Conditions for solvability

Different ratios indicate different slopes, so the lines intersect. Intersecting lines give one solution.

Here (1/3=(-2)/(-6)=6/18), so the lines are coincident. Coincident lines have infinitely many common points.

Here, \(a_1/a_2=4/8=1/2\), \(b_1/b_2=1/2\), but \(c_1/c_2=(-3)/(-5)=3/5\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the two lines are distinct and parallel. Therefore, the pair has no solution and is inconsistent. Option C is incorrect because infinitely many solutions require all three ratios to be equal. Exam tip: if \(a_1/a_2=b_1/b_2\ne c_1/c_2\), the pair is inconsistent.

Here (6/12 \ne 7/15), so there is one unique solution. Different coefficient ratios indicate intersecting lines.

An inconsistent pair has no common solution for the two linear equations. Graphically, the two lines are distinct and parallel. One solution occurs when the lines intersect at one point, while infinitely many solutions occur when the lines coincide. A pair of linear equations cannot have exactly two solutions. Exam tip: remember that inconsistent always means “no solution.”

A pair of linear equations is called consistent when the two equations have at least one common solution. Geometrically, their graphs either meet at one point or lie on the same line. Thus, consistency does not require exactly one solution; a pair may also have infinitely many solutions.

For two lines, intersecting lines give one common point, while coincident lines give infinitely many common points. Both cases are consistent. Parallel distinct lines have no common point and are inconsistent. Therefore, option A is the complete and correct condition. The ratio statement in option D is not a general definition and cannot identify every consistent pair.

Write the equations as a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0. Here, a₁/a₂ = 9/3 = 3, b₁/b₂ = (−3)/(−1) = 3, and c₁/c₂ = 6/2 = 3. Since all three ratios are equal, both equations represent the same, or coincident, line. Every point on that line satisfies both equations, so the pair has infinitely many solutions. Options A and D describe different conditions.

Here (2/4=1/2) but (5/12) is different, so the lines are parallel. Distinct parallel lines have no solution.

The second equation is (2) times the first, so both are the same line. The same line gives infinitely many solutions.

For the first equation, a₁=1 and b₁=3, while for the second equation, a₂=2 and b₂=5. Here, \(a₁/a₂=1/2\) and \(b₁/b₂=3/5\), so the ratios are unequal. Therefore, the two lines intersect at one point and the pair has a unique solution. In fact, the solution is \(x=-8, y=5\). Exam tip: If \(a₁/a₂ \ne b₁/b₂\), a pair of linear equations has a unique solution.

Coincident lines are two equations that represent exactly the same straight line. Every point on that line satisfies both equations, so the pair has infinitely many common solutions. Because at least one common solution exists, the pair is consistent. Because the equations do not determine one unique point, it is also called dependent.

This gives the classification “consistent and dependent,” which is option B. An inconsistent pair would represent distinct parallel lines and have no common solution. A consistent independent pair would represent intersecting lines and have exactly one solution. Therefore, neither A nor C describes coincident lines, and option B follows directly.

When two lines intersect at exactly one point, they have one common solution. Therefore, the pair is called consistent and independent. An inconsistent pair has no solution, whereas dependent or coincident lines have infinitely many solutions. Exam tip: one point of intersection means one unique solution and hence a consistent independent pair.

For a unique solution, (k/3 \ne 2/1), so (k \ne 6). In exams, make sure the (a) and (b) ratios are not equal.

A pair of linear equations has infinitely many solutions when both equations represent exactly the same line. In standard form, this happens when the ratios of corresponding coefficients and constants are equal: 
m a_1/a_2=b_1/b_2=c_1/c_2
m. This means one equation is a constant multiple of the other, so every point on one line also lies on the other. The equations here are already arranged in comparable standard form.

For 
m 2x+ky=4
m and 
m 6x+9y=12
m, the ratio from the x-coefficients is 
m 2/6=1/3
m, and the constant ratio is 
m 4/12=1/3
m. Therefore the y-coefficient must satisfy 
m k/9=1/3
m. Multiplying by 9 gives 
m k=3
m. Indeed, multiplying the first equation by 3 produces the second, so option C is correct.

Write the equations in standard form and compare coefficient ratios. Here a₁/a₂ = 4/2 = 2 and b₁/b₂ = 6/3 = 2. For two linear equations to have no solution, the ratios of the x- and y-coefficients must be equal, but the ratio of constant terms must be different: a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Thus we need 10/k ≠ 2. Equality would occur when 10/k = 2, which gives k = 5. At k = 5, the second equation multiplied by 2 becomes the first, so the lines coincide and there are infinitely many solutions. Therefore k must not be 5, making option A correct.

The second equation is exactly twice the first: multiplying \\(x+2y=3\\) by 2 gives \\(2x+4y=6\\). Hence, both equations represent the same line and have infinitely many common solutions. Exam tip: check \\(a_1/a_2=b_1/b_2=c_1/c_2\\); when all three ratios are equal, the lines are coincident. Therefore, the parallel-lines option is not correct here because the two equations do not represent distinct lines.

(2/4=3/6) but (6/15) is different, so the lines are parallel. Such a graph has no intersection.

Compare the ratios of the corresponding coefficients. Here, 5/10 = 1/2, whereas 2/3 is different. Thus, a₁/a₂ ≠ b₁/b₂, so the two lines intersect at exactly one point and the pair has a unique solution. Parallel lines require the first two coefficient ratios to be equal, while coincident lines require all three ratios to be equal. Exam tip: If a₁/a₂ ≠ b₁/b₂, the lines intersect at one point.

For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, a unique solution occurs when the two corresponding lines intersect at exactly one point. Algebraically, this happens when a₁/a₂ ≠ b₁/b₂. The unequal ratios mean the lines have different slopes and cannot be parallel or coincident, so they meet once. Option A describes coincident lines and gives infinitely many solutions. Option B describes parallel distinct lines and gives no solution. Option D is incomplete and is not a valid general condition. Hence option C is the only correct condition.

No solution occurs when the first two ratios are equal and the third is different. This is the condition for parallel lines.