Conditions for solvability

For two linear equations to have infinitely many solutions, the ratios of the corresponding coefficients and constants must be equal: \(\frac{p}{12}=\frac{3}{6}=\frac{15}{30}=\frac{1}{2}\). Thus, \(\frac{p}{12}=\frac{1}{2}\), giving \(p=6\). Therefore, option C is correct. Exam tip: For infinitely many solutions, check whether \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

For two linear equations to have a unique solution, the ratios of the corresponding coefficients must be unequal. Thus, \(\frac{3}{7}\ne\frac{q}{14}\). On cross-multiplying, this gives \(42\ne 7q\), or \(q\ne 6\). If \(q=6\), the ratios of the coefficients of \(x\) and \(y\) become equal, so the system does not have a unique solution. Exam tip: For a unique solution, remember \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\).

The first equation must be (5) times the second, so (k=20). A common multiplier makes both lines coincident.

For two linear equations to have no solution, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\) must hold. Here, \(\frac{8}{2}=4\) and \(\frac{40}{13}\ne4\). Therefore, \(\frac{a}{6}=4\), giving \(a=24\). Hence, option D is correct. Exam tip: equal ratios of the variable coefficients but an unequal ratio of the constants indicate no solution.

For the two equations, \\(a_1/a_2=10/2=5\\) and \\(b_1/b_2=15/3=5\\), but \\(c_1/c_2=35/9\\). Thus, \\(a_1/a_2=b_1/b_2\\ne c_1/c_2\\), so the two lines are distinct and parallel. They neither intersect nor coincide. Exam tip: the condition \\(a_1/a_2=b_1/b_2\\ne c_1/c_2\\) represents distinct parallel lines.

All the coefficients and constants in the first equation are 4 times those in the second: \(12=4\times3\), \(8=4\times2\), and \(44=4\times11\). Hence, both equations represent the same line and the pair has infinitely many solutions. Exam tip: if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the lines are coincident; for distinct parallel lines, only the first two ratios are equal and the third is different.

The ratios of the coefficients of x and y are 3/7 and 8/13, respectively, and they are unequal. Thus, a₁/a₂ ≠ b₁/b₂, so the two lines intersect at exactly one point and the pair has a unique solution. Option B is incorrect because distinct parallel lines require a₁/a₂ = b₁/b₂. Exam tip: Compare the ratios of the coefficients of x and y first to determine the relationship between the lines.

In slope-intercept form, a line is represented by y = mx + c, where m is the slope and c is the y-intercept. If two lines have the same slope and the same intercept, their equations describe exactly the same set of points. Thus the lines are coincident, not merely parallel. Every point on that common line satisfies both equations, so the pair has infinitely many solutions. Such a pair is called consistent because solutions exist, and dependent because one equation is a scalar multiple of the other. Therefore option C is correct. Options A and D describe no common solution, while option B would require intersecting lines and exactly one solution.

For a line y = mx + c, the value m fixes its direction and c fixes its vertical position. Equal slopes mean the two lines point in the same direction. Because their intercepts are different, they occupy different positions and cannot be the same line. They are therefore distinct parallel lines. Distinct parallel lines do not intersect, so there is no ordered pair (x, y) satisfying both equations at once. This is the condition of an inconsistent pair, and option B is correct. A unique solution would result from two lines with different slopes. Infinitely many solutions and coincident lines would require both slope and intercept to be equal, not just the slopes.

For the first equation, \(a_1=3, b_1=5, c_1=-20\), and for the second, \(a_2=9, b_2=15, c_2=-60\). Thus, \(\frac{a_1}{a_2}=\frac{3}{9}=\frac{1}{3}\), \(\frac{b_1}{b_2}=\frac{5}{15}=\frac{1}{3}\), and \(\frac{c_1}{c_2}=\frac{-20}{-60}=\frac{1}{3}\). Since all three ratios are equal, the two lines are coincident and the pair has infinitely many solutions. Exam tip: the condition \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) indicates infinitely many solutions.

For two linear equations, the condition \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\) represents distinct parallel lines, so the pair has no solution. Here, \(\frac{6}{12}=\frac{-7}{-14}=\frac{1}{2}\), whereas \(\frac{9}{25}\ne\frac{1}{2}\). Therefore, option A is correct. Exam tip: compare the ratios of the coefficients of \(x\) and \(y\) first, and then compare them with the constant-term ratio.

Here (7/14 \ne 4/9), 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 coefficients of x and y must be equal, while the ratio of the constant terms must be different. Thus, \(\frac{a}{10}=\frac{6}{15}\neq\frac{18}{47}\). Since \(\frac{6}{15}=\frac{2}{5}\), we get \(a=10\times\frac{2}{5}=4\). Option 3 is incorrect because it would make \(a/10=3/10\), which is not equal to \(6/15\). Exam tip: for inconsistent equations, check equal coefficient ratios but unequal constant-term ratios.

