Conditions for solvability

For the pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the ratios of corresponding coefficients classify the graphs. If a₁/a₂ is not equal to b₁/b₂, the coefficients of x and y are not proportional in the same way. Consequently, the two lines have different slopes and must meet at exactly one point, producing a unique solution. Therefore option C is correct. Parallel lines occur when a₁/a₂ = b₁/b₂ but this common ratio is not equal to c₁/c₂. Coincident lines occur when all three ratios are equal. The condition given here rules out both of those cases, so the lines are intersecting rather than parallel or coincident.

To determine the number of solutions, compare the two equations. Multiplying every term of 2x + 5y = 20 by 2 gives 4x + 10y = 40, which is exactly the second equation. Thus both equations represent the same line, not two distinct lines. Every point on that line satisfies both equations, so the pair has infinitely many solutions and option C is correct. In coefficient form, a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 5/10 = 1/2, and c₁/c₂ = 20/40 = 1/2; equality of all three ratios confirms coincident lines. No solution would correspond to distinct parallel lines, while exactly one solution would require intersecting lines. Two solutions cannot occur for two distinct linear equations in this setting.

For a pair a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0, a unique solution occurs when a₁/a₂ is not equal to b₁/b₂. This means the two lines have different slopes and intersect at exactly one point. In option C, the coefficient ratios are 1/3 and 3/1, which are unequal, so the lines intersect uniquely. In option A, the second equation becomes the first divided by 2, so the lines coincide. In option B, the left side doubles but the constant does not, producing parallel distinct lines. In option D, the first equation is four times the second, so the lines coincide. Therefore C is the only correct choice.

For the pair of linear equations a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0, the lines are coincident when a₁/a₂ = b₁/b₂ = c₁/c₂. Here, 4/2 = 2, 6/3 = 2, and 10/5 = 2. Since all three ratios are equal, the second equation is exactly one-half of the first equation. Both equations therefore represent the same line, so infinitely many points satisfy the pair. Option B, coincident, is correct. Parallel distinct lines would require the first two ratios to be equal but different from the constant ratio; intersecting lines would require unequal coefficient ratios. Thus A and C do not fit, and D is not a valid classification here.

Use the consistency criterion for two linear equations. Compute each ratio: a₁/a₂ = 6/2 = 3, b₁/b₂ = 9/3 = 3, and c₁/c₂ = 12/4 = 3. Since all three ratios are equal, the second equation is obtained by multiplying the first equation by 1/3. Thus both equations represent the same geometric line. The lines are coincident and the system has infinitely many solutions, so option B is correct. Distinct parallel lines would require the first two ratios to be equal but different from the constant-term ratio. Intersecting lines would require unequal ratios for the coefficients, and perpendicularity is not established by this test.

The governing concept is the coefficient-ratio test for the pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0. Compute the three ratios: a₁/a₂ = 7/1 = 7, b₁/b₂ = 14/2 = 7, and c₁/c₂ = 21/3 = 7. Since all three ratios are equal, the second equation is obtained by multiplying the first equation by 1/7, or equivalently both equations represent the same geometric line. Hence the lines are coincident and the system has infinitely many solutions. Parallel distinct lines would require the coefficient ratios for a and b to be equal but the constant ratio to be different. Intersecting lines would have unequal ratios for a and b, while perpendicularity is not established by this test. Therefore option B is correct.

The governing concept is the consistency test for a pair of linear equations. Write the equations as 9x − 6y − 21 = 0 and 3x − 2y − 7 = 0. Here, a1/a2 = 9/3 = 3, b1/b2 = (−6)/(−2) = 3, and c1/c2 = (−21)/(−7) = 3. Since all three corresponding ratios are equal, the two equations represent the same straight line. Therefore every point on one line also lies on the other, so the pair has infinitely many solutions. Hence option C, coincident lines, is correct. Equal coefficients only for x and y with unequal constants would instead indicate distinct parallel lines.

The governing concept is the condition for infinitely many solutions. The two equations must represent the same line, so every coefficient and constant in the second equation must be the same multiple of the corresponding quantity in the first. Since 8x is twice 4x and 6y is twice 3y, the second equation must equal 2 times the first equation. Multiplying 4x + 3y = p by 2 gives 8x + 6y = 2p. Comparing this with 8x + 6y = 42 yields 2p = 42, so p = 21. Therefore option C is correct. If p had any other value, the lines would have equal variable coefficients but different constants, making them distinct parallel lines with no solution.

The governing concept is the ratio test for a pair of linear equations. Put both equations in standard form: 12x + 15y − 30 = 0 and 4x + 5y − 10 = 0. Their corresponding ratios are a1/a2 = 12/4 = 3, b1/b2 = 15/5 = 3, and c1/c2 = (−30)/(−10) = 3. Because all three ratios are equal, the equations represent the same line rather than two separate lines. Consequently, every point common to one equation is also common to the other, giving infinitely many solutions. Hence option C, coincident lines, is correct. If only the first two ratios had matched while the constant ratio differed, the lines would have been distinct parallel lines.

