Conditions for solvability

Correct answer: A. (k=\frac{3\pm\sqrt{41}}{2})

For non-unique solutions, the determinant must be zero. From ((k-3)k-8=0), (k=\frac{3\pm\sqrt{41}}{2}).

Correct answer: B. (k=0)

Equating coefficient ratios gives (k=0). The constant ratio (\frac{3}{8}) is different, so there is no solution.

Correct answer: C. \(p\neq -1\)

A pair of linear equations has a unique solution when the ratios of the coefficients of x and y are unequal. Here, the condition is \(\frac{p+4}{5}\neq\frac{-9}{-15}\). Since \(\frac{-9}{-15}=\frac{3}{5}\), we need \(\frac{p+4}{5}\neq\frac{3}{5}\), which gives \(p\neq -1\). At \(p=-1\), the x- and y-coefficient ratios become equal, so the solution cannot be unique. Exam tip: for a unique solution, first compare the ratios of the coefficients of x and y.

Correct answer: B. \(r=2\)

For infinitely many solutions, the two linear equations must represent the same line; hence \(\frac{r}{8}=\frac{\frac{3}{2}}{6}=\frac{6}{24}\). The last two ratios are \(\frac{1}{4}\), so \(\frac{r}{8}=\frac{1}{4}\) gives \(r=2\). If \(r=4\), then \(\frac{r}{8}=\frac{1}{2}\), which is not equal to the other ratios. Exam tip: for infinitely many solutions, verify \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Correct answer: B. (\alpha\neq3)

For a unique solution, (\frac{\alpha}{9}\neq\frac{4}{12}) must hold. Thus (\alpha\neq3) is correct.

Correct answer: C. (\beta=3)

Equating coefficient ratios gives (\beta=3). The constant ratio (\frac{5}{9}) is different, so there is no solution.

Correct answer: C. (\gamma=9)

The coefficient ratio is (\frac{1}{4}). For infinitely many solutions, (\frac{\gamma}{36}=\frac{1}{4}), so (\gamma=9).

Correct answer: B. \(s^2+s-17\ne 0\)

A pair of linear equations has a unique solution when its coefficient determinant is non-zero. Here, \(D=(s+2)(s-1)-3\times5=s^2+s-17\). Therefore, the required condition is \(s^2+s-17\ne 0\). In option A, the determinant becomes zero, so a unique solution is not possible. Exam tip: use \(a_1b_2-a_2b_1\ne0\) to test for a unique solution.

Correct answer: B. (t\ne192)

The direct answer is option B: \(t\ne192\). For two equations \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\), inconsistency occurs when the coefficient ratios are equal but the constant ratio is different. Here the second left side is exactly three times the first: \(15=3\times5\) and \(27=3\times9\). Therefore, to represent the same line, its right side would also have to be \(3\times64=192\). If \(t=192\), the equations are the same line and have infinitely many solutions. If \(t\ne192\), they have the same left side direction but different positions, so they are parallel and never meet. Option A is wrong because it gives the dependent case, not inconsistency. Option B is correct because every value other than 192 makes the constants incompatible. Option C is wrong because 64 is not the required threefold value. Option D is wrong because 128 is also not 192. Memory cue: when one complete equation is multiplied by 3, multiply its constant by 3 too; any other constant gives parallel lines.

Correct answer: C. 11

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, \(\frac{6}{18}=\frac{42}{126}=\frac{1}{3}\). Therefore, \(\frac{a}{33}=\frac{1}{3}\), giving \(a=11\). Hence, option C is correct. Exam tip: For infinitely many solutions, the ratios of the corresponding coefficients and constant terms must all be equal; checking only one ratio is not sufficient.

Correct answer: B. 4

For two linear equations to have no solution, the ratios of the coefficients of x and y must be equal, while the ratio of the constant terms must be different. Thus, p/14 = 10/35 = 2/7, giving p = 4. Also, 50/122 = 25/61, which is not equal to 2/7, so the equations are inconsistent. Exam tip: Remember the condition a₁/a₂ = b₁/b₂ ≠ c₁/c₂ for no solution.

