Conditions for solvability

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 \(b_1/b_2\).

The slope of a line determines its direction, while the y-intercept determines where it crosses the y-axis. Two lines with the same slope are parallel in direction. If their y-intercepts are different, they are distinct parallel lines, not the same line. Distinct parallel lines never meet at a common point, so there is no ordered pair (x, y) satisfying both equations simultaneously. Therefore the pair has no solution, and option B is correct. If both slope and y-intercept were the same, the lines would coincide and have infinitely many solutions; if the slopes differed, they would intersect once and have a unique solution.

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.

A line can be written in slope-intercept form as 
m y=mx+c
m, where 
m m
m is the slope and 
m c
m is the intercept on the y-axis. The slope determines the direction of the line, while the intercept determines its position. If two lines have equal slopes but different intercepts, they are distinct parallel lines. If both slope and intercept are equal, their equations describe the very same set of points, so the lines coincide.

Here both lines have the same value of 
m m
m and the same value of 
m c
m. Consequently, every point satisfying the first line also satisfies the second, and no separate intersection point can be identified. The lines are therefore coincident and the pair has infinitely many common solutions. Thus option B is correct. Distinct parallel lines would require equal slopes but unequal intercepts.

For infinitely many solutions, (1/2=k/6=2/4), so (k=3). Keeping ratios equal makes the equations dependent.

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.

Two linear equations in two variables usually represent two lines. If the lines have different slopes, they meet at exactly one point, producing one unique ordered pair that satisfies both equations. In coefficient form, unequal ratios of the x- and y-coefficients indicate different slopes. Equal ratios of all corresponding coefficients would instead indicate coincident lines, while equal coefficient ratios with a different constant ratio would indicate parallel lines.

For 
m 2x+5y=1
m and 
m 3x+7y=4
m, compare the coefficient ratios: 
m 2/3
m is not equal to 
m 5/7
m. Thus the slopes are different, so the lines intersect at one point. For an additional check, elimination gives 
m 6x+15y=3
m and 
m 6x+14y=8
m; subtracting yields 
m y=-5
m and then 
m x=13
m. Hence option C is correct.

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

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

Here (3/6=4/8) but (11/25) is different so there is no solution. In exams equal first two ratios and a different third ratio mean inconsistent.

Here (1/2 \ne 7/9) so the lines intersect at one point. Different coefficient ratios give one unique solution.