Conditions for solvability

For two linear equations, compare the ratios of the coefficients of x, y, and the constant terms. If the x and y coefficient ratios are equal but the constant ratio is different, the equations represent distinct parallel lines. Parallel lines do not meet at any point, so the pair has no solution.

Here, \\(8/2=4\\) and \\(12/3=4\\), so the ratios of the variable coefficients are equal. However, \\(24/7\\) is not equal to 4. Thus the constant terms do not have the same ratio as the variable coefficients. The equations cannot represent the same line; they represent parallel lines. Therefore option C, no solution, is correct.

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.

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

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.

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.

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.

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

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.

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.

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.

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.

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.

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.

Two linear equations \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\) have a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(a_1=q, b_1=6, a_2=5, b_2=3\), so the required condition is \(\frac{q}{5}\ne\frac{6}{3}\), or \(q\ne10\). Option A gives equal ratios, so it does not represent a unique solution. Exam tip: for a unique solution, the ratios of the coefficients of the two variables must be unequal.

For two linear equations to have infinitely many solutions, the ratios of corresponding coefficients and constants must be equal: a₁/a₂ = b₁/b₂ = c₁/c₂. Here, 2/4 = 9/18 = 1/2, so 3/k = 1/2, which gives k = 6. Therefore, option C is correct. Exam tip: For infinitely many solutions, always check a₁/a₂ = b₁/b₂ = c₁/c₂.

The first equation is 3x+2y=11. The second equation has coefficients 6 and 4, which are both twice the corresponding coefficients in the first equation. Therefore its left side is twice the first left side. If the equations represented the same line, its right side would also have to be twice 11, namely 22.

Thus t=22 makes the second equation 2(3x+2y)=22, exactly the same equation, giving infinitely many solutions. For every other listed value, the coefficient ratios remain equal but the constant ratio is different, so the lines are distinct and parallel and there is no solution. Therefore t must not be 22, and option C is correct.

When every term of one linear equation is obtained by multiplying every term of the other equation by the same nonzero number, both equations represent the same line. Such a pair is called consistent and dependent. Every point on that line satisfies both equations, so there are infinitely many common solutions, although the graph is visually one line.

Multiplying \\(x+2y=6\\) by 3 gives \\(3x+6y=18\\), which is exactly the second equation. Therefore the two equations are not different intersecting or parallel lines. Their graphs coincide completely. The graph will appear as one line, so option B is correct. The phrase “no solution” would be wrong because every point on the common line is a solution.

Multiplying the second equation by 2 gives 2x+4y=14, whereas the first equation is 2x+4y=10. Thus, the ratios of the coefficients of x and y are equal, but the ratio of the constant terms is different. Therefore, the two lines are distinct and parallel. Option A is incorrect because coincident lines require all three corresponding ratios to be equal. Exam tip: if a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel and distinct.

The ratios of the coefficients of x and y are not equal: 5/3 ≠ 8/4. Therefore, the lines are neither coincident nor parallel; they intersect at one unique point. Solving the equations gives the intersection point (3, 0). Exam tip: If a₁/a₂ ≠ b₁/b₂, the pair of linear equations has a unique solution, so the lines intersect at one point.

In the fourth option (1/2 \ne 3/5) so there is one unique solution. This identifies a consistent and independent pair.