Conditions for solvability

When all three ratios are equal both equations give the same line. This is called a consistent and dependent pair.

Equal first two ratios give the same slope and a different third ratio keeps the lines separate. Such lines have no solution.

When the ratios of (a) and (b) are different the lines intersect. Intersecting lines give exactly one common point.

The second equation is (2) times the first so both lines are the same. All points on the same line are solutions.

Here (4/8=1/2) but (6/15) is different so the lines are parallel. Distinct parallel lines give no solution.

The first equation is (3) times the second so both are the same line. In such questions spotting the common multiplier is the fastest method.

Comparing the coefficients, \(a_1/a_2=2/4=1/2\) and \(b_1/b_2=(-3)/(-6)=1/2\), but \(c_1/c_2=5/7\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), which is the condition for distinct parallel lines. Therefore, option C is correct. Option B is incorrect because coincident lines require all three ratios to be equal. Exam tip: If \(a_1/a_2=b_1/b_2\ne c_1/c_2\), the lines are distinct and parallel.

Here (7/5 \ne 2/4) so the lines meet at one point. One intersection means one unique solution.

Lines that never meet are parallel so there is no common point. No common point means no solution.

The governing concept is the graphical classification of a pair of linear equations. When two lines completely overlap, they are called coincident lines. Every point on one line also lies on the other, so every such point satisfies both equations simultaneously. A line contains infinitely many points; consequently, the pair has infinitely many ordered-pair solutions. Thus option C is correct. Distinct parallel lines never meet and give no solution, while intersecting lines share exactly one point and give one solution. Option D is not a general description of a coincident pair because the common points need not lie on y = 0; they can lie anywhere on the common line.

When two lines intersect at exactly one point, the pair has one unique solution. Therefore, it is a consistent and independent pair. An inconsistent pair is represented by distinct parallel lines, whereas a consistent and dependent pair has coincident lines and infinitely many solutions. Exam tip: One intersection point = one unique solution = consistent and independent.

The second equation is (3) times the first so both lines are coincident. Coincident lines give infinitely many solutions.

Here (9/3=6/2) but (21/8) is different so the lines are parallel. Distinct parallel lines have no solution.

The governing concept is the ratio test for a pair of linear equations, a x + b y + c = 0 and a₁x + b₁y + c₁ = 0. Here a/a₁ = 3/6 = 1/2, but b/b₁ = 5/11, and these ratios are unequal. Therefore the two lines have different slopes and intersect at exactly one point. A pair with one intersection has one unique solution, so option A is correct. Infinitely many solutions would require all three corresponding ratios to be equal, while no solution would require a/a₁ = b/b₁ but not c/c₁. Since the coefficient ratios already differ, neither of those cases applies.

Having no common solution identifies an inconsistent pair. In a graph it appears as distinct parallel lines.

Infinitely many common solutions occur only when both equations represent the same line. In exams this can be written as coincident lines.

Exactly one common solution means that the graphs of the two linear equations share only one point. Therefore, the lines intersect at exactly one point. Coincident lines have infinitely many common points, whereas distinct parallel lines have no common point. Exam tip: one solution means intersecting lines, no solution means distinct parallel lines, and infinitely many solutions mean coincident lines.

The governing condition for infinitely many solutions is a/a₁ = b/b₁ = c/c₁ for the two equations in standard form. Comparing 2x + ay = 7 with 4x + 10y = 14 gives 2/4 = a/10 = 7/14. The first and third ratios both equal 1/2. Hence a/10 must also equal 1/2, so a = 5. With this value, the second equation is exactly twice the first: 4x + 10y = 14, so both equations represent the same line. Therefore option C is correct. The other values do not make all three ratios equal and would not produce infinitely many common solutions.

A pair of linear equations has no solution when the ratios of the coefficients of x and y are equal, but the ratio of the constant terms is different. Thus, k/8 = 3/6 gives k = 4. Also, 6/15 is not equal to 3/6, so the two lines are distinct and parallel. Therefore, 4 is correct. Exam tip: For no solution, check a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

For a pair of linear equations to have infinitely many solutions, a₁/a₂ = b₁/b₂ = c₁/c₂ must hold. Here, 5/10 = 9/18 = 1/2, so 2/b = 1/2, which gives b = 4. Therefore, option C is correct. Exam tip: For infinitely many solutions, all three corresponding ratios must be equal, not just two of them.