Conditions for solvability

A pair of linear equations has a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{8}{2}=4\) and \(\frac{p}{3}\), so the condition is \(4\ne\frac{p}{3}\), which gives \(p\ne12\). Therefore, option B is correct. Notice that when \(p=12\), the coefficient ratios become equal, but the constant ratio \(\frac{24}{7}\) is different; hence the equations have no solution rather than a unique solution. Exam tip: compare the ratios of the coefficients first to test for a unique solution.

For infinitely many solutions, both linear equations must represent the same line. The coefficients on the left side of the second equation are twice those in the first equation: 10 = 2×5 and 4 = 2×2. Therefore, the constant on the right side must also be doubled, so n = 2×16 = 32. Hence, option D is correct. Exam tip: For infinitely many solutions, the ratios of the corresponding coefficients and constants must be equal.

Write the equations in comparable coefficient form: 6x − 2y = 18 and 3x − y = t. The coefficient ratios are 6/3 = 2 and (−2)/(−1) = 2. For the pair to be inconsistent, the constant ratio must differ from these equal coefficient ratios. The constant ratio is 18/t, so inconsistency requires 18/t ≠ 2. If t = 9, then 18/t = 2 and the second equation is exactly half of the first, giving infinitely many solutions rather than inconsistency. Therefore t must not equal 9, so option B is correct. The other numerical choices do not express the required general condition.

Every term of the second equation is three times the corresponding term of the first: (9,6,24)=3(3,2,8) . Hence both equations represent the same line, so they have infinitely many common points and infinitely many solutions. Therefore, option C is correct. For exams, check the ratios: if a_1/a_2=b_1/b_2=c_1/c_2 , the pair has infinitely many solutions.

Here, \(a_1/a_2=4/1=4\) and \(b_1/b_2=8/2=4\), but \(c_1/c_2=12/5\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), which is the condition for an inconsistent pair of linear equations. In fact, the first equation simplifies to \(x+2y=3\), while the second is \(x+2y=5\); therefore, the lines are parallel and there is no solution. Exam tip: In the ratio test, always compare the constant-term ratio as well.

For equations written as \\(a_1x+b_1y+c_1=0\\) and \\(a_2x+b_2y+c_2=0\\), unequal ratios \\(a_1/a_2\\) and \\(b_1/b_2\\) mean that the two lines have different slopes. Lines with different slopes meet at exactly one point, so the pair has a unique solution. Equality of the first two ratios is not required here.

In this pair, \\(7/2=3.5\\), while \\(3/1=3\\). Hence \\(7/2\\ne3/1\\). The lines therefore have different directions and intersect once. The constant ratio is not needed to establish uniqueness after the first two ratios are found unequal. Thus option C correctly states that there is one unique solution.

The governing concept is the classification of a pair of linear equations by comparing the ratios of corresponding coefficients and constants. Write the equations as 12x+18y-30=0 and 2x+3y-5=0. For the first equation, each coefficient is six times the corresponding coefficient in the second: 12=6(2), 18=6(3), and 30=6(5). Thus the first equation is exactly six times the second, not a different equation. Both equations represent the same straight line, so every point on that line satisfies both equations. The pair has infinitely many solutions and is therefore consistent and dependent. Hence option A is correct. An inconsistent pair has parallel distinct lines, while an independent consistent pair has intersecting lines and exactly one solution; neither situation applies here.

For the two equations, \(\frac{a_1}{a_2}=\frac{10}{2}=5\) and \(\frac{b_1}{b_2}=\frac{15}{3}=5\), but \(\frac{c_1}{c_2}=\frac{25}{8}\). Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the two lines are distinct and parallel. Hence, the pair is inconsistent and has no solution. Option B would be correct only if all three ratios were equal. Exam tip: Compare all three coefficient ratios in such questions.