The governing concept is the condition for infinitely many solutions of a pair of linear equations. Infinitely many solutions occur when both equations represent the same, or coincident, line. The coefficients of x and y in the second equation satisfy 15 = 3 × 5 and 12 = 3 × 4. Therefore, for the constant terms to have the same proportionality, 96 must also equal 3p. Solving 3p = 96 gives p = 96/3 = 32. Hence option C is correct. If p were 30, 31, or 33, the constant-term ratio would not equal the coefficient ratio, so the two equations would represent distinct parallel lines rather than one common line. Thus the algebraic ratio test and the graphical interpretation agree.

The governing idea is the condition for dependent linear equations. Multiplying 3x+2y=20 by 2 gives 6x+4y=40, exactly the first equation. Thus both equations represent the same line, and every point on that line satisfies both. Therefore infinitely many ordered pairs are solutions, so B is correct; the system is neither inconsistent nor uniquely determined.

Compare the equations by multiplying the second equation by 3: 3(x+2y=8) gives 3x+6y=24, which is exactly the first equation. Hence the two equations describe one identical straight line rather than two intersecting lines. Every point on that line is a common solution, so there are infinitely many solutions. Therefore option C is correct.

Multiply the second equation x−2y=3 by 5. It becomes 5x−10y=15, exactly the first equation. Therefore both equations represent the same straight line, not two distinct intersecting or parallel lines. Every point on that line satisfies both equations, so the system has infinitely many solutions. Hence option D is correct.

For infinitely many solutions, both linear equations must represent the same line. Therefore, the ratios of corresponding coefficients must be equal: \(\frac{2}{4}=\frac{p}{6}=\frac{10}{20}\). Since \(\frac{2}{4}=\frac{1}{2}\), we get \(\frac{p}{6}=\frac{1}{2}\), so \(p=3\). If \(p=6\), the coefficients are not in the same ratio, so the equations do not represent the same line. Exam tip: for infinitely many solutions, check \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

The first equation is obtained by multiplying the second equation by 7: 7(x−2y=4) gives 7x−14y=28. Thus the equations are identical and represent the same line. A line contains infinitely many points, and each of those points satisfies both equations. Therefore option C is correct; the alternatives incorrectly assume inconsistency, a single intersection, or two isolated solutions.

For two linear equations to have infinitely many solutions, the ratios of corresponding coefficients and constants must all be equal: a₁/a₂ = b₁/b₂ = c₁/c₂. For 2x + ay = 7 and 4x + 10y = 14, the known ratios are 2/4 = 1/2 and 7/14 = 1/2. Therefore the remaining ratio must satisfy a/10 = 1/2. Multiplying by 10 gives a = 5. With a = 5, the first equation is 2x + 5y = 7, and twice it is 4x + 10y = 14, exactly the second equation. Thus both equations represent the same line and have infinitely many common points. Hence option C is correct; the other values fail the required equality.

Write the equations in standard form if needed and compare the ratios of corresponding coefficients. The x-coefficient ratio is 13/4, while the y-coefficient ratio is 2/5. Since 13/4 and 2/5 are unequal, the coefficients are not proportional. In geometric terms, the slopes are −13/2 and −4/5, which are different. Lines with different slopes cannot be parallel or coincident; they intersect at exactly one point. Therefore the pair is consistent and independent and has one unique solution. Option C is correct. Option A and option D describe distinct parallel lines, while option B describes coincident lines. The unequal coefficient ratios are sufficient to identify the unique intersection without calculating its coordinates.

For two linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the condition a₁/a₂ = b₁/b₂ = c₁/c₂ means that all coefficients, including the constants, are in the same proportion. One equation is therefore a scalar multiple of the other, so both represent the same line. Since every point on that line satisfies both equations, the pair has infinitely many solutions. Unequal constant ratios would instead indicate distinct parallel lines.

The first equation must be (2) times the second, so (k=9). For infinitely many solutions, keep all three ratios equal.

A pair of linear equations has infinitely many solutions when the corresponding coefficient ratios are equal, namely a₁/a₂ = b₁/b₂ = c₁/c₂. For 3x + ay = 12 and 6x + 8y = 24, the x-coefficient ratio is 3/6 = 1/2 and the constant ratio is 12/24 = 1/2. Hence the y-coefficient ratio must also equal 1/2: a/8 = 1/2. Multiplying both sides by 8 gives a = 4. Substituting this value makes the first equation 3x + 4y = 12; multiplying it by 2 gives 6x + 8y = 24, the second equation. Thus the lines coincide and infinitely many points satisfy both equations. Option B is correct.

For two linear equations to have no solution, the ratios of the coefficients of the variables must be equal, but this common ratio must differ from the ratio of the constants. Here, \(k/6=5/10=1/2\), which gives \(k=3\). Also, \(20/45=4/9\), not \(1/2\), so the two lines are distinct and parallel. Therefore, option B is correct. Exam tip: For ‘no solution,’ check \(a_1/a_2=b_1/b_2\ne c_1/c_2\).

The first two ratios are equal, so for inconsistency the constant ratio must differ. Hence (m \ne 14).

The second equation is (2) times the first, so both lines are coincident. Coincident lines have infinitely many solutions.

The ratios of the corresponding coefficients are 7/14 = (-3)/(-6) = 1/2. However, the ratio of the constant terms is 10/25 = 2/5, which is not equal to 1/2. Thus, the two lines are distinct and parallel, so they do not intersect at any point. Therefore, the pair has no solution. Exam tip: If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair of linear equations has no solution.

Here (3/6 \ne 4/7), so the lines intersect at one point. Different coefficient ratios give one unique solution.