Conditions for solvability

In the two equations, \(a_1=3, b_1=1\) and \(a_2=1, b_2=-2\). Since \(\frac{a_1}{a_2}=\frac{3}{1}=3\) and \(\frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\) are unequal, the two lines intersect at exactly one point. Therefore, the pair has one unique solution. Option B would apply only when the lines are parallel, which requires \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\). Exam tip: retain the negative sign while comparing ratios of coefficients.

Here, \(\frac{3}{6}=\frac{4}{8}=\frac{22}{44}=\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), so the two linear equations represent coincident lines and have infinitely many solutions. Exam tip: when all three ratios are equal, the lines are coincident; when only the coefficient ratios are equal, they are parallel; and when the first two ratios are equal but the third differs, the system is inconsistent.

Compare the ratios of corresponding coefficients: \(\frac{8}{2}=4\) and \(\frac{12}{3}=4\), whereas \(\frac{40}{12}=\frac{10}{3}\). Hence, \(\frac{8}{2}=\frac{12}{3}\ne\frac{40}{12}\), so the two lines are parallel and distinct and the pair has no solution. Option B is incorrect because the ratio of the constant terms is not equal to the other two ratios. Exam tip: If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the pair of linear equations has no solution.

Here, \(a_1=6, a_2=12, b_1=7, b_2=13\). Thus, \(\frac{a_1}{a_2}=\frac{6}{12}=\frac{1}{2}\), whereas \(\frac{b_1}{b_2}=\frac{7}{13}\); the two ratios are unequal. For a pair of linear equations, the condition \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\) indicates a unique solution. In fact, solving the equations gives \(y=1\) and \(x=8/3\), so the value \(y=2\) in option A is incorrect. In an exam, compare the ratios of the coefficients of x and y first; if they are unequal, the pair has a unique solution.

For a pair of consistent and dependent linear equations, the corresponding coefficients and constants must be proportional, giving infinitely many solutions. Here, multiplying the first equation by 2 gives \(2(2x+5y=17)\), or \(4x+10y=34\). Therefore, \(s=34\), so option C is correct. If \(s\) were 33 or 35, the equations would not represent the same line and would not be dependent. Exam tip: For dependent equations, check \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

The first two ratios are equal, so for inconsistency the constant ratio must be different. Hence, (r \ne 58).

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, \