Conditions for solvability

For infinitely many solutions, all three ratios are equal. From (\frac{m-1}{3}=\frac{6}{9}=\frac{10}{15}), (m=3).

A pair of linear equations has a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, we need \(\frac{p+2}{6}\ne\frac{5}{10}\). Since \(\frac{5}{10}=\frac12\), this gives \(\frac{p+2}{6}\ne\frac12\), so \(p+2\ne3\), or \(p\ne1\). At \(p=1\), the ratios of the coefficients of \(x\) and \(y\) are equal, but the constant ratio \(\frac{11}{13}\) is different; hence there is no solution then. Exam tip: For a unique solution, first compare the ratios of the coefficients of \(x\) and \(y\).

For distinct parallel lines, coefficient ratios are equal and constant ratio is different. Solving gives (q=\frac{3}{2}).

In infinitely many solutions, (\frac{4}{10}=\frac{r-2}{15}=\frac{14}{35}). This gives (r=8).

For a pair of linear equations to have a unique solution, the determinant of the coefficient matrix must be non-zero. Here, \(D=ab-(7\times5)=ab-35\). Therefore, a unique solution exists when \(ab\neq 35\). If \(ab=35\), the determinant is zero, so the pair has either no solution or infinitely many solutions, not a unique solution. Exam tip: Check whether \(a_1b_2-a_2b_1\neq0\) for a unique solution.

For infinitely many solutions, all ratios must be equal. From (\frac{6}{s}=\frac{-8}{-12}=\frac{16}{24}), (s=9).

The coefficient ratio is (\frac{1}{4}). For no solution, (\frac{t}{20}\neq\frac{1}{4}), so (t\neq5).

For a unique solution, (\frac{11}{22}\neq\frac{2}{n}) is required. Hence (n\neq4) is the correct exam condition.

The second equation is obtained by multiplying every term of the first equation by 3: 3(2x − 5y = 1) gives 6x − 15y = 3. Therefore, both equations represent the same line. In ratio form, 6/2 = (−15)/(−5) = 3/1 = 3, confirming that all three ratios are equal. Coincident lines have infinitely many common points, so the pair has infinitely many solutions. Two linear equations cannot have exactly two isolated intersection points.

Coefficient ratios are equal but the constant ratio is different. Hence the lines are parallel and have no solution.

The governing concept is the consistency condition for a pair of linear equations in two variables. Write the equations in the form a₁x+b₁y+c₁=0 or compare the corresponding coefficients directly. For the x-coefficients, 4/12=1/3, and for the y-coefficients, (-7)/(-21)=1/3. Thus the two left-hand sides are proportional. However, the constants have ratio 5/17, which is not equal to 1/3. Therefore the two equations represent distinct parallel lines, so they do not intersect and have no common solution. Hence option C is correct. Infinitely many solutions would require the constant ratio to match as well; a unique solution would require unequal coefficient ratios.

Multiplying 2x + 3y = 5 by 2 gives 4x + 6y = 10, exactly the second equation. Thus the two equations are equivalent and represent coincident lines. The ratio test confirms this: 4/2 = 6/3 = 10/5 = 2. Every point on the common line satisfies both equations, so the pair has infinitely many solutions. The value x = 0 is not the only possibility; for example, x = 1 gives y = 1, which satisfies both equations.

The coefficient ratio is equal but the constant ratio is not equal. In exams, treat this as distinct parallel lines.

Option A is correct. When all three ratios are equal, both equations represent the same line, so every point on it is a solution. Option C gives distinct parallel lines. Exam tip: always compare the ratios in the order \(a,b,c\).

The governing concept is the no-solution condition for two linear equations: a₁/a₂=b₁/b₂ but c₁/c₂ must be different. Here the coefficient ratio for y is 4/8=1/2. To make the x-coefficient ratio equal to it, require (k+1)/6=1/2. Multiplying by 6 gives k+1=3, so k=2. For this value, the equations have proportional x- and y-coefficients, but their constants have ratio 9/15=3/5, which is not 1/2. Hence the corresponding lines are distinct and parallel, producing no solution. Therefore option B is correct. The other choices do not make the two variable-coefficient ratios equal, so they do not satisfy the required condition.

For two linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 to have infinitely many solutions, the corresponding coefficients and constants must be proportional: a1/a2 = b1/b2 = c1/c2. Here, 7/14 = 13/26 = 1/2. Therefore, (m − 4)/6 must also equal 1/2. Solving gives 2(m − 4) = 6, so m − 4 = 3 and m = 7. Thus option B is correct. The other values do not make all three ratios equal; they would represent either intersecting or parallel distinct lines, not the same line.

Two linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\) have a unique solution when \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\). Here, we need \(\frac{p-2}{5}\neq\frac{9}{15}=\frac{3}{5}\). The ratios become equal only when \(p-2=3\), i.e. \(p=5\). Therefore, the pair has a unique solution for \(p\neq 5\). At \(p=2\), the ratios are still unequal, so it does not prevent a unique solution. Exam tip: compare the ratios of the coefficients of \(x\) and \(y\) first.

For infinitely many solutions, both equations must represent the same line. Hence, \(\frac{3}{9}=\frac{a}{12}=\frac{6}{b}\). Since \(\frac{3}{9}=\frac{1}{3}\), we get \(a=4\) and \(b=18\). In \((4,12)\), \(\frac{6}{b}=\frac{1}{2}\), so it cannot give infinitely many solutions. Exam tip: for infinitely many solutions, equate all three coefficient ratios.

The coefficient ratio is (\frac{1}{3}). For no solution, (\frac{c}{21}\neq\frac{1}{3}), hence (c\neq7).

A pair of linear equations has a unique solution when \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\). Here, \(\frac{r}{8}\neq\frac{5}{10}=\frac12\), which gives \(r\neq4\). If \(r=4\), the two equations become proportional and have infinitely many solutions, not a unique solution. Exam tip: compare the ratios of the coefficients of \(x\) and \(y\) first.