Correct answer: B. \(k\ne\frac{44}{5}\)

Two linear equations have a unique solution when the ratios of the coefficients of the variables are unequal. Here, the required condition is \(\frac{11}{5}\ne\frac{k}{4}\). On simplifying, this gives \(44\ne5k\), or \(k\ne\frac{44}{5}\). Therefore, option B is correct. Exam tip: for a unique solution, use \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). In option A, the two ratios become equal, so it does not represent a unique solution.

Correct answer: C. No solution

Here, \(a_1/a_2=12/36=1/3\) and \(b_1/b_2=(-7)/(-21)=1/3\), but \(c_1/c_2=29/91\), which is not equal to \(1/3\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the two lines are parallel and distinct. Therefore, the pair has no solution. Exam tip: First compare the ratios of the coefficients of x and y, and then compare them with the ratio of the constant terms.

Correct answer: C. Parallel lines with no solution

Here, \(a_1/a_2=10/20=1/2\) and \(b_1/b_2=3/6=1/2\), but \(c_1/c_2=470/955\neq1/2\). Thus, the two lines have the same slope but different intercepts, so they are parallel and have no point of intersection. Therefore, the system is inconsistent. A common mistake is to conclude that equal first two ratios imply infinitely many solutions; the third ratio must also be equal for coincident lines. Exam tip: if \(a_1/a_2=b_1/b_2\neq c_1/c_2\), the system is inconsistent.

Correct answer: B. \(k\neq2\)

A pair of linear equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\) has a unique solution when \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\). Here, \(\frac{k}{6}\neq\frac{5}{15}=\frac{1}{3}\), which gives \(k\neq2\). At \(k=2\), the ratios of the coefficients on the left-hand sides are equal, so the solution cannot be unique. Exam tip: for a unique solution, compare the ratios of the coefficients of \(x\) and \(y\).

Correct answer: C. (m=6)

The direct answer is option C, m=6. For two equations to have no solution, their coefficient ratios must be equal but their constant ratio must be different: \(a_1/a_2=b_1/b_2\ne c_1/c_2\). Here \(4/10=2/5\). Set \(m/15=2/5\): multiplying by 15 gives \(m=6\). The constant ratio is \(8/21\), which is not \(2/5\), so the lines are parallel and distinct. Option A, m=4, gives \(4/15\), so the coefficient ratios are not equal. Option B, m=5, gives \(1/3\), also not equal to \(2/5\). Option C gives equal coefficient ratios but unequal constants, so it is correct. Option D, m=8, gives \(8/15\), again wrong. Memory cue: no solution means parallel lines: coefficients proportional, constants not proportional.

Correct answer: C. (p=5)

The direct answer is option C: \(p=5\). Infinitely many solutions mean both equations describe exactly the same line. For equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\), we need \(a_1/a_2=b_1/b_2=c_1/c_2\). Compare the second equation with the first: \(7/21=1/3\) and \(14/42=1/3\). Thus the first coefficient must satisfy \((p-2)/9=1/3\). Multiplying by 9 gives \(p-2=3\), so \(p=5\). Option A, \(p=3\), gives \((p-2)/9=1/9\), so the lines are not the same. Option B, \(p=4\), gives \(2/9\), also incorrect. Option C gives the required ratio and makes the first equation one-third of the second. Option D, \(p=7\), gives \(5/9\), so it fails. Memory cue: for infinitely many solutions, all three ratios must match, not just the two coefficient ratios.

Correct answer: B. \(a=4\)

For infinitely many solutions, the two linear equations must represent the same line; hence \(\frac{8}{20}=\frac{a}{10}=\frac{12}{30}\). Here, \(\frac{8}{20}=\frac{12}{30}=\frac{2}{5}\). Therefore, \(\frac{a}{10}=\frac{2}{5}\), which gives \(a=4\). For \(a=3,5\), or \(6\), the ratios of the corresponding coefficients are not equal. Exam tip: for infinitely many solutions, check that \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Correct answer: B. (c\neq20)

The coefficient ratio is (\frac{1}{4}). For no solution, (\frac{5}{c}\neq\frac{1}{4}), so (c\neq20).

Correct answer: C. (n=5)

Equating coefficient ratios, (\frac{2}{6}=\frac{n}{15}) gives (n=5). The constant ratio is different, so the pair is inconsistent.

Correct answer: B. \(s\neq6\)

A pair of linear equations has a unique solution when the ratios of corresponding coefficients are unequal: \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\). Here, \(\frac{11}{22}=\frac12\), which equals \(\frac{3}{s}\) only when \(s=6\). Therefore, the pair has a unique solution for \(s\neq6\). At \(s=6\), the lines are parallel because the constant-term ratio \(\frac{4}{10}\) is not \(\frac12\). Exam tip: use the determinant condition \(11s-22\times3\neq0\) for a quick check.

Correct answer: C. (t=9)

The direct answer is option C: \(t=9\). For two linear equations to have infinitely many solutions, both equations must represent the same line. For equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\), the condition is \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), provided the coefficients are written in the same order. Here the ratios are \(\frac{6}{14}=\frac{3}{7}\) and \(\frac{18}{42}=\frac{3}{7}\). The middle ratio must also be \(\frac{3}{7}\): \(\frac{-t}{-21}=\frac{t}{21}=\frac{3}{7}\). Multiplying by 21 gives \(t=9\). Thus option C works. Option A, \(t=6\), gives \(t/21=2/7\), not \(3/7\), so the lines are not identical. Option B, \(t=8\), gives \(8/21\), again not the required ratio. Option C gives all three equal ratios, so the equations are multiples of one another and have infinitely many common points. Option D, \(t=12\), gives \(4/7\), which also fails. A useful check is to multiply the first equation by \(7/3\): it becomes \(14x-\frac{7t}{3}y=42\). To match the second equation, \(-\frac{7t}{3}=-21\), giving \(t=9\). Memory cue: infinitely many solutions means same line, so all three coefficient-to-constant ratios must match.

Correct answer: C. Inconsistent

The direct answer is option C: inconsistent. Compare the two equations. The second equation has coefficients three times those of the first: \(21=3\times7\) and \(6=3\times2\). If its constant were also three times 9, it would be 27. Instead, it is 25. Thus the coefficient ratios are equal, \(7/21=2/6=1/3\), but the constant ratio is different, \(9/25\ne1/3\). Therefore the lines have the same slope but different intercepts; they are distinct parallel lines and have no common solution. Option A is wrong because equal coefficient ratios do not give an independent intersection. Option B is wrong because dependent lines require the constant ratio to match too. Option C is correct. Option D is wrong because no pair is always unique merely from these coefficients; in fact this pair has no solution. Exam cue: equal coefficient ratios plus unequal constant ratio means inconsistent.

Correct answer: C. Infinitely many solutions

Two linear equations represent the same line when every coefficient and the constant in one equation is multiplied by the same nonzero number to obtain the other equation. In that case, every point on the line satisfies both equations, so the pair has infinitely many solutions. Therefore option C is correct. The equations do not represent two separate intersecting lines.

Multiplying the first equation by 2 gives \(2(9x-5y)=2(13)\), which is \(18x-10y=26\), exactly the second equation. Thus the two equations are dependent and describe coincident lines. Every solution of the first is also a solution of the second, giving infinitely many ordered pairs rather than one, none, or exactly two solutions.

Correct answer: C. One unique solution

For the two equations, the coefficient ratios are \(\frac{10}{20}=\frac{1}{2}\) and \(\frac{3}{9}=\frac{1}{3}\), which are unequal. Hence, the lines are not parallel and intersect at exactly one point. Therefore, the pair has one unique solution. In the no-solution case, the first two ratios are equal but the ratio of constants is different. Exam tip: first compare \(\frac{a_1}{a_2}\) and \(\frac{b_1}{b_2}\).