Conditions for solvability

Correct answer: A. Inconsistent

Here (4/8=3/6) but (10/21) is different so there is no solution. Equal first two ratios and a different third ratio mean parallel lines.

Correct answer: D. One unique solution

The direct answer is option D: one unique solution. Compare the coefficients of the equations \(x+5y=12\) and \(3x+14y=31\). We have \(a_1/a_2=1/3\) and \(b_1/b_2=5/14\). These are unequal, so the lines have different slopes and intersect at exactly one point. Option A, no solution, would require distinct parallel lines, represented by equal coefficient ratios but a different constant ratio. Option B, infinitely many solutions, and option C, coincident lines, describe the same situation: both equations represent one line, so all three ratios would be equal. They are not equal here. For an additional check, multiply the first equation by 3: \(3x+15y=36\). Comparing it with \(3x+14y=31\) shows the equations are not identical; subtracting gives \(y=5\), then \(x=-13\), confirming one solution. Thus D is correct.

Correct answer: B. Intersecting

For a pair of linear equations, if \(a_1/a_2 \ne b_1/b_2\), the two lines have different slopes. Hence, they intersect at exactly one point, and the pair has a unique solution. Parallel lines require \(a_1/a_2 = b_1/b_2 \ne c_1/c_2\). Exam tip: compare the coefficient ratios first to determine the position of the lines and the number of solutions.

Correct answer: C. No solution

For a pair of linear equations, if \(a_1/a_2=b_1/b_2\ne c_1/c_2\), the two lines have the same slope but different y-intercepts. Hence, they are distinct parallel lines and never intersect, so the pair has no solution. Exam tip: remember that the pattern \(a_1/a_2=b_1/b_2\ne c_1/c_2\) represents distinct parallel lines and an inconsistent pair.

Correct answer: C. Same line

The direct answer is option C: the same line, or coincident lines. Write the equations in the standard form \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\). If \(a_1/a_2=b_1/b_2=c_1/c_2\), every corresponding coefficient is in the same proportion. One equation is therefore just a nonzero multiple of the other. Multiplying an equation does not change its set of solutions, so both equations represent exactly the same line. Every point on that line satisfies both equations; hence there are infinitely many common solutions. Option A, two intersecting lines, occurs when the first two coefficient ratios are unequal. Option B, distinct parallel lines, occurs when the first two ratios are equal but the constant ratio is different. Option D, no line, is not the result of a normal pair of linear equations with valid coefficients; each equation represents a line. Thus C is correct. Memory cue: all three ratios equal means same line and infinitely many solutions.

Correct answer: C. Consistent and dependent

Compare the two equations directly. Multiplying every term of 5x-y=9 by 2 gives 10x-2y=18, which is exactly the second equation. Thus both equations represent the same straight line rather than two different lines. Every point on this common line satisfies both equations, so there are infinitely many common solutions. Such a pair is called consistent and dependent: it is consistent because solutions exist, and dependent because one equation is a scalar multiple of the other. Option A would require one intersection, option B would indicate parallel distinct lines, and option D is not the relevant classification.

Correct answer: C. No solution

Dividing the first equation by 2 gives 3x+y=4, whereas the second equation is 3x+y=5. The ratios of the coefficients of x and y are equal, \(6/3=2/1\), but the ratio of the constant terms is different, \(8/5\ne 2\). Thus, the two lines are parallel and distinct, so the pair has no solution. Infinitely many solutions would require both equations to represent the same line. Exam tip: if \(a_1/a_2=b_1/b_2\ne c_1/c_2\), the pair has no solution.

Correct answer: A. One unique solution

The ratios of the coefficients of x and y are 2/5 and 7/9, respectively. Since 2/5 ≠ 7/9, the two lines intersect at exactly one point, giving one unique solution. Therefore, option A is correct. Exam tip: For a pair of linear equations, if a₁/a₂ ≠ b₁/b₂, the pair has a unique solution; hence it has neither no solution nor infinitely many solutions.

Correct answer: D. Inconsistent

Distinct parallel lines never intersect, so they have no common point and no common solution. Therefore, the pair is called inconsistent. In contrast, coincident lines have infinitely many solutions. Exam tip: intersecting lines indicate one solution, coincident lines indicate infinitely many solutions, and distinct parallel lines indicate no solution.

Correct answer: B. One unique solution

In the graphical method, a solution of a pair of linear equations is represented by a common point of the two graphs. If the two lines intersect at exactly one point, only the coordinates of that point satisfy both equations simultaneously. Therefore the pair has exactly one solution, called a unique solution. In standard terminology, the pair is consistent and independent. No solution occurs when distinct lines are parallel, while infinitely many solutions occur when the two equations represent the same coincident line. “Inconsistent” is another name for the no-solution case, so it cannot describe a pair that intersects once. Hence option B is correct.

