Conditions for solvability

In option A, \(a_1/a_2=1/2\) and \(b_1/b_2=4/8=1/2\), but \(c_1/c_2=6/13\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the two lines are distinct and parallel and the pair has no solution. Hence, it is inconsistent. In option C, all three ratios are equal, so the pair is dependent and consistent. Exam tip: For an inconsistent pair, check whether \(a_1/a_2=b_1/b_2\ne c_1/c_2\).

In the second option the second equation is (3) times the first. Therefore both are the same line and the pair is dependent.

All terms are multiplied by (2) so the equations give the same line. In such questions identifying the multiplier is the fastest method.

In standard form, the coefficients are \(a_1=3,b_1=6,c_1=12\) and \(a_2=1,b_2=2,c_2=5\). Thus, \(\frac{a_1}{a_2}=3\) and \(\frac{b_1}{b_2}=3\), but \(\frac{c_1}{c_2}=\frac{12}{5}\). Therefore, the first two ratios are equal while the third is different. The lines are distinct and parallel, so the pair has no solution. Exam tip: If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the pair is inconsistent.

For the equations 4x+y=9 and 7x+2y=13, compare the ratios of the coefficients of x and y. We obtain a₁/a₂=4/7 and b₁/b₂=1/2. Since 4/7 is not equal to 1/2, the two coefficient ratios are unequal. This is the condition for two non-parallel lines, so their graphs intersect at exactly one point and the pair has one unique solution. The constant-term ratio 9/13 need not be compared to establish this conclusion once the first two ratios differ. Option A contains a false equality, while C and D would describe coincident lines. Therefore B is correct.

Two lines with different slopes intersect at exactly one point. Hence, the pair of linear equations has one unique solution and is called consistent and independent. Lines with the same slope but different intercepts are inconsistent, while equations representing the same line have infinitely many solutions. Exam tip: different slopes always imply a unique solution.

A line written as y=mx+c is determined by its slope m and y-intercept c. If two lines have the same value of m and the same value of c, their equations are identical: y=m₁x+c₁ and y=m₁x+c₁. Therefore they are not two separate parallel lines; they lie exactly on top of each other and are called coincident lines. Every point on this common line satisfies both equations, so the corresponding pair has infinitely many solutions and is consistent and dependent. Distinct parallel lines have the same slope but different intercepts. Intersecting lines have different slopes, and perpendicular lines have slopes whose product is -1 when defined. Hence option C is correct.

For the two equations, 13/26 = 5/10 = 1/2, but 22/45 is not equal to 1/2. Thus, a₁/a₂ = b₁/b₂, while c₁/c₂ is different. Therefore, the two lines are parallel and distinct, so they have no common solution and the pair is inconsistent. Exam tip: When a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair is inconsistent.

For the two equations, \(a_1=4, b_1=-7, a_2=8, b_2=-13\). Here, \(\frac{a_1}{a_2}=\frac{4}{8}=\frac{1}{2}\), whereas \(\frac{b_1}{b_2}=\frac{-7}{-13}=\frac{7}{13}\); the two ratios are unequal. Therefore, the lines intersect at exactly one point, so the system has one unique solution. Exam tip: If \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), a pair of linear equations has a unique solution.

Every term of the first equation is 5 times the corresponding term of the second: \(15=5\times3\), \(10=5\times2\), and \(5=5\times1\). Thus, both equations represent the same line and have infinitely many common solutions. Therefore, option C is correct. Option A would be correct only if the two lines were distinct and parallel. Exam tip: If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), a pair of linear equations has infinitely many solutions.

Dividing the first equation by 8 gives 2x-y=3, whereas the second equation is 2x-y=4. The ratios of the coefficients of x and y are equal, 16/2 = (-8)/(-1), but the ratio of the constant terms is different. Thus, the two lines are parallel and distinct, so they have no common solution. Exam tip: If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair of linear equations is inconsistent and has no solution.

Here, \(a_1=6, b_1=11\) and \(a_2=3, b_2=5\). We have \(a_1/a_2=6/3=2\), whereas \(b_1/b_2=11/5\); the two ratios are unequal. Therefore, the pair is consistent and independent and has a unique solution, so the lines intersect at one point. For parallel or coincident lines, the relevant coefficient ratios must be equal, while perpendicular lines require the product of their slopes to be \(-1\). Exam tip: whenever \(a_1/a_2 \ne b_1/b_2\), conclude that the pair has a unique solution.

For two linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the condition a₁/a₂ = b₁/b₂ = c₁/c₂ means that all coefficients, including the constants, are in the same proportion. One equation is therefore a scalar multiple of the other, so both represent the same line. Since every point on that line satisfies both equations, the pair has infinitely many solutions. Unequal constant ratios would instead indicate distinct parallel lines.

For two equations a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0, if a₁/a₂=b₁/b₂ but this common ratio is not equal to c₁/c₂, the coefficients of x and y are proportional while the constants are not. Consequently, one equation cannot be a complete scalar multiple of the other. The graphs therefore have the same slope but different positions, so they are distinct parallel lines. Parallel distinct lines do not share any point, which means the pair has no solution and is called inconsistent. Coincident lines require all three ratios to be equal, whereas intersecting lines require the first two ratios to be unequal. Thus option C is correct.

Here, \(a_1/a_2=17/9\) and \(b_1/b_2=4/2=2\), so \(a_1/a_2\ne b_1/b_2\). Therefore, the two lines intersect at exactly one point, giving one unique solution. Option A would require all three ratios to be equal, while option B would require the coefficient ratios to be equal but the constant-term ratio to be different. Exam tip: first compare \(a_1/a_2\) and \(b_1/b_2\).

The first equation is (6) times the second so both lines are coincident. Coincident lines have infinitely many solutions.

Here (5/1=10/2) but (25/6) is different so the lines are parallel and distinct. Such lines have no common solution.

The first equation must be (2) times the second, so (k=9). For infinitely many solutions, keep all three ratios equal.

For a pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, infinitely many solutions occur when a₁/a₂ = b₁/b₂ = c₁/c₂. Here the corresponding ratios are 3/6, a/8, and 12/24. Since 3/6 = 12/24 = 1/2, we must have a/8 = 1/2. Multiplying by 8 gives a = 4. Therefore, option B is correct. Option A, C, and D do not make all three ratios equal, so they would not represent coincident lines and cannot produce infinitely many common solutions.

For two linear equations to have no solution, the ratios of the coefficients of the variables must be equal, but this common ratio must differ from the ratio of the constants. Here, \(k/6=5/10=1/2\), which gives \(k=3\). Also, \(20/45=4/9\), not \(1/2\), so the two lines are distinct and parallel. Therefore, option B is correct. Exam tip: For ‘no solution,’ check \(a_1/a_2=b_1/b_2\ne c_1/c_2\).