Conditions for solvability

Every coefficient and the constant in the first equation is 5 times the corresponding term in the second: \(15x+10y=40=5(3x+2y=8)\). Hence, both equations have the same solution set, so their graphs coincide as the same line. Option B is incorrect because distinct parallel lines require the coefficients to be proportional but the constants not to be proportional. Exam tip: when one complete linear equation is a non-zero multiple of the other, the lines are coincident and the pair has infinitely many solutions.

For the two lines, \(\frac{a_1}{a_2}=\frac{2}{5}\) and \(\frac{b_1}{b_2}=\frac{11}{19}\). Since \(\frac{2}{5}\ne\frac{11}{19}\), the lines are neither parallel nor coincident; hence, they intersect at exactly one point and the pair has a unique solution. Perpendicular lines would require the product of their slopes to be \(-1\), which is not the required condition here. Exam tip: Compare \(\frac{a_1}{a_2}\) and \(\frac{b_1}{b_2}\) first to identify the number of solutions.

For the first equation, \(a_1=4, b_1=9, c_1=-31\), and for the second, \(a_2=12, b_2=27, c_2=-93\). Thus, \(\frac{a_1}{a_2}=\frac{4}{12}=\frac{1}{3}\), \(\frac{b_1}{b_2}=\frac{9}{27}=\frac{1}{3}\), and \(\frac{c_1}{c_2}=\frac{-31}{-93}=\frac{1}{3}\). Since all three ratios are equal, the two lines are coincident and the pair has infinitely many solutions. Exam tip: To identify infinitely many solutions, compare all three ratios; equality of only the first two is not sufficient.

For the first equation, \(a_1=8, b_1=-3, c_1=22\), and for the second equation, \(a_2=16, b_2=-6, c_2=47\). Thus, \(\frac{a_1}{a_2}=\frac{8}{16}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\), whereas \(\frac{c_1}{c_2}=\frac{22}{47}\), which is not equal to \(\frac{1}{2}\). Therefore, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the 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 (9/18 \ne 5/11), so the lines intersect at one point. In this case, one unique solution is obtained.

A pair of linear equations has infinitely many solutions when a₁/a₂ = b₁/b₂ = c₁/c₂. Here, 3/12 = 24/96 = 1/4. Therefore, b/20 = 1/4, which gives b = 5. Hence, option C is correct. Exam tip: all three coefficient-to-constant ratios must be equal; checking only one ratio is insufficient.

A pair of linear equations has a unique solution when the ratios of the coefficients of the variables are unequal: \(\frac{11}{4}\ne\frac{p}{2}\). Cross-multiplication gives \(22\ne4p\), so \(p\ne\frac{11}{2}\). Hence, option B is correct. Option A is the boundary value at which the lines become parallel and no solution exists. Exam tip: for a unique solution, check that \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\).

Dividing every term of the second equation by 3 gives \(6x+7y=41\), which is exactly the first equation. Hence, both equations represent the same line and have infinitely many common solutions. Therefore, option C is correct. Option B would apply if the two lines were distinct parallel lines. For exams, remember: if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the pair has infinitely many solutions.

Here, 12/3=4 and 16/4=4, so the ratios of the coefficients of x and y are equal. However, 36/11 is not equal to 4. Thus, a₁/a₂=b₁/b₂≠c₁/c₂, so the two lines are distinct and parallel and the pair has no solution. Exam tip: In this case, the lines have equal slopes but do not intersect.

The equations in this item are the same as in the preceding comparison, and the relevant rule is unchanged. When the ratio of the x coefficients differs from the ratio of the y coefficients, the two lines have different slopes. Two lines with different slopes intersect once, so the pair has exactly one solution. Equality of both variable ratios would be needed for parallel or coincident-line tests.

Calculate the ratios: \\(11/5=2.2\\) and \\(6/3=2\\). Since these values are unequal, \\(11/5\\ne6/3\\). The lines cannot be parallel with the same slope, and they cannot be the same line. They meet at one point, so the system has a unique solution. Therefore option C agrees with the mathematics.

Use the consistency criterion for two linear equations. The second equation is 2x+3y=11. Multiplying every term by 8 gives 16x+24y=88, which is precisely the first equation. Thus the coefficient ratios and constant ratio are equal: 16/2=24/3=88/11=8. Geometrically, both equations represent the same straight line, so they have infinitely many common points. Algebraically, there is only one independent equation for the two variables, leaving one variable free; this is why the pair is dependent. Because at least one common solution exists—in fact infinitely many—the pair is consistent as well. Therefore option A is the only correct answer. Inconsistent would mean no common solution, and independent would mean two non-proportional lines meeting at one point. Neither description fits these proportional equations.

For the first equation, \(a_1=18, b_1=27, c_1=63\), and for the second, \(a_2=2, b_2=3, c_2=8\). Here, \(a_1/a_2=18/2=9\) and \(b_1/b_2=27/3=9\), but \(c_1/c_2=63/8\), which is not 9. Thus, the two lines are distinct and parallel, so they have no common solution. Therefore, the pair is inconsistent. Exam tip: if \(a_1/a_2=b_1/b_2\ne c_1/c_2\), the pair is inconsistent.

The ratios of the coefficients of x and y are unequal: 13/6 ≠ 4/2. Therefore, the two lines intersect at exactly one point, giving a unique solution. Hence, the pair is consistent and independent. Options A and D describe parallel-line cases, while option B would require infinitely many solutions. Exam tip: If a₁/a₂ ≠ b₁/b₂, the pair of linear equations is consistent and independent.

Here, \(a_1/a_2=5/15=1/3\), \(b_1/b_2=6/18=1/3\), and \(c_1/c_2=32/96=1/3\). Thus, all three ratios are equal. Therefore, the two equations represent the same line and have infinitely many solutions. Option B would apply if only the first two ratios were equal, but here the constant-term ratio is also equal. Exam tip: if \(a_1/a_2=b_1/b_2=c_1/c_2\), the lines are coincident.

Here, \(\frac{10}{2}=5\) and \(\frac{15}{3}=5\), whereas \(\frac{50}{13}\ne 5\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the two lines are distinct and parallel and the pair has no solution. Option B is incorrect because \(\frac{50}{13}\) is not equal to 5. Exam tip: Recognize this ratio condition directly to conclude that the pair is inconsistent.

The coefficient ratios are unequal: \(\frac{7}{14}=\frac{1}{2}\), whereas \(\frac{8}{15}\ne\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), so the two lines intersect at exactly one point and the pair has a unique solution. Options C and D describe the inconsistent and coincident-line cases, respectively, neither of which applies here. Exam tip: When \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), a pair of linear equations in two variables has a unique solution.

For a pair of consistent and dependent linear equations, the ratios of the corresponding coefficients and constant terms must be equal. Here, \(\frac{4}{8}=\frac{5}{10}=\frac{1}{2}\), so \(\frac{29}{s}=\frac{1}{2}\). Therefore, \(s=58\). The nearby value 57 is incorrect because it does not preserve the same ratio for the constant terms. Exam tip: when one equation is multiplied by \(k\), its constant term must also be multiplied by \(k\).

The first two ratios are equal, so the constant ratio must be different for inconsistency. Hence, (r \ne 62).

\(\frac{3}{6}=\frac{-2}{-4}=\frac12\), but \(\frac{5}{12}\ne\frac12\). Hence, the lines are parallel and have no solution. Exam tip: always compare all three ratios.

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, \(\frac{8}{2}=\frac{72}{18}=4\), so \(\frac{k}{7}=4\), giving \(k=28\). Therefore, option C is correct. Exam tip: Find the ratio of the constant terms first and make the corresponding coefficient ratio equal to it.