Conditions for solvability

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.

The second equation is exactly three times the first: multiplying 6x+7y=41 by 3 gives 18x+21y=123. Thus both equations represent the same line, so the pair has infinitely many solutions. It does not represent two distinct parallel lines, which would give no solution; the lines are also not perpendicular because their slopes are equal. Exam tip: for equations in standard form, if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the pair has infinitely many solutions.

For the two equations in standard form, \(a_1/a_2=12/3=4\) and \(b_1/b_2=16/4=4\), whereas \(c_1/c_2=36/11\), which is not 4. Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the lines are distinct and parallel and the pair has no solution. Exam tip: when the first two coefficient ratios are equal but the constant ratio is different, the pair has no solution.

The coefficient-ratio test determines how two linear equations are related. If the ratios of the x and y coefficients are different, the lines have different slopes. Different slopes guarantee one intersection point, which means the pair is consistent and independent and has a unique solution. The constant terms do not change this conclusion once the first two ratios are unequal.

For these equations, \\(11/5=2.2\\), but \\(6/3=2\\). Hence \\(11/5\\ne6/3\\). The lines therefore have different slopes and intersect at exactly one point. They cannot be coincident, because coincident lines would have equal ratios for all corresponding coefficients. Thus option C is correct.

The coefficients of the first equation are 9 times those of the second: \(18/2=27/3=9\). However, the ratio of the constants is \(63/8\), which is not 9. Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the two lines are parallel but distinct and the pair is inconsistent; it has no solution. For coincident lines, all three ratios must be equal. Exam tip: always compare the constant-term ratio along with the coefficient ratios.

For a pair of linear equations, if \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\), the pair has a unique solution and is called consistent and independent. Here, \(\frac{13}{6} \ne \frac{4}{2}\), so the two lines intersect at exactly one point. Therefore, option C is correct. Option B would require infinitely many solutions, which occurs only when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). Exam tip: first compare \(a_1/a_2\) and \(b_1/b_2\).

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

Here, \(a_1/a_2=5/10=1/2\) and \(b_1/b_2=4/8=1/2\), but \(c_1/c_2=180/370=18/37\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the two lines are distinct and parallel and do not intersect. Therefore, the system is inconsistent and has no solution. Exam tip: Compare the ratios of the coefficients of \(x\) and \(y\) first, and then compare the ratio of the constants.

The determinant of the coefficients is \\(4(-3)-1(1)=-13\\ne0\\). Therefore, the two lines intersect at exactly one point, so the pair has one unique solution. Option B would apply only when the lines are parallel, with equal ratios of the corresponding coefficients. Exam tip: if \\(a_1b_2-a_2b_1\\ne0\\), a pair of linear equations has a unique solution.

Here, a₁/a₂ = 5/15 = 1/3, b₁/b₂ = 6/18 = 1/3, and c₁/c₂ = 32/96 = 1/3. Hence, all three ratios are equal, so the two equations represent coincident lines and have infinitely many solutions. In an exam, compare a₁/a₂, b₁/b₂, and c₁/c₂ in that order for equations written as ax + by + c = 0.

In option A, the second equation is exactly twice the first: multiplying \(2x+3y=6\) by 2 gives \(4x+6y=12\). Thus, both represent the same line and have infinitely many solutions. In B, the constant ratio differs, so the lines are parallel. Exam tip: compare all three ratios.

For the pair, \(a_1/a_2=7/14=1/2\) and \(b_1/b_2=8/15\), and these ratios are unequal. Thus, \(a_1/a_2\ne b_1/b_2\), so the two lines intersect at one point and the equations have a unique solution. Options A and C incorrectly treat the ratios as equal; coincident lines require all corresponding ratios to be equal. Exam tip: first compare \(a_1/a_2\) and \(b_1/b_2\).