Conditions for solvability

The first two ratios are equal, so for inconsistency the constant ratio must differ. Hence (m \ne 14).

The second equation is (2) times the first, so both lines are coincident. Coincident lines have infinitely many solutions.

The ratios of the corresponding coefficients are 7/14 = (-3)/(-6) = 1/2. However, the ratio of the constant terms is 10/25 = 2/5, which is not equal to 1/2. Thus, the two lines are distinct and parallel, so they do not intersect at any point. Therefore, the pair has no solution. Exam tip: If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair of linear equations has no solution.

Here (3/6 \ne 4/7), so the lines intersect at one point. Different coefficient ratios give one unique solution.

When all three ratios are equal, both equations represent the same line. Such a pair has infinitely many solutions.

If the first two ratios are equal and the third differs, the lines are distinct parallel. Therefore, there is no common solution.

For two linear equations to have infinitely many solutions, the ratios of their corresponding coefficients and constant terms must be equal. Here, p/9 = 2/6 = 6/18 = 1/3. Therefore, p/9 = 1/3, giving p = 3, so option C is correct. For example, if p = 2, the ratio of the x-coefficients would be 2/9, not 1/3; hence the equations would not have infinitely many solutions. Exam tip: For infinitely many solutions, always check the equality of all three corresponding ratios.

For two linear equations to have a unique solution, the ratios of the corresponding coefficients must not be equal. Here, the determinant is \(2\times10-5\times q=20-5q\). A unique solution requires \(20-5q\ne0\), which gives \(q\ne4\). When \(q=4\), the two lines have the same slope but different constants, so they are parallel and have no solution. Exam tip: for a unique solution, check that the coefficient determinant is non-zero.

The first equation must be (4) times the second, so (k=12). A common multiplier makes the lines coincident.

Two linear equations have no solution when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). Here, \(\frac{6}{2}=3\) must equal \(\frac{a}{5}\), so \(\frac{a}{5}=3\) and hence \(a=15\). Also, \(\frac{30}{11}\ne3\), confirming that the two lines are distinct and parallel. Therefore, 15 is correct. Exam tip: equate the ratios of the coefficients first, then verify that the ratio of constants is different.

For the two equations, \(a_1/a_2=8/2=4\) and \(b_1/b_2=12/3=4\), whereas \(c_1/c_2=20/6=10/3\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the lines are distinct and parallel; they neither intersect nor coincide. Exam tip: compare the three coefficient ratios—if the first two are equal but the third is different, the lines are distinct parallel lines.

Dividing every term of \(9x+6y=15\) by 3 gives \(3x+2y=5\). Hence, both equations represent the same line and have infinitely many common solutions. Option B is incorrect because distinct parallel lines require the coefficients of \(x\) and \(y\) to be proportional but the constants not to be proportional. Exam tip: if \(a_1/a_2=b_1/b_2=c_1/c_2\), the two lines are coincident, or the same line.

The slope of a non-vertical line describes its steepness, and the y-intercept tells where it crosses the y-axis. If two lines have the same slope, they are parallel or identical. If their y-intercepts are also the same, they pass through the same point on the y-axis and have the same direction. Consequently, they are not merely parallel; they coincide completely.

Coincident lines have every point in common. Therefore the pair has infinitely many solutions and is called consistent and dependent. Option C is correct. An inconsistent pair has no common point, as with distinct parallel lines. A consistent independent pair has one common point, which does not apply here.

A line written as y = mx + c has slope m and y-intercept c. If two lines have the same m, they have the same direction and are parallel unless their intercepts are also equal. Different y-intercepts place the lines at different vertical positions, so they are distinct parallel lines. Distinct parallel lines never meet at a common point; therefore the associated pair of linear equations has no solution. A unique solution would require intersecting lines, while infinitely many solutions would require coincident lines with both the same slope and the same intercept. Hence option B is correct.

When two lines have different slopes, they cannot be parallel or coincident. Hence, they intersect at exactly one point, giving the pair of linear equations one unique solution. Therefore, option C is correct. Option A represents parallel lines, while option B represents coincident lines. Exam tip: different slopes always indicate a unique solution.

For the first equation, a₁=2, b₁=3 and c₁=-11; for the second, a₂=6, b₂=9 and c₂=-33. Thus, \(\frac{a_1}{a_2}=\frac{2}{6}=\frac{1}{3}\), \(\frac{b_1}{b_2}=\frac{3}{9}=\frac{1}{3}\), and \(\frac{c_1}{c_2}=\frac{-11}{-33}=\frac{1}{3}\). Therefore, option C gives the correct relation. The second equation is three times the first, so the two lines are coincident and the pair has infinitely many solutions. Exam tip: Equal values of all three ratios indicate infinitely many solutions.

For equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the relation a₁/a₂ = b₁/b₂ ≠ c₁/c₂ indicates distinct parallel lines and hence no solution. In this question, 4/8 = 1/2 and (−5)/(−10) = 1/2, while 7/9 is not 1/2. Thus the first two ratios are equal but the constant ratio is different. The lines have the same slope but different positions, so they do not intersect. Therefore option A is correct; option C would indicate coincident lines, and the other options state an incorrect ratio pattern.

Here (5/10 \ne 6/9), so the lines intersect at one point. In this case, the third ratio need not be checked.

For two linear equations to have no solution, the ratios of the coefficients of both variables must be equal, while the ratio of the constant terms must be different. Thus, \(\frac{a}{6}=\frac{4}{8}=\frac{1}{2}\), which gives \(a=3\). Also, \(\frac{10}{25}=\frac{2}{5}\), which is not equal to \(\frac{1}{2}\); therefore, the two lines are parallel and inconsistent. Exam tip: For ‘no solution,’ first check that the variable-coefficient ratios are equal and then verify that the constant-term ratio is 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, \(\frac{3}{9}=\frac{15}{45}=\frac{1}{3}\). Therefore, \(\frac{b}{12}=\frac{1}{3}\), which gives \(b=4\). Hence, option C is correct. Exam tip: For infinitely many solutions, check that all three corresponding coefficient and constant ratios are equal.