Conditions for solvability

The governing concept is the graphical classification of a pair of linear equations. If both graphs lie exactly on the same line, they are coincident lines. Every point on that common line satisfies the first equation and the second equation, so the pair has infinitely many common ordered pairs and therefore infinitely many solutions. Option B is correct. Intersecting lines meet at one point and produce one solution, whereas parallel distinct lines never meet and produce no solution. Two solutions are not possible for a pair of linear equations in two variables when the lines coincide; the common points are not limited to two. Algebraically, the equations are dependent and one equation is a scalar multiple of the other.

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.

For two lines to be coincident, the ratios of their corresponding coefficients and constants must be equal: \(\frac{3}{3}=\frac{a}{4}=\frac{24}{24}\). Thus, \(\frac{a}{4}=1\), giving \(a=4\). Therefore, option B is correct. Exam tip: coincident lines satisfy \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

For two lines to be coincident, the ratios of their corresponding coefficients and constant terms must be equal. Multiplying \(2x+3y=10\) by 2 gives \(4x+6y=20\), which must match \(4x+ay=20\). Hence, \(a=6\), so option A is correct. In option B, the coefficient of \(y\) would remain 3, and the two equations would not represent the same line. Exam tip: For coincident lines, check \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

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 infinitely many solutions, both equations must represent the same line. Therefore, the ratios of corresponding coefficients must be equal: \(\frac{p}{2}=\frac{-6}{-3}=\frac{18}{9}=2\). Hence \(\frac{p}{2}=2\), so \(p=4\). If \(p=2\), the ratio of the coefficients of \(x\) would not match the other ratios. Exam tip: for infinitely many solutions, check \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Dividing the first equation by 5 gives \(x+2y=4\). On putting \(k=1\), the second equation becomes \(x+2y=7\). The coefficients of \(x\) and \(y\) are the same, but the constants are different; hence the lines are parallel and have no solution. For \(k=2\) or \(k=5\), the coefficient ratios are not equal, so the lines intersect at one point. Exam tip: for no solution, check \(a_1/a_2=b_1/b_2\ne c_1/c_2\).