When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), one equation is a multiple of the other. Hence, both equations represent the same line and have infinitely many solutions. For distinct parallel lines, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so option C is not correct. Exam tip: If all three ratios are equal, identify coincident lines with infinitely many solutions.

Two linear equations have a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{9}{3}=3\) and the second ratio is \(\frac{p}{5}\). Thus, \(3\ne\frac{p}{5}\), which gives \(p\ne15\). Therefore, option B is correct. If \(p=15\), the coefficient ratios become equal, so the equations cannot have a unique solution. Exam tip: For a unique solution, compare the ratios of the coefficients of \(x\) and \(y\).

For two linear equations to have infinitely many solutions, one equation must be a constant multiple of the other. Here, the coefficients of x and y in the second equation are twice those in the first: 12=2×6 and 10=2×5. Therefore, the constant term must also be doubled, so n=2×31=62. Hence, option C is correct. Exam tip: For infinitely many solutions, check a₁/a₂=b₁/b₂=c₁/c₂.

The standard condition for an inconsistent pair is a₁/a₂ = b₁/b₂ but c₁/c₂ unequal to that common ratio. For 8x − 4y = 24 and 2x − y = t, the x-coefficient ratio is 8/2 = 4 and the y-coefficient ratio is (−4)/(−1) = 4. Thus the two lines are parallel or coincident. Coincidence occurs when 24/t = 4, which gives t = 6. In that case, the second equation is one-fourth of the first and there are infinitely many solutions. To obtain distinct parallel lines and therefore inconsistency, t must not be 6. Hence option B is correct; the other listed values do not state the required exclusion condition.

Every term in the second equation is three times the corresponding term in the first: 15x+9y=3(5x+3y) and 51=3×17. Thus, both equations represent the same line; they are dependent and consistent, so they have infinitely many solutions. No solution or distinct parallel lines would result if the coefficients had the same ratio but the constant terms had a different ratio. Exam tip: if \(a_1/a_2=b_1/b_2=c_1/c_2\), the pair has infinitely many solutions.

Compare the coefficients in standard form: \(a_1/a_2=6/1=6\), \(b_1/b_2=12/2=6\), but \(c_1/c_2=18/4=9/2\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the two lines are distinct and parallel, and the pair has no solution. Option A is incorrect because all three ratios are not equal. Exam tip: remember the no-solution condition as \(a_1/a_2=b_1/b_2\ne c_1/c_2\).

For a pair of linear equations, if the ratios of the coefficients of x and y are unequal, the two equations represent lines with different slopes. Such lines intersect at exactly one point. Consequently, the system is consistent and has one unique solution. Infinite solutions would require both variable-coefficient ratios and the constant ratio to agree, while no solution would require equal variable ratios but a different constant ratio.

Here, \\(9/4=2.25\\), whereas \\(4/2=2\\). Therefore \\(9/4\\ne4/2\\), so the lines are not parallel and cannot be the same line. They meet once, giving a unique ordered pair \\(x,y\\). Thus option C is the correct conclusion from the comparison.

For the two equations, \(a_1/a_2=12/2=6\) and \(b_1/b_2=18/3=6\), but \(c_1/c_2=42/8=21/4\). Thus, \(a_1/a_2=b_1/b_2\neq c_1/c_2\), so the two lines are distinct and parallel. Hence, the pair has no solution and is inconsistent. A coincident or dependent pair requires all three ratios to be equal. Exam tip: compare \(a_1/a_2\), \(b_1/b_2\), and \(c_1/c_2\) systematically.

When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the two lines have the same slope but are not the same line. Hence, they are distinct parallel lines and never intersect. Therefore, the pair has no solution. Infinitely many solutions occur only when all three ratios are equal, so option C is incorrect. Exam tip: if the first two ratios are equal but the third is different, choose ‘no solution’.

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

Here, \(\frac{4}{8}=\frac{3}{6}=\frac{1}{2}\), but \(\frac{120}{250}=\frac{12}{25}\), which is not equal to \(\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). The two lines are therefore parallel and have no common solution, so the system is inconsistent. Exam tip: Compare these three ratios to identify a ‘no solution’ case quickly.

In the two equations, \(a_1=3, b_1=1\) and \(a_2=1, b_2=-2\). Since \(\frac{a_1}{a_2}=\frac{3}{1}=3\) and \(\frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\) are unequal, the two lines intersect at exactly one point. Therefore, the pair has one unique solution. Option B would apply only when the lines are parallel, which requires \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\). Exam tip: retain the negative sign while comparing ratios of coefficients.