Conditions for solvability

For two linear equations to have infinitely many solutions, the condition is \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). Here, \(\frac{3}{6}=\frac{12}{24}=\frac{1}{2}\), so \(\frac{k}{8}=\frac{1}{2}\), giving \(k=4\). Therefore, option B is correct. Exam tip: infinitely many solutions require all three corresponding ratios to be equal, not just two of them.

For two linear equations to have no solution, the condition is \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). Here, \(\frac{a}{6}=\frac{2}{4}=\frac{1}{2}\), while \(\frac{5}{13}\ne\frac{1}{2}\). Hence, \(a=3\). With the other options, the coefficient ratios are unequal, so the equations have a unique solution. Exam tip: first equate the ratios of the coefficients of x and y, then compare that ratio with the constants.

The first equation must be (2) times the second so (m=10). A common multiplier makes both lines coincident.

Two linear equations have a unique solution when the ratios of the corresponding coefficients are unequal. Here, \(\frac{p}{5}\ne\frac{6}{3}\), which gives \(p\ne10\). Therefore, option B is correct. Exam tip: For a unique solution, check that \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\).

The first equation must be (7) times the second so (a=14). Then both equations represent the same line.

For two linear equations to have no solution, the condition is \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). Here, \(\frac{2}{k}=\frac{3}{6}=\frac{1}{2}\), which gives \(k=4\). Also, \(\frac{6}{15}=\frac{2}{5}\), which is not equal to \(\frac{1}{2}\); hence the two lines are distinct parallel lines and the system is inconsistent. In an exam, first equate the ratios of the variable coefficients, then verify the condition using the constants.

Dividing every term of the second equation by 3 gives the first equation: 3x+6y=24 becomes x+2y=8. Hence both equations represent the same line and have infinitely many common solutions. Exam tip: when the ratios of the coefficients of x, y, and the constants are all equal, the lines are coincident, not distinct parallel lines.

For the two equations, \(\frac{a_1}{a_2}=\frac{2}{6}=\frac{1}{3}\) and \(\frac{b_1}{b_2}=\frac{1}{3}\), but \(\frac{c_1}{c_2}=\frac{4}{10}=\frac{2}{5}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\), which is the condition for an inconsistent pair represented by distinct parallel lines. Therefore, option C is correct. Exam tip: if all three ratios are equal, the lines are coincident; if only the first two are equal, they are distinct parallel lines.

The graph of each linear equation in two variables is a straight line. Two such lines intersect at exactly one point when their slopes are different. In standard form, the slope of \(ax+by=c\) is \(-a/b\), provided \(b\ne0\). Equal coefficient ratios would indicate parallel or coincident lines, but unequal ratios give different slopes.

For the first equation, the slope is \(-4/7\). For the second, it is \(-8/13\). These are not equal, since \(4/7\ne8/13\). Therefore the lines cannot be parallel or identical; they meet at one and only one point. The correct description is option C, lines intersecting at one point.

In the first option the second equation is (2) times the first. Therefore the lines are coincident and have infinitely many solutions.

In the second option (1/2=2/4) but (7/15) is different. Therefore the lines are parallel and have no solution.

The governing concept is that a pair has one unique solution when the two lines intersect, which occurs when a₁/a₂ ≠ b₁/b₂. In option C, the x-coefficient ratio is 4/2 = 2, while the y-coefficient ratio is 1/3; these are unequal, so the lines have different slopes and meet at exactly one point. Therefore option C is correct. In option A, both equations are equivalent, giving infinitely many solutions. In option D, the same dependence occurs because the first equation is five times the second. In option B, the coefficient ratios are equal, but the constant ratios differ, so the lines are distinct parallel lines and there is no solution.

The governing concept is consistency of a pair of linear equations. An inconsistent pair has no ordered pair (x, y) that satisfies both equations simultaneously. Graphically, its two lines are distinct and parallel: they have the same slope but different intercepts, so they never meet. Hence option B, “It has no solution,” is correct. Infinitely many solutions occur when the lines are coincident, not inconsistent. A pair of linear equations cannot generally be described as having two solutions; two intersections belong to other types of curves. Coincident lines give infinitely many common points, whereas distinct parallel lines give none. These distinctions make the classification unambiguous.

A consistent pair may have one or infinitely many solutions. The key point is that at least one common solution must exist.

The governing concept is the classification of a pair of linear equations by the relationship between their graphs. A consistent pair has at least one common solution, while an independent pair represents two different lines. Two different non-parallel lines intersect at exactly one point, and the coordinates of that intersection satisfy both equations simultaneously. Hence the pair has one and only one solution, called a unique solution. Option A describes parallel distinct lines, option B describes coincident lines, and option D is not a possible general classification for two linear equations in two variables.

In a consistent and dependent pair both equations represent the same line. Therefore every common point is a solution.

Multiplying every term of the second equation by 4 gives 4(2x-y)=4(3), or 8x-4y=12. Thus, both equations represent the same line and have infinitely many solutions. Option B is incorrect because the multiplying factor is 4, not 2. Exam tip: when the coefficients and constant terms of one linear equation are in the same ratio as those of the other, the lines are coincident and the system has infinitely many solutions.

Comparing the coefficients, \(10/2=5\) and \(5/1=5\), but the ratio of the constants is \(30/7\), which is not 5. Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the two lines are distinct and parallel and the pair has no solution. Infinitely many solutions would require all three ratios to be equal. Exam tip: Apply this ratio test directly to determine the solution status quickly.

For the two linear equations, \(\frac{a_1}{a_2}=\frac{11}{22}=\frac{1}{2}\), while \(\frac{b_1}{b_2}=\frac{3}{7}\). Since these ratios are unequal, the two lines intersect at exactly one point, so the pair has a unique solution. Therefore, option C is correct. Exam tip: if \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), a pair of linear equations has one unique solution.

The second equation is exactly twice the first: multiplying x-5y=1 by 2 gives 2x-10y=2. Hence both equations represent the same line, and every point on that line satisfies both equations. Therefore, the pair has infinitely many solutions. Option A would apply if the two lines were distinct and parallel. Exam tip: when a₁/a₂ = b₁/b₂ = c₁/c₂, the pair has infinitely many solutions.