For the given equations, \\(a_1/a_2=11/5\\) and \\(b_1/b_2=4/2=2\\). Since \\(11/5 \ne 2\\), the two lines intersect at one point and the pair has a unique solution. Therefore, it is a consistent and independent pair. For a consistent dependent pair, all three ratios are equal. Exam tip: Compare \\(a_1/a_2\\) and \\(b_1/b_2\\) first; if they are unequal, there is exactly one solution.

The second equation is (2) times the first. Therefore, both conditions give the same information and have infinitely many solutions.

When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), both equations represent the same line. Hence, every point on that line satisfies both equations, giving infinitely many solutions. In option B, the lines are distinct and parallel, so there is no solution. Exam tip: For infinitely many solutions, all three ratios must be equal.

For the first equation, \(a_1=1, b_1=1\), and for the second equation, \(a_2=2, b_2=-1\). Here, \(\frac{a_1}{a_2}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=-1\), so the two ratios are unequal. Therefore, the corresponding lines intersect at exactly one point, giving one unique solution. Option B would apply if the relevant ratios were equal and the lines were parallel. Exam tip: if \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), a pair of linear equations has a unique solution.

Here, a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 3/6 = 1/2, and the ratio of the constants is 13/26 = 1/2. Hence, all three ratios are equal. This means that both equations represent the same line, so the pair has infinitely many solutions. Exam tip: When a₁/a₂ = b₁/b₂ = c₁/c₂, the two lines are coincident and the pair has infinitely many solutions.

Compare the ratios of the corresponding coefficients and constants: \(\frac{6}{2}=3\), \(\frac{9}{3}=3\), but \(\frac{30}{12}=2.5\). Therefore, \(\frac{6}{2}=\frac{9}{3}\ne\frac{30}{12}\), so option C is correct. This condition means that the pair of linear equations has no solution: the variable-coefficient ratios are equal, but the constant-term ratio is different. Exam tip: when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the pair is inconsistent and has no solution.

For two linear equations, if \(a_1/a_2 \ne b_1/b_2\), the corresponding lines intersect at exactly one point, giving a unique solution. Here, \(7/14=1/2\), whereas \(5/11\ne1/2\); hence \(7/14\ne5/11\), so option B is correct. Options C and D use incorrect equalities among the ratios. Exam tip: first compare \(a_1/a_2\) and \(b_1/b_2\) to determine the number of solutions.

For a pair of consistent and dependent linear equations, one equation must be an exact multiple of the other. Here, the second equation is twice the first: \(2(x+4y=9)\), so its right-hand side must be \(2\times9=18\). Therefore, \(s=18\). Values such as 17 or 19 would not produce the same equation or the same line. Exam tip: For dependent equations, check whether \(a_1/a_2=b_1/b_2=c_1/c_2\).

The first two ratios are equal, so for inconsistency the constant ratio must be different. Hence (r \ne 28).

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, \(\frac{2}{4}=\frac{1}{2}\). Thus, \(\frac{3}{a}=\frac{1}{2}\), which gives \(a=6\). Also, \(\frac{5}{11}\ne\frac{1}{2}\), so the two lines are distinct and parallel and have no common solution. Exam tip: For ‘no solution,’ first equate the ratios of the variable coefficients and then verify that the constants’ ratio is different.

For two linear equations to have infinitely many solutions, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) must hold. Here, \(\frac{5}{1}=\frac{25}{5}=5\). Therefore, \(\frac{k}{3}=5\), giving \(k=15\). Hence, option B is correct. Exam tip: In the infinitely many solutions case, the corresponding coefficients and constants are proportional.

For a pair of linear equations to have no solution, the ratios of the coefficients of x and y must be equal, while the ratio of the constant terms must be different. Here, l/6 = 7/14 = 1/2, so l = 3. Also, 21/40 is not equal to 1/2, so the two lines are parallel and distinct. Exam tip: no solution occurs when a₁/a₂ = b₁/b₂ ≠ c₁/c₂.