Conditions for solvability

When two lines intersect at exactly one point, they have one common solution. Therefore, the pair is called consistent and independent. An inconsistent pair has no solution, whereas dependent or coincident lines have infinitely many solutions. Exam tip: one point of intersection means one unique solution and hence a consistent independent pair.

For a unique solution, (k/3 \ne 2/1), so (k \ne 6). In exams, make sure the (a) and (b) ratios are not equal.

A pair of linear equations has infinitely many solutions when both equations represent exactly the same line. In standard form, this happens when the ratios of corresponding coefficients and constants are equal: 
m a_1/a_2=b_1/b_2=c_1/c_2
m. This means one equation is a constant multiple of the other, so every point on one line also lies on the other. The equations here are already arranged in comparable standard form.

For 
m 2x+ky=4
m and 
m 6x+9y=12
m, the ratio from the x-coefficients is 
m 2/6=1/3
m, and the constant ratio is 
m 4/12=1/3
m. Therefore the y-coefficient must satisfy 
m k/9=1/3
m. Multiplying by 9 gives 
m k=3
m. Indeed, multiplying the first equation by 3 produces the second, so option C is correct.

For two linear equations, no solution occurs when the ratios of the coefficients of x and y are equal but the ratio of the constants is different. Here, 4/2 = 2 and 6/3 = 2, while the constant ratio is 10/k. If 10/k = 2, then 10 = 2k and k = 5. At k = 5, multiplying the second equation by 2 gives 4x + 6y = 10, so both equations represent the same line and have infinitely many solutions, not no solution. Therefore k must not be 5, making option A correct. For k = 6, 8, or 10, the lines have equal coefficient ratios but unequal constants after comparison, so they are distinct parallel lines with no common solution.

The second equation is exactly twice the first: multiplying \\(x+2y=3\\) by 2 gives \\(2x+4y=6\\). Hence, both equations represent the same line and have infinitely many common solutions. Exam tip: check \\(a_1/a_2=b_1/b_2=c_1/c_2\\); when all three ratios are equal, the lines are coincident. Therefore, the parallel-lines option is not correct here because the two equations do not represent distinct lines.

(2/4=3/6) but (6/15) is different, so the lines are parallel. Such a graph has no intersection.

Compare the ratios of the corresponding coefficients. Here, 5/10 = 1/2, whereas 2/3 is different. Thus, a₁/a₂ ≠ b₁/b₂, so the two lines intersect at exactly one point and the pair has a unique solution. Parallel lines require the first two coefficient ratios to be equal, while coincident lines require all three ratios to be equal. Exam tip: If a₁/a₂ ≠ b₁/b₂, the lines intersect at one point.

Consider the pair a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0. A unique solution means that the corresponding lines intersect at exactly one point. This occurs when their slopes are different, which is represented algebraically by a₁/a₂ ≠ b₁/b₂. Since the coefficient ratios are unequal, the lines cannot be parallel or coincident, so one common ordered pair satisfies both equations. Option A, where all three ratios are equal, represents coincident lines and infinitely many solutions. Option B represents distinct parallel lines and no solution. Option D is incomplete and is not a general criterion. Therefore option C is the only correct condition.

The direct answer is option A. Write the pair as a_1x+b_1y+c_1=0 and a_2x+b_2y+c_2=0. No solution occurs when the two lines are distinct and parallel. The condition is a_1/a_2=b_1/b_2 but c_1/c_2 is different. Equal first two ratios mean the x- and y-coefficients have the same proportionality, so the slopes are equal. A different constant ratio means the lines are not the same line; therefore they never meet. Option A is exactly this condition. Option B, a_1/a_2≠b_1/b_2, normally indicates unequal slopes, so the lines intersect at one point and there is one solution. Option C, all three ratios equal, indicates coincident lines and infinitely many solutions. Option D says only that the second and third ratios differ; without knowing the first ratio, this alone does not guarantee no solution. Thus it is not a sufficient general condition. Memory cue: equal coefficient ratios but unequal constant ratio means parallel distinct lines: no solution.

For two linear equations a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0, infinitely many solutions occur when the two equations represent exactly the same line. The condition for this is a₁/a₂=b₁/b₂=c₁/c₂. Every point on that common line then satisfies both equations, so there are infinitely many common points.

Option A describes intersecting lines and therefore gives one solution. Option B gives parallel distinct lines because the coefficient ratios agree but the constant ratio does not, so it gives no solution. Option C has all three ratios equal and is therefore correct. Option D is not the standard algebraic condition and does not establish coincident lines.

Writing the equations in standard form gives \\(a_1=1,b_1=-1,c_1=-2\\) and \\(a_2=2,b_2=-2,c_2=-5\\). Thus, \\(a_1/a_2=1/2\\) and \\(b_1/b_2=(-1)/(-2)=1/2\\), but \\(c_1/c_2=(-2)/(-5)=2/5\\). Since \\(a_1/a_2=b_1/b_2\\) but \\(c_1/c_2\\) is different, the two lines are parallel and distinct, so they have no solution. Exam tip: for an inconsistent pair, the first two coefficient ratios are equal while the constant-term ratio is different.