Correct answer: D. Infinitely many solutions

The direct answer is D: infinitely many solutions. A pair of linear equations represents two lines. If both equations give exactly the same line, every point on that line satisfies the first equation and also the second equation. Since a line contains infinitely many points, the pair has infinitely many ordered-pair solutions. Option A, one solution, is wrong because one solution occurs when two different lines meet at exactly one point. Option B, no solution, is wrong because parallel distinct lines never meet. Option C, three solutions, is wrong because two linear equations cannot have exactly three intersection points. Option D is correct because coincident lines share all their points. Memory cue: same line means infinitely many; parallel different lines mean none; intersecting lines mean one.

Correct answer: C. There are infinitely many solutions

Multiply the first equation, 7x+4y=20, by 2. The result is 14x+8y=40, exactly the second equation. Hence both equations have the same set of solutions and their graphs are coincident, not merely parallel. Every point on this common line satisfies both equations, so the pair has infinitely many solutions. In the ratio test, a₁/a₂=7/14=1/2, b₁/b₂=4/8=1/2, and c₁/c₂=20/40=1/2; equality of all three ratios confirms a consistent and dependent pair. Therefore option C is correct; a unique solution would require unequal coefficient ratios.

Correct answer: C. Both equations have no solution.

Rewrite \(8x+12y=24\) by dividing by \(4\): \(2x+3y=6\). But the second equation is \(2x+3y=7\). The same left side cannot be equal to two different numbers, so the system has no solution.

Correct answer: A. One unique solution

For the two linear equations, \(\frac{a_1}{a_2}=\frac{9}{4}\) and \(\frac{b_1}{b_2}=\frac{5}{2}\). Since \(\frac{9}{4}\ne\frac{5}{2}\), the two lines intersect at exactly one point, so the pair has a unique solution. Therefore, option A is correct. Exam tip: when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), a pair of linear equations always has a unique solution.

Correct answer: C. When (a_1/a_2 \ne b_1/b_2)

A consistent and independent pair has one unique solution. For this the ratios of (a) and (b) must be different.

Correct answer: A. When (a₁/a₂=b₁/b₂=c₁/c₂)

For the pair a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0, the condition a₁/a₂=b₁/b₂=c₁/c₂ means that every coefficient, including the constant term, is in the same proportion. Consequently, one equation is a non-zero scalar multiple of the other, so both equations represent the same line. The two lines have infinitely many common points, making the pair consistent; because the equations describe the same line and one depends on the other, it is dependent. If only the first two ratios are equal but the constant ratio differs, the lines are distinct parallel and inconsistent. Unequal first two ratios give one intersection. Thus A is correct.

Correct answer: C. No solution

Here, \(\frac{1}{2}=\frac{-3}{-6}\), but \(\frac{4}{9}\) is different. Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the two lines are parallel and distinct and have no common point. Therefore, the pair has no solution. Exam tip: When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), a pair of linear equations has no solution.

Correct answer: C. Coincident lines and infinitely many solutions

The second equation is 4 times the first: \\(12x-8y=24=4(3x-2y=6)\\). Hence both equations represent the same line, so the pair has infinitely many solutions. Using the coefficient test, \\(a_1/a_2=b_1/b_2=c_1/c_2=1/4\\). Therefore, the lines are coincident, not distinct parallel lines. Exam tip: when all three corresponding ratios are equal, the pair has infinitely many solutions.

Correct answer: A. One unique solution

Here (11/5 \ne 6/3) so the lines will intersect. In exams write unique solution when the first two ratios are different.

Correct answer: B. 3

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, \(\frac{2}{6}=\frac{5}{15}=\frac{1}{3}\). Therefore, \(\frac{a}{9}=\frac{1}{3}\), giving \(a=3\). Hence, option B is correct. Exam tip: infinitely many solutions require all three coefficient-to-constant ratios to be equal; equality of only two ratios is not sufficient.

Correct answer: C. 6

For two linear equations to have no solution, the condition is a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Here, k/3 = 4/2 = 2, so k = 6. Also, 8/9 is not equal to 2; therefore, the two lines are distinct and parallel. Hence, option C is correct. Exam tip: For ‘no solution’, first equate the ratios of the coefficients of the variables and then check that the ratio of the constants is different.

Correct answer: B. 3

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, \(\frac{5}{10}=\frac{20}{40}=\frac{1}{2}\), so \(\frac{b}{6}=\frac{1}{2}\), giving \(b=3\). If \(b=4\), then \(\frac{b}{6}\neq\frac{1}{2}\), so the equations would not have infinitely many solutions. Exam tip: For infinitely many solutions, verify that all three corresponding ratios are equal.

Correct answer: B. \(2\)

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, \(\frac{p}{4}=\frac{5}{10}=\frac{1}{2}\), which gives \(p=2\). Also, \(\frac{10}{23}\ne\frac{1}{2}\), so the two lines are parallel and distinct. Exam tip: first equate the ratios of the coefficients of x and y, then verify that the ratio of the constant terms is different.

Correct answer: C. 14

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, \(\frac{7}{1}=\frac{28}{4}=7\), so \(\frac{m}{2}=7\), giving \(m=14\). Therefore, option C is correct. Exam tip: infinitely many solutions occur when the coefficients and constants of both equations are proportional.

Correct answer: A. \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

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: equality of all three ratios means infinitely many solutions; if only the first two ratios are equal and the third differs, there is no solution.