Conditions for solvability

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.

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.

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.

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.

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.

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 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.

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.

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.

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.

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.

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.

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.

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.

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.