The second equation is (3) times the first, so both lines are coincident. Such a pair has infinitely many solutions.

Compare the coefficient ratios after writing the equations in standard form: \(\frac{a_1}{a_2}=\frac{4}{8}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-5}{-9}=\frac{5}{9}\). Since these ratios are unequal, the two lines intersect at exactly one point, so the pair has a unique solution. Infinitely many solutions or the same line require all corresponding ratios to be equal, while no solution requires \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). Exam tip: first compare \(a_1/a_2\) and \(b_1/b_2\).

In the slope-intercept form y = mx + c, m is the slope and c is the y-intercept. Equal slopes mean that two lines have the same direction. If their y-intercepts are different, they occupy different positions and are distinct parallel lines. Distinct parallel lines never meet, so there is no point (x, y) that lies on both lines and the corresponding pair has no solution. Thus option B is correct. If both the slopes and y-intercepts were equal, the lines would coincide and there would be infinitely many solutions. If the slopes were unequal, the lines would intersect at one point and give a unique solution. Three solutions are impossible for two distinct non-coincident lines.

When two lines have different slopes, they are not parallel and intersect at exactly one point. The coordinates of this intersection satisfy both linear equations, so the pair has exactly one unique solution. Option B applies to distinct parallel lines with equal slopes. Exam tip: different slopes imply one unique solution.

In the form \(y=mx+c\), the slope is \(m\) and the intercept is \(c\). If both \(m\) and \(c\) are the same for two lines, then both equations represent the exact same line. So the lines are coincident.

For infinitely many solutions, the equations must be proportional. Compare \(2x\) with \(x\): multiplying \(x+ky=2\) by \(2\) gives \(2x+2ky=4\). To match \(2x+6y=4\), we need \(2k=6\Rightarrow k=3\).

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, \(\frac{2}{4}=\frac{7}{14}=\frac{1}{2}\), so \(\frac{3}{k}=\frac{1}{2}\). Cross-multiplication gives \(k=6\), so option B is correct. Exam tip: for infinitely many solutions, all three coefficient-to-constant ratios must be equal; equality of only two ratios is not sufficient.

Two linear equations have no solution when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\). Here, \(\frac{k}{4}=\frac{1}{2}\), giving \(k=2\). The first equation then becomes \(2x+y=3\); multiplying it by 2 gives \(4x+2y=6\), whereas the second equation is \(4x+2y=8\). Thus, the two lines have the same slope but different intercepts, so they do not intersect and have no solution. Exam tip: equate the ratios of the coefficients of \(x\) and \(y\), then verify that the constants have a different ratio.

For two linear equations to have no solution, the condition is \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\). Here, \(\frac{3}{6}=\frac{k}{8}\), which gives \(k=4\). Also, \(\frac{9}{20}\neq\frac{1}{2}\), so the two lines are parallel and inconsistent. Exam tip: first equate the ratios of the variable coefficients, then check that the ratio of the constants is different.

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, \(\frac{5}{10}=\frac{a}{6}=\frac{15}{30}=\frac{1}{2}\). Thus, \(\frac{a}{6}=\frac{1}{2}\), giving \(a=3\). Therefore, option B is correct. Exam tip: infinitely many solutions require all three corresponding ratios to be equal, not just two of them.

For a unique solution, (a/3 \ne 4/2), so (a \ne 6). Equal (a) and (b) ratios prevent a unique solution.

Compare every term of the two equations. Multiplying the first equation, 2x − 3y = 4, by 2 gives 4x − 6y = 8, which is exactly the second equation. Thus the two equations are equivalent and represent the same straight line, called coincident lines. They do not represent two separate lines; instead, every point on the common line satisfies both equations. Consequently, the pair has infinitely many solutions. Option A correctly states the algebraic relationship. Option C would require proportional x- and y-coefficients but a non-proportional constant, while option D would require unequal coefficient ratios.

For the two equations, \(\frac{a_1}{a_2}=\frac{7}{14}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{1}{2}\), whereas \(\frac{c_1}{c_2}=\frac{10}{25}=\frac{2}{5}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), which represents distinct parallel lines. Hence, the pair has no solution; option D states the algebraic consequence, but not the geometrical situation asked here. Exam tip: remember the condition \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\) for distinct parallel lines.

Multiply \(2x+5y=1\) by \(3\): \(6x+15y=3\).
Multiply \(3x+7y=4\) by \(2\): \(6x+14y=8\).
Subtract: \((6x+15y)-(6x+14y)=3-8\Rightarrow y=-5\).
Substitute into \(2x+5y=1\): \(2x+5(-5)=1\Rightarrow 2x-25=1\Rightarrow x=13\).
Since we get a definite \((x,y)=(13,-5)\), there is exactly one solution.