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, \(\frac{-10}{-5}=\frac{30}{15}=2\), so \(\frac{p}{3}=2\), giving \(p=6\). Therefore, option C is correct. Exam tip: For infinitely many solutions, all three corresponding coefficient and constant ratios must be equal; matching only two ratios is not sufficient.

Two linear equations have infinitely many solutions when the ratios of their corresponding coefficients and constants are equal. Multiplying the second equation by 3 gives \(6x+9y=27\), which must coincide with the first equation \(ax+9y=27\). Therefore, \(a=6\). Exam tip: for infinitely many solutions, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Multiply the first equation by 2: 2(4x+7y=31) gives 8x+14y=62, exactly the second equation. Hence the two equations are dependent and represent the same straight line. Since every point on that line satisfies both equations, the pair has infinitely many ordered-pair solutions. Therefore option D is correct, not a unique or inconsistent solution.

For two linear equations to have infinitely many solutions, the ratios of the corresponding coefficients and constants must be equal: 4/12 = a/21 = 32/96. Since 4/12 and 32/96 are both 1/3, a/21 = 1/3, giving a = 7. Therefore, option C is correct. Exam tip: all three ratios must be equal; equality of only two ratios is not sufficient for infinitely many solutions.

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, \(\frac{p}{20}=\frac{9}{30}=\frac{3}{10}\), so \(p=6\). Also, \(\frac{45}{103}\ne\frac{3}{10}\), confirming that the lines are distinct and parallel. Exam tip: first equate the ratios of the coefficients of x and y, then verify that the ratio of constants is different.

Equal ratios of \(a\) and \(b\) give the two lines the same slope. Since the \(c\)-ratios differ, the lines cannot coincide; they are distinct parallel lines with no solution. Exam tip: always compare all three ratios.

Here, \(\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{5}{10}=\frac{1}{2}\), but \(\frac{c_1}{c_2}=\frac{41}{85}\neq\frac{1}{2}\). Thus, the two lines are parallel and distinct, so the pair has no solution. Option B is incorrect because infinitely many solutions require all three ratios to be equal. Exam tip: when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\), the pair is inconsistent and has no solution.

Here (11/17 \ne 14/22), so the lines intersect at one point. If the first two ratios are different, one unique solution is obtained.

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, \(\frac{14}{21}=\frac{42}{63}=\frac{2}{3}\), so \(\frac{k}{18}=\frac{2}{3}\), giving \(k=12\). Therefore, option C is correct. In an exam, check the constant-term ratio as well as the coefficient ratios; matching only the variable coefficients is not sufficient for infinitely many solutions.

A pair of linear equations has infinitely many solutions when the ratios of corresponding coefficients and constants are equal. Here, the second equation must be twice the first because 26=2×13 and 104=2×52. Therefore, 18=2q, giving q=9. Hence, option C is correct. Exam tip: For infinitely many solutions, verify a₁/a₂=b₁/b₂=c₁/c₂.

For two linear equations to have a unique solution, the condition is \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{12}{3}=4\) and \(\frac{p}{5}\), so \(4\ne\frac{p}{5}\), which gives \(p\ne20\). Hence, option B is correct. If \(p=20\), then \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=4\), but \(\frac{c_1}{c_2}=\frac{60}{16}\), so the lines are parallel and there is no solution. Exam tip: For a unique solution, ensure that the ratios of the corresponding coefficients of the variables are unequal.

Two distinct non-parallel lines intersect at exactly one point. This intersection point satisfies both linear equations, so the pair has exactly one solution. Distinct parallel lines have no solution, whereas coincident lines have infinitely many solutions. Exam tip: Count the intersection points in the graph to identify the number of solutions quickly.

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, \(\frac{b}{14}=\frac{16}{28}=\frac{4}{7}\), so \(b=8\). Also, \(\frac{64}{131}\ne\frac{4}{7}\), confirming that the two lines are parallel and inconsistent. Exam tip: For “no solution,” first equate the ratios of the variable coefficients and then verify that the constant-term ratio is different.

For two linear equations, compare the ratios of the coefficients of x and y with the ratio of the constant terms. If the x-coefficient ratio and y-coefficient ratio are equal, but the constant ratio is different, the two equations represent different parallel lines. Such lines never meet, so the pair has no common solution. This is the idea behind option C, distinct parallel lines.

Here, the ratios are \(22/2=11\) and \(33/3=11\), while \(99/12=8.25\). Thus the first two ratios are equal, but the third is not. Therefore, the lines have the same direction but different positions. They are not coincident, because coincident lines would have all three ratios equal; they are also not perpendicular or intersecting.

Every coefficient in the first equation is 6 times the corresponding coefficient in the second: 30=6×5, 18=6×3, and 126=6×21. Dividing the first equation by 6 gives exactly 5x+3y=21. Hence, both equations represent the same line, and the pair has infinitely many solutions. Exam tip: if a₁/a₂ = b₁/b₂ = c₁/c₂, the two lines are coincident and the pair has infinitely many solutions.

For the coefficients of the two equations, \(\frac{5}{12} \ne \frac{17}{41}\). Therefore, the lines are neither parallel nor coincident; they intersect at exactly one point, giving a unique solution. This can also be verified using the determinant: \(5\times41-12\times17=1\ne0\). Exam tip: if \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\), the pair of lines intersects at one point.

All three ratios are equal: \(\frac{10}{30}=\frac{13}{39}=\frac{-71}{-213}=\frac{1}{3}\). Thus, every coefficient in the second equation is 3 times the corresponding coefficient in the first equation, so both equations represent the same line. The pair therefore has infinitely many solutions. Options A and D are incorrect because they wrongly treat one ratio as unequal. Exam tip: always compare the ratios of the coefficients of \(x\), \(y\), and the constant term, including their signs.

Here, \(a_1=16, b_1=-9, c_1=55\) and \(a_2=32, b_2=-18, c_2=113\). We get \(\frac{a_1}{a_2}=\frac{16}{32}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-9}{-18}=\frac{1}{2}\), whereas \(\frac{c_1}{c_2}=\frac{55}{113}\ne\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the two lines are parallel and the pair has no solution. Exam tip: when the first two coefficient ratios are equal but the constant-term ratio is different, the pair is inconsistent.

Here (15/30 \ne 8/17), so the lines intersect at one point. In this case, one unique solution is obtained.

For a pair of linear equations to have no solution, the ratios of the coefficients of x and y must be equal, but this ratio must differ from the ratio of the constants: \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). Here, \(\frac{c}{24}=\frac{18}{48}=\frac{3}{8}\), so \(c=9\). Also, \(\frac{72}{145}\ne\frac{3}{8}\), confirming that the lines are distinct and parallel. Exam tip: First equate the ratios of the x- and y-coefficients to find the parameter, then verify the constants’ ratio.

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, \(\frac{7}{28}=\frac{63}{252}=\frac{1}{4}\). Therefore, \(\frac{d}{36}=\frac{1}{4}\), giving \(d=9\). Option 8 is a close distractor, but it gives \(\frac{d}{36}=\frac{2}{9}\), which is not equal to \(\frac{1}{4}\). Exam tip: For infinitely many solutions, all three corresponding coefficient-to-constant ratios must be equal.

Two linear equations have a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{17}{8}\ne\frac{p}{3}\) is required. Cross-multiplication gives \(51\ne8p\), or \(p\ne\frac{51}{8}\). Hence, option B is correct. Exam tip: For a unique solution, the ratios of corresponding coefficients must be unequal; in option A, the ratios become equal, so the equations do not have a unique solution.

For two linear equations to have infinitely many solutions, all corresponding coefficients and constants must be proportional. Here, the coefficients of x and y in the second equation are three times those in the first: 33=3×11 and 21=3×7. Therefore, the constant term must also be three times 59, so n=3×59=177. Hence, option C is correct. Exam tip: infinitely many solutions require
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Dividing the first equation by 8 gives \(2x-y=8\). Thus, the left-hand sides of the two equations are identical. If \(t=8\), both equations represent the same line and have infinitely many solutions. For the equations to be inconsistent, their right-hand sides must differ; hence, \(t\ne 8\). Exam tip: identical left-hand sides with different right-hand sides imply no solution.

The second equation is exactly 3 times the first: multiplying 8x+15y=73 by 3 gives 24x+45y=219. Hence, both equations represent the same line, so the system has infinitely many solutions. Exam tip: if \(a_1/a_2=b_1/b_2=c_1/c_2\), the lines are coincident and infinitely many solutions exist.