Conditions for solvability

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.

The direct answer is option B. For equations a1x+b1y=c1 and a2x+b2y=c2, no solution occurs when a1/a2=b1/b2 but c1/c2 is different. In option A, doubling x+2y=7 gives 2x+4y=14, so both equations are the same line and have infinitely many solutions, not none. In option B, the first two coefficient ratios are 1/2=2/4, but the constants give 7/15, which is different. The lines therefore have the same direction but different positions: they are parallel and never meet. Option C has unequal coefficient ratios 1/3 and 2/5, so the lines intersect once. Option D also gives a normal pair: solving 2x-y=3 and x+y=4 gives x=7/3 and y=5/3, so it has one solution. Memory cue: equal coefficient ratios but unequal constant ratio means 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.

The direct answer is option A: a consistent pair has at least one solution. In a pair of linear equations, consistency means that the two equations share at least one ordered pair (x,y). There are two consistent possibilities: intersecting lines have exactly one common point, while coincident lines have infinitely many common points. Option A correctly includes both cases. Option B describes an inconsistent pair, whose lines are parallel and distinct. Option C is not correct because distinct parallel lines are always inconsistent, not consistent. Option D is wrong in two ways: a pair need not have zero solutions, and the phrase only (0) solutions is not a standard result; even if the origin is one solution, there may be another or infinitely many. A useful exam cue is that consistent means ‘common solution exists’; independent usually means one solution, and dependent usually means infinitely many.

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.

Compare the ratios of the corresponding coefficients and constants: \(\frac{12}{3}=\frac{4}{1}=4\), but \(\frac{16}{5}\neq 4\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\), so the two lines are distinct and parallel and have no common solution. Therefore, the pair is inconsistent. Exam tip: equal ratios for the first two terms but an unequal ratio for the constant terms indicate an inconsistent pair.

A line is determined by its slope and its y-intercept. If two lines have the same slope and the same y-intercept, they occupy exactly the same position on the coordinate plane. They are not two different lines; they are coincident lines. Consequently, every point on that common line satisfies both equations.

Since every point of the line is a common solution, the pair does not have just one or two solutions. It has infinitely many solutions. In the usual classification, this is called a consistent and dependent pair. “No solution” would describe distinct parallel lines, which have the same slope but different intercepts. Therefore option C, infinitely many solutions, is correct.

The governing graphical idea is that lines with the same slope are parallel, unless they also have the same intercept. Different intercepts mean the lines are distinct: they never meet at a common point. Therefore no ordered pair (x,y) can satisfy both equations at once, so the pair has no solution. A pair with no common solution is called inconsistent. Option A and option D refer to coincident lines, which have infinitely many common points, while option C refers to two non-parallel lines that intersect once. Thus the correct classification is inconsistent, option B.

The slope of a nonvertical line describes how much its y-value changes when its x-value changes. Lines with different slopes have different directions, so they cannot remain parallel or overlap completely. In a plane, two such lines meet at exactly one point. That point gives one ordered pair satisfying both linear equations. Therefore, a pair represented by lines with unequal slopes is called consistent and independent and has one unique solution.

To see the algebraic meaning, different slopes imply that the equations are not scalar multiples with equal coefficient ratios. The two lines therefore cannot describe the same line, and they cannot be distinct parallel lines, because parallel lines have equal slopes. They must intersect once. Hence the number of common solutions is one, which is option C. Infinite solutions correspond to coincident lines, while no solution corresponds to distinct parallel lines.

Compare the coefficients and constants in the two equations. Multiplying 5x + 3y = 2 by 3 gives 15x + 9y = 6, which is exactly the first equation. Thus the two equations are dependent and represent one coincident line rather than two different lines. Every point on this common line is a solution of the pair, so infinitely many solutions exist. A unique solution would require intersecting lines, and distinct parallel lines would have no common point.

For the coefficients, \(\frac{a_1}{a_2}=\frac{14}{2}=7\) and \(\frac{b_1}{b_2}=\frac{-7}{-1}=7\), but \(\frac{c_1}{c_2}=\frac{28}{3}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\), which represents an inconsistent pair of equations and two distinct parallel lines. Therefore, there is no solution. Option B would be correct only if all three ratios were equal, while option A requires the first two ratios to be unequal. Exam tip: Compare the corresponding ratios in the same order.

The direct answer is option C: infinitely many solutions. First compare the two equations. Multiplying every term of 3x+2y=7 by 2 gives 6x+4y=14, which is exactly the second equation. Thus both equations describe the same straight line, not two different lines. Every point satisfying the first equation also satisfies the second, so there are infinitely many common points. Option A, no solution, is wrong because the lines are not parallel distinct lines. Option B, one unique solution, would occur when two different lines intersect at one point. Option C is correct because coincident lines overlap completely. Option D, exactly two solutions, is impossible for two linear equations representing lines: two distinct lines meet once, identical lines meet infinitely often, and parallel lines never meet. Memory cue: if one complete equation is a non-zero multiple of the other, the lines coincide and solutions are infinite.