/b₂ = c₁/c₂. Option 3 is incorrect because 3/10 is not equal to 2/5.

For infinitely many solutions, the two linear equations must represent the same line, so one equation must be a multiple of the other. Here, \(14x+10y=58\) must be twice the first equation. Thus, \(2q=10\), giving \(q=5\). With a close option such as \(q=4\), the coefficients of \(y\) would not be in the same ratio, so infinitely many solutions would not occur. Exam tip: for infinitely many solutions, check \(a_1/a_2=b_1/b_2=c_1/c_2\).

Two linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\) have a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{6}{3}=2\) and the other coefficient ratio is \(\frac{k}{5}\). Thus, for a unique solution, \(2\ne\frac{k}{5}\), which gives \(k\ne10\). If \(k=10\), the coefficient ratios are equal, but \(\frac{24}{17}\ne2\), so the equations have no solution. Exam tip: compare the ratios of the coefficients of \(x\) and \(y\) first.

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, \(\frac{5}{15}=\frac{35}{105}=\frac{1}{3}\), so \(\frac{a}{18}=\frac{1}{3}\), giving \(a=6\). Therefore, option C is correct. Exam tip: infinitely many solutions require the ratios of both variable coefficients and the constant terms to be equal; equality of only one ratio is insufficient.

Two linear equations have no solution when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). Here, \(\frac{b}{12}=\frac{8}{24}=\frac{1}{3}\), which gives \(b=4\). Also, \(\frac{32}{71}\ne\frac{1}{3}\), so the two lines are parallel and have no common solution. Exam tip: for no solution, the ratios of the coefficients must be equal, but the ratio of the constants must be 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 lines have the same slope but different positions, so they are distinct parallel lines. Exam tip: when a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair represents distinct parallel lines.

Dividing every term of the first equation by 5 gives the second equation: 15x+10y=40 ⇒ 3x+2y=8. Thus, both equations represent the same line and have infinitely many common solutions. Exam tip: when the corresponding coefficients and constants of two linear equations are in the same ratio, their graphs coincide as one line.

For the two equations, \,\(a_1/a_2=2/5\) and \,\(b_1/b_2=11/19\). Since \,\(2/5 \ne 11/19\), the two lines have different slopes and intersect at exactly one point. Hence, the pair has a unique solution. Distinct parallel lines require \,\(a_1/a_2=b_1/b_2\), which is not true here. Exam tip: first compare \,\(a_1/a_2\) and \,\(b_1/b_2\).

For the first equation, a₁=4, b₁=9, c₁=-31, and for the second equation, a₂=12, b₂=27, c₂=-93. Therefore, \(\frac{a_1}{a_2}=\frac{4}{12}=\frac{1}{3}\), \(\frac{b_1}{b_2}=\frac{9}{27}=\frac{1}{3}\), 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. Hence, option C is correct; options A and D incorrectly make one ratio unequal, while B incorrectly states that the first two ratios are unequal. Exam tip: When comparing the three ratios, check \(c_1/c_2\) with its signs included.

Here, \(\frac{a_1}{a_2}=\frac{8}{16}=\frac{1}{2}\) 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 \(\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the two lines are parallel and inconsistent; consequently, the pair has 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 (9/18 \ne 5/11) 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 corresponding coefficients must be equal, while the ratio of the constants must be different. Thus, \(\frac{a}{14}=\frac{9}{21}=\frac{3}{7}\), giving \(a=6\). Also, \(\frac{27}{64}\ne\frac{3}{7}\), so the equations are inconsistent and have no solution. Exam tip: first equate the ratios of the coefficients to find the parameter, then verify that the constants are not in the same ratio.

For two linear equations to have infinitely many solutions, the dependent-pair condition \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) must hold. Here, \(\frac{3}{12}=\frac{24}{96}=\frac{1}{4}\), so \(\frac{b}{20}=\frac{1}{4}\), giving \(b=5\). 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 a unique solution, the ratios of the coefficients of the two variables must be unequal: \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{a_1}{a_2}=\frac{11}{4}\) and \(\frac{b_1}{b_2}=\frac{p}{2}\). Thus, \(\frac{11}{4}\ne\frac{p}{2}\), which gives \(p\ne\frac{11}{2}\). Therefore, option B is correct. In option A, the two ratios become equal, so the system cannot have a unique solution. Exam tip: For a unique solution, check that the ratios of the corresponding variable coefficients are unequal.

For two linear equations to have infinitely many solutions, the ratios of the corresponding coefficients and constant terms must be equal. Here, the coefficients of x and y in the second equation are three times those in the first: 21=3×7 and 12=3×4. Therefore, the constant term must also be three times 37, so n=3×37=111. Hence, option C is correct. Exam tip: Infinitely many solutions mean that both equations represent the same line.

For a pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the equations are inconsistent when a₁/a₂ = b₁/b₂ but c₁/c₂ is different. Here, 10x − 5y = 30 can be compared with 2x − y = t. The coefficient ratios are 10/2 = 5 and (−5)/(−1) = 5, so the two lines have the same slope. They will be distinct parallel lines, and hence have no solution, only when 30/t ≠ 5, which is equivalent to t ≠ 6. If t = 6, the second equation multiplied by 5 becomes the first, giving infinitely many solutions rather than inconsistency. Therefore option B is correct; the other numerical values do not express the required condition.