Conditions for solvability

Correct answer: B. \(k\ne 10\)

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.

Correct answer: C. 6

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.

Correct answer: B. 4

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.

Correct answer: C. Lines are distinct parallel

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.

Correct answer: A. Same line

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.

Correct answer: C. Lines intersecting at one point

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 \,\(b_1/b_2\).

Correct answer: C. \(\frac{4}{12}=\frac{9}{27}=\frac{-31}{-93}\)

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.

Correct answer: A. \(\frac{8}{16}=\frac{-3}{-6}\ne\frac{22}{47}\)

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.

Correct answer: C. One unique solution

Here (9/18 \ne 5/11) so the lines intersect at one point. In this case one unique solution is obtained.

Correct answer: C. (6)

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.

Correct answer: C. 5

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.

Correct answer: B. \(p\ne\frac{11}{2}\)

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.

Correct answer: C. 111

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.

Correct answer: B. t ≠ 6

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.

Correct answer: C. There are infinitely many solutions

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.

Correct answer: B. The first two ratios are equal, but the ratio of the constants is different

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.

Correct answer: C. (11/5 \ne 6/3) so one unique 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.

Correct answer: C. Parallel but distinct lines; inconsistent

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.

Correct answer: C. Consistent and independent

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 \(b_1/b_2\).

Correct answer: C. Infinitely many solutions

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

Correct answer: C. Inconsistent

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.

Correct answer: A. One unique solution

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.

Correct answer: A. All three ratios are equal

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.

Correct answer: A. \(2x+3y=6\) और \(4x+6y=12\)

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.

Correct answer: B. 7/14 ≠ 8/15, so there is a unique solution

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 \(b_1/b_2\).