Conditions for solvability

Correct answer: C. s = 6

For infinitely many solutions, the ratios of corresponding coefficients and constants must be equal. In the equations 12x + sy = 18 and 20x + 10y = 30, the ratio of x-coefficients is 12/20 = 3/5, and the ratio of constants is 18/30 = 3/5. Therefore the y-coefficient ratio must also be 3/5: s/10 = 3/5. Multiplying by 10 gives s = 6. With s = 6, the first equation is three-fifths of the second, so the two equations represent the same line and have infinitely many common solutions. Hence option C is correct. If s were 4, 5, or 8, the y ratio would differ from 3/5, so the equations would not be coincident and could not have infinitely many solutions.

Correct answer: A. Infinitely many solutions

For a pair of linear equations, compare the corresponding coefficients and constants. Multiplying x+2y=3 by 4 gives 4x+8y=12, exactly the second equation. Thus both equations represent the same straight line, and every point on that line satisfies both equations. Therefore infinitely many ordered pairs are solutions. A unique solution would require intersecting lines, while parallel distinct lines would give no solution.

Correct answer: C. It has no solution

Put both equations into standard form: 3x − y − 4 = 0 and 6x − 2y − 9 = 0. The ratios of the x- and y-coefficients are 3/6 = 1/2 and (−1)/(−2) = 1/2. However, the constant ratio is (−4)/(−9) = 4/9, which is not 1/2. Thus the coefficient pairs are proportional, but the constants are not. Geometrically, the two equations represent distinct parallel lines: they have the same slope but different intercepts. Distinct parallel lines never intersect, so there is no ordered pair satisfying both equations. Therefore option C is correct. A unique solution would require intersecting lines, and infinitely many solutions would require all three ratios to be equal.

Correct answer: C. m = 7

For infinitely many solutions, both equations must represent the same straight line. Therefore, the ratios of corresponding coefficients must be equal: 4/12 = 9/27 = 1/3. The coefficient of y must follow the same ratio, so (m − 5)/6 = 1/3. Multiplying by 6 gives m − 5 = 2, and hence m = 7. Thus option C is correct. The other options make the y-coefficient ratio different from 1/3, so the three ratios are not equal. In that case the equations do not describe coincident lines and cannot possess infinitely many common solutions.

Correct answer: C. 7/13 ≠ 19/35, therefore one unique solution

For the pair a1x + b1y = c1 and a2x + b2y = c2, the relationship between a1/a2 and b1/b2 determines whether the lines have the same slope. Here, 7/13 is not equal to 19/35: cross-multiplication gives 7 × 35 = 245, whereas 19 × 13 = 247. Thus the first two ratios are unequal. The lines have different slopes and must intersect at exactly one point, so the pair has a unique solution. Option C is correct. Infinite solutions require all three ratios to be equal, while no solution requires the first two ratios equal but the constant ratio different.

Correct answer: A. q ≠ 6

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.

Correct answer: B. r = 5

For infinitely many solutions, the two equations must represent the same line. Therefore, the ratios of corresponding coefficients and constants must be equal. Here, 5/15 = 1/3 and 11/33 = 1/3. The remaining ratio must satisfy (r + 1)/18 = 1/3. Multiplying both sides by 18 gives r + 1 = 6, so r = 5. Substituting r = 5 makes the first equation 5x + 6y = 11, while multiplying it by 3 gives 15x + 18y = 33, exactly the second equation. Thus the equations are coincident and have infinitely many common solutions. Option B is correct. Every other listed value breaks the proportionality condition and cannot produce infinitely many solutions.

Correct answer: C. Unique solution

The governing concept is the solvability condition for two linear equations. For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, a unique solution exists when a₁/a₂ is different from b₁/b₂. Rewriting the equations gives 12x − 7y − 5 = 0 and 18x − 11y − 8 = 0. Their coefficient ratios are 12/18 = 2/3 and (−7)/(−11) = 7/11. Since 2/3 is not equal to 7/11, the corresponding lines have different slopes and must intersect at exactly one point. Therefore, the pair has a unique solution, so option C is correct. No solution would require parallel distinct lines, while infinitely many solutions would require coincident lines.

Correct answer: A. u = 1

Infinitely many solutions occur when both equations represent the same straight line. Thus, the ratios of corresponding coefficients and constants must be equal. For the second equation and the constants, 8/20 = 2/5 and 16/40 = 2/5. Therefore the x-coefficient ratio must also satisfy (u + 3)/10 = 2/5. Multiplying by 10 gives u + 3 = 4, so u = 1. A direct check confirms that the first equation becomes 4x + 8y = 16, which is obtained by multiplying 10x + 20y = 40 by 2/5. Hence both equations are coincident and have infinitely many common points. Option A is correct. The other listed values fail the proportionality condition.

Correct answer: B. α ≠ 5

For two linear equations, a unique solution exists when the ratios of the x-coefficients 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)/8 = 1/2, so α − 1 = 4 and α = 5. At that exceptional value, the coefficient ratios are equal, but the constant ratio 2/6 = 1/3 is different; hence the lines would be parallel and there would be no solution. For every α other than 5, the coefficient ratios differ and the lines intersect once. Therefore option B is correct.

Correct answer: C. d = 5

The condition for infinitely many solutions is that both equations represent the same line. Therefore, corresponding coefficients and constants must have equal ratios. Comparing the equations gives 4/12 = d/15 = 7/21. The first ratio simplifies to 1/3, and the constant ratio also simplifies to 1/3. Hence d/15 = 1/3, so d = 5. Substitution verifies the result: 4/12 = 5/15 = 7/21 = 1/3. Thus the first equation is exactly one-third of the second equation, and every point on either line satisfies both equations. Option C is correct. Values 3, 4, and 6 fail the proportionality condition and cannot produce infinitely many common solutions.

Correct answer: C. k = 4

A pair of linear equations has no solution when the corresponding coefficient ratios are equal but the constant ratio is different: a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Here, the y-coefficient ratio is 3/6 = 1/2. To make the x-coefficient ratio equal to it, 2/k = 1/2, which gives k = 4. Now check the constants: 4/11 is not equal to 1/2. Thus the equations have proportional x- and y-coefficients but non-proportional constants. Geometrically, their lines have the same slope but different intercepts, so they are parallel and never meet. Therefore no solution exists, and option C is correct. Other values do not make the coefficient ratios equal.

Correct answer: C. n = 6

Infinitely many solutions require the two equations to be proportional, so the ratios of corresponding coefficients and constants must be equal. We write 3/5 = n/10 = 18/30. The x-coefficient ratio is 3/5, and the constant ratio 18/30 also simplifies to 3/5. Therefore n/10 = 3/5. Multiplying both sides by 10 gives n = 6. Verification gives 3/5 = 6/10 = 18/30, so all three ratios agree. Hence the equations describe the same line and have infinitely many common solutions. Option C is correct. If n were 4, 5, or 8, the y-coefficient ratio would differ from the other ratios, so the equations would not be coincident.

Correct answer: A. p = 4

For no solution, the two equations must represent distinct parallel lines. In ratio form, this requires a₁/a₂ = b₁/b₂ but c₁/c₂ must be different. The y-coefficient ratio is (−5)/(−15) = 1/3. Therefore p/12 must equal 1/3, which gives p = 4. The constant ratio is 7/19, and 7/19 is not equal to 1/3. Thus the x- and y-coefficients are proportional, while the constants are not; the lines have equal slopes but different intercepts. They never intersect, so there is no solution. Hence option A is correct. The other listed values do not make p/12 equal to 1/3.

Correct answer: C. q = 6

For infinitely many solutions, the two equations must be proportional and therefore represent the same line. Comparing corresponding coefficients and constants gives q/18 = 11/33 = 22/66. The known ratios are 11/33 = 1/3 and 22/66 = 1/3. Hence q/18 must also equal 1/3. Multiplying by 18 gives q = 6. Substitution verifies the result: qx + 11y = 22 becomes 6x + 11y = 22, while multiplying this equation by 3 gives 18x + 33y = 66, exactly the second equation. Thus the equations are coincident and have infinitely many common solutions. Option C is correct. Values 4, 5, and 7 do not preserve proportionality.