Conditions for solvability

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).

The direct answer is A: n is not equal to 4. For two linear equations, a unique solution requires the lines not to be parallel or coincident. Using the coefficient test, we need \(11/22\ne2/n\). Since \(11/22=1/2\), equality would mean \(1/2=2/n\), so n=4. Therefore every value except n=4 gives a unique solution. If n=4, the first two coefficient ratios are equal, while the constant ratio is \(3/8\), not equal to 1/2; the lines are parallel and there is no solution. Option A states exactly the required condition. Options B, C and D give particular values n=4, 2 and 8; B is the exceptional no-solution value, while C and D do give unique solutions but are not the complete condition. Exam cue: state the full restriction, not just one example.

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

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 \(a,b,c\).

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.

Equating coefficient ratios, (\frac{u}{4}=\frac{9}{6}) gives (u=6). The constant ratio is different, so there is no solution.

A pair of linear equations has a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{13}{26}=\frac12\) and \(\frac{-t}{-10}=\frac{t}{10}\). Therefore, a unique solution requires \(\frac12\ne\frac{t}{10}\), which gives \(t\ne5\). At \(t=5\), the coefficients on the left-hand sides are proportional, so the solution cannot be unique. Exam tip: include the negative signs while comparing the coefficients of \(y\).

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, \(\frac{-3}{-5}=\frac{6}{10}=\frac{3}{5}\). Therefore, \(\frac{v+2}{10}=\frac{3}{5}\), giving \(v+2=6\) and hence \(v=4\). Option \(v=3\) is a close distractor, but it makes the first ratio \(\frac{1}{2}\), not \(\frac{3}{5}\). Exam tip: for infinitely many solutions, compare all three coefficient ratios.

The coefficient ratio is (\frac{1}{7}). For no solution, (\frac{d}{28}\neq\frac{1}{7}), so (d\neq4).

The determinant of the coefficient matrix is \(D=a(3b)-(2a)b=ab\). A pair of linear equations has a unique solution only when \(D\neq 0\). Hence, the required condition is \(ab\neq 0\), meaning that neither \(a\) nor \(b\) can be zero. Conditions such as \(a+b=0\) or \(a=b\) do not by themselves guarantee a unique solution. Exam tip: for a unique solution, check whether \(a_1b_2-a_2b_1\neq0\).

Equating coefficient ratios, (\frac{l}{9}=\frac{6}{18}) gives (l=3). The constant ratio is different.

The direct answer is option C: h=7. For infinitely many solutions, the ratios of corresponding coefficients and constants must be equal: 6/15=(h+1)/20=12/30. Compute 6/15=2/5 and 12/30=2/5, so we need (h+1)/20=2/5. Multiply by 20: h+1=8, hence h=7. With this value, the second equation is 5/2 times the first: 6×5/2=15, 8×5/2=20, and 12×5/2=30, so both equations represent the same line. Option A, h=5, gives h+1=6 and breaks the required ratio. Option B gives 7, and option D gives 9; neither equals 8, so their lines are distinct and parallel rather than coincident. Memory cue: infinite solutions require all three coefficient ratios to match.

Compare the ratios of the coefficients: \(\frac{a_1}{a_2}=\frac{4}{8}=\frac{1}{2}\), whereas \(\frac{b_1}{b_2}=\frac{5}{9}\). Since \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), the two lines intersect at exactly one point, so the pair has a unique solution. Infinitely many solutions would require all three ratios to be equal. Exam tip: compare \(a_1/a_2\) and \(b_1/b_2\) first.

No solution occurs when lines are parallel and distinct. Its ratio form is (\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}).

For a pair of linear equations, if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the two lines have the same slope but different intercepts. Hence, they are distinct parallel lines and never intersect, so the pair has no solution and is inconsistent. Exam tip: equal coefficient ratios with an unequal constant-term ratio indicate “no solution” immediately.

A pair of linear equations in two variables represents two lines. If the ratios of the coefficients of x and y are unequal, the lines do not have the same slope. Lines with different slopes meet at exactly one point, so the pair has one and only one ordered pair satisfying both equations. This is called a unique solution, which is option C.

The condition is \(a_1/a_2\ne b_1/b_2\). Because the coefficient ratios differ, the lines cannot be parallel or coincident. The constants may affect the exact location of the intersection, but they do not change this conclusion when the two coefficient ratios are unequal. Therefore, “no solution” and “infinitely many solutions” are not suitable choices here. The correct conclusion is an unique, or अद्वितीय, solution.

Coincident lines represent the same line. Hence all three ratios are equal and infinitely many solutions occur.

Two lines with the same slope are either coincident or parallel. Different (y)-intercepts show that the lines are not coincident, so they are distinct parallel lines and never intersect. Hence, the pair of equations has no solution. Exam tip: equal slopes with different intercepts indicate no solution, whereas equal slopes with equal intercepts indicate infinitely many solutions.

The slope tells us how steep a line is. If two lines have different slopes, they cannot remain parallel and they cannot be the same line. Their directions are different, so they meet at one and only one point. Therefore, the pair of linear equations is consistent and independent, and it has a unique solution. The correct choice is C, “Unique solution.”

In the graph, the common point gives the values of both variables simultaneously. Choice A would apply to distinct parallel lines, which have equal slopes and never meet. Choice B and choice D describe coincident, or dependent, lines; such lines have the same slope and share infinitely many points. Since the slopes here are different, exactly one intersection is possible, so C follows directly.