Conditions for solvability

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.

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.

Every term of the second equation is twice the corresponding term of the first: 24x+10y=82 = 2(12x+5y=41). Hence, both equations represent the same line and the system has infinitely many solutions. Options A or D would apply only when the corresponding ratios are unequal. Exam tip: if \(a_1/a_2=b_1/b_2=c_1/c_2\), the lines coincide and the system has infinitely many solutions.

The ratios of the corresponding coefficients are \(14/7=2\) and \((-10)/(-5)=2\). However, the ratio of the constant terms is \(36/19\), which is not equal to 2. Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the two lines are parallel and distinct. Therefore, the pair has no solution. Option B would be correct only if all three ratios were equal, while option C requires the first two ratios to be unequal. Exam tip: For consistency questions, compare the three ratios in the order \(a_1/a_2\), \(b_1/b_2\), and \(c_1/c_2\).

For a pair of linear equations, if \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), the two lines intersect at exactly one point and have a unique solution. Here, \(\frac{17}{8}\ne\frac{6}{3}\), so the correct status is one unique solution. Infinitely many solutions would require all three corresponding ratios to be equal. Exam tip: first compare the ratios of the coefficients of \(x\) and \(y\).

The first two ratios are equal, so the constant ratio must be different for inconsistency. Hence, (s \ne 78) is correct.

A pair of linear equations has infinitely many solutions when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). Here, \(\frac{5}{15}=\frac{15}{45}=\frac{1}{3}\), so \(\frac{m}{18}=\frac{1}{3}\), giving \(m=6\). Therefore, option C is correct. For instance, if \(m=5\), then \(\frac{m}{18}=\frac{5}{18}\), which is not equal to \(\frac{1}{3}\). Exam tip: For infinitely many solutions, all three corresponding ratios must be equal; equality of only two ratios is not sufficient.

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, \(\frac{7}{14}=\frac{1}{2}\), whereas \(\frac{20}{47}\ne\frac{1}{2}\). Thus, the required condition is \(\frac{a}{10}\ne\frac{20}{47}\), which gives \(a\ne\frac{200}{47}\). Therefore, option A is correct. In option B, all three ratios become equal, giving infinitely many solutions. Although \(a=5\) is one particular value that produces no solution, it is not the complete condition. Exam tip: first compare the ratios of the coefficients of x and y, and then compare them with the ratio of the constants.

Two linear equations have a unique solution when the determinant of their coefficient matrix is non-zero. Here, the determinant is \(2k-15\), so \(2k-15\ne0\), which gives \(k\ne\frac{15}{2}\). Therefore, option D is correct. For option A, the determinant becomes zero, so the equations do not have a unique solution. Exam tip: For parameter-based questions, apply \(a_1b_2-a_2b_1\ne0\) directly.

Here, \(a_1=6, b_1=-4, c_1=11\) and \(a_2=15, b_2=-10, c_2=28\). We have \(\frac{a_1}{a_2}=\frac{6}{15}=\frac{2}{5}\) and \(\frac{b_1}{b_2}=\frac{-4}{-10}=\frac{2}{5}\), but \(\frac{c_1}{c_2}=\frac{11}{28}\), which is not equal to \(\frac{2}{5}\). Thus, the two lines are distinct and parallel, so they do not intersect. Therefore, the pair has no solution. Exam tip: If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the lines are parallel and the pair has no solution.

Compare the second equation with the first. Multiplying 3x+8y=27 by 3 gives 9x+24y=81, exactly the second equation. Since multiplication by a nonzero constant does not change the solutions, both equations describe the same relationship between x and y.

Consequently, their graphs are not two different lines. They lie on top of one another and have infinitely many common points. Therefore option B, one same line, is correct. Two distinct parallel lines would have proportional coefficients but different proportional constants; here every term, including the right-hand side, has been multiplied by 3.

For equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, a unique solution occurs when a₁/a₂ ≠ b₁/b₂. Here the ratios of the x- and y-coefficients are 11/4 and 7/3. They are unequal because 11 × 3 = 33, whereas 7 × 4 = 28. Hence the corresponding lines have different slopes and intersect at exactly one point. A pair with one common solution is called consistent and independent. It is not dependent, because dependent equations represent the same line and require all three coefficient ratios to be equal. It is not inconsistent or parallel, because those descriptions apply when the lines have the same slope but are distinct. Therefore option C is correct.