Conditions for solvability

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

A pair of linear equations a1x + b1y = c1 and a2x + b2y = c2 has a unique solution when a1/a2 is not equal to b1/b2. This means the two lines have different slopes and intersect at exactly one point. For the given equations, q/15 must not equal 4/10, which simplifies to 2/5. Therefore q/15 ≠ 2/5, so q ≠ 15 × 2/5 = 6. Hence option A is correct. If q = 6, the coefficient ratios become equal; depending on the constants, the pair would then be either parallel distinct or coincident, but it would not satisfy the unique-solution condition. The values 15 and the equality q = 15 are irrelevant to the required ratio.

For a pair of linear equations to have infinitely many solutions, the ratios of corresponding coefficients and constants must be equal: a1/a2 = b1/b2 = c1/c2. Here, 5/15 = 1/3 and 11/33 = 1/3. Therefore, (r + 1)/18 must also equal 1/3. Hence r + 1 = 6, so r = 5. Option B is correct. The other values do not make the coefficient ratio equal to the constant ratio, so they would produce either a unique solution or no solution rather than coincident lines.

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

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.

For infinitely many solutions, (\frac{6}{14}=\frac{-t}{-21}=\frac{18}{42}) must hold. This gives (t=9).

Coefficient ratios are equal but the constant ratio is different. Hence the lines are parallel and the pair is inconsistent.

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.

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 \(\frac{b_1}{b_2}\).

Compare the coefficients of the two equations. The second equation has coefficients 8 and 18, which are twice 4 and 9, the corresponding coefficients in the first equation. However, twice the first right-hand side is \\(2\\times2=4\\), not 7. Thus, the left sides are proportional but the constants are not.

The equations therefore have the same slope but different positions. Their graphs are distinct parallel lines, so they never meet and no ordered pair satisfies both equations simultaneously. This is an inconsistent pair, giving option B. A unique solution would require different slopes, while infinitely many solutions would require the constant ratio to match as well.

Here, \(\frac{a_1}{a_2}=\frac{13}{26}=\frac{1}{2}\), \(\frac{b_1}{b_2}=\frac{4}{8}=\frac{1}{2}\), and \(\frac{c_1}{c_2}=\frac{6}{12}=\frac{1}{2}\). Therefore, both equations represent the same line, so the pair has infinitely many solutions. Option B would be correct only if the first two ratios were equal but the third ratio were different. Exam tip: when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the pair has infinitely many solutions.

For equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, a unique solution exists when a1/a2 is not equal to b1/b2. In the given equations, the coefficient ratios are 12/18 = 2/3 and (-7)/(-11) = 7/11. Since 2/3 and 7/11 are unequal, the two lines have different slopes and intersect at exactly one point. Therefore, the pair has a unique solution, so option C is correct. The no-solution and infinitely-many-solution cases require parallel or coincident lines, respectively.

Infinitely many solutions occur when the two equations represent the same line. Thus, the ratios of corresponding coefficients and constants must be equal. We have 8/20 = 2/5 and 16/40 = 2/5. Therefore, (u + 3)/10 must also equal 2/5. Solving gives u + 3 = 4, so u = 1. Hence option A is correct. If u had any other listed value, the first coefficient ratio would not match the other ratios, and the equations would not be coincident equations with infinitely many common solutions.

Equating coefficient ratios gives (v-2=3). Thus (v=5), and the different constant ratio gives no solution.

A pair of linear equations has a unique solution when the determinant of its coefficient matrix is non-zero. Here, the determinant is \(w(-24)-16(-6)=96-24w=24(4-w)\). It is non-zero when \(w\ne4\). At \(w=4\), the coefficients of \(x\) and \(y\) are in the same ratio, but the constants are not, so the lines are parallel and there is no solution. Exam tip: for a unique solution, check that \(a_1b_2-a_2b_1\ne0\).

The coefficient ratio is (\frac{1}{6}). For infinitely many solutions, (\frac{\lambda}{48}=\frac{1}{6}), so (\lambda=8).

(\frac{14}{21}=\frac{2}{3}). Equating coefficient ratios, (\frac{\mu}{6}=\frac{2}{3}) gives (\mu=4).

A pair of linear equations has a unique solution when the ratios of the x- and y-coefficients are unequal: a1/a2 ≠ b1/b2. Here, the condition is (α - 1)/8 ≠ 5/10. Since 5/10 = 1/2, equality would occur when α - 1 = 4, or α = 5. At α = 5, the coefficient ratios are equal; because the constant ratio 2/6 = 1/3 is different, the equations would have no solution. Therefore, every value except α = 5 gives a unique solution. Option B is correct.

For infinitely many solutions, (\frac{9}{27}=\frac{\beta+4}{18}=\frac{15}{45}) must hold. This gives (\beta=2).

Equating coefficient ratios, (\frac{5}{25}=\frac{\gamma}{15}) gives (\gamma=3). The constant ratio is different.