Conditions for solvability

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.

Here, \(a_1/a_2=6/12=1/2\), while \(b_1/b_2=5/10=1/2\) and \(c_1/c_2=240/490=24/49\). Thus, \(a_1/a_2=b_1/b_2\neq c_1/c_2\), so the two lines are distinct and parallel and have no common solution. Therefore, the system is inconsistent. It is not consistent and dependent, because that case requires all three ratios to be equal and gives infinitely many solutions. Exam tip: Compare all three corresponding ratios in such questions.

When \(a_1/a_2\ne b_1/b_2\), the two lines have different slopes and intersect at exactly one point, so the solution is unique. Option B represents parallel lines. Exam tip: compare the ratios of the \(a\) and \(b\) coefficients first.

Here, a₁/a₂ = 8/2 = 4, b₁/b₂ = 12/3 = 4, and c₁/c₂ = 40/10 = 4. Hence, a₁/a₂ = b₁/b₂ = c₁/c₂, so all three ratios are equal. This means that the two equations represent the same line and have infinitely many solutions. Option B is incorrect because the third ratio is not different; it is also equal to 4. Exam tip: when all three ratios are equal, the pair of linear equations represents coincident lines and has infinitely many solutions.

Dividing the first equation by 4 gives \(4x-3y=11\), whereas the second equation is \(4x-3y=12\). The ratios of the coefficients of \(x\) and \(y\) are equal, but the ratio of the constant terms is different: \(16/4=(-12)/(-3)=4\), while \(44/12=11/3\). Hence, the two lines are parallel and distinct, so they have no common solution. Option C would be correct only if all three ratios were equal. Exam tip: if \(a_1/a_2=b_1/b_2\ne c_1/c_2\), the pair of linear equations is inconsistent and has no solution.

Here, \(\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{2}{5}\). Since \(\frac{1}{2}\ne\frac{2}{5}\), we have \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), so the two lines intersect at exactly one point. Therefore, the pair has a unique solution. Options A and C incorrectly assert equality of ratios, while coincident lines require all three ratios to be equal. In an exam, first compare \(\frac{a_1}{a_2}\) and \(\frac{b_1}{b_2}\) to identify a unique solution quickly.

For two linear equations to have no solution, the ratios of the coefficients of x and y must be equal, but this common ratio must differ from the ratio of the constants: c/9 = 10/15 ≠ 50/80. Since 10/15 = 2/3, c/9 = 2/3 gives c = 6. Also, 50/80 = 5/8, which is different from 2/3, so the lines are parallel and distinct. Hence, the correct answer is (6). Exam tip: First equate the ratios of the x- and y-coefficients, then verify that the constants have a different ratio.

For two linear equations to have infinitely many solutions, the condition is 4/12 = a/15 = 28/84. Since both 4/12 and 28/84 equal 1/3, we get a/15 = 1/3, giving a = 5. Hence, option B is correct. Exam tip: For infinitely many solutions, all three corresponding coefficient and constant ratios must be equal; checking 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, \(\frac{6}{21}=\frac{2}{7}\), so \(\frac{p}{14}=\frac{2}{7}\), giving \(p=4\). Also, \(\frac{18}{50}=\frac{9}{25}\), which is not equal to \(\frac{2}{7}\); hence the lines are parallel and inconsistent. Exam tip: First equate the ratios of the variable coefficients to find the parameter, then verify that the ratio of constants is different.

The first two ratios are equal, so the constant ratio must be different for inconsistency. Hence, (m \ne 75) is correct.

Multiplying every term of 6x − 5y = 17 by 3 gives 18x − 15y = 51, which is the second equation. Hence the two equations are not independent; they are different algebraic forms of the same line. In ratio form, 18/6 = (−15)/(−5) = 51/17 = 3, confirming coincidence. A coincident pair has every point of the common line as a solution, so the number of solutions is infinite, not one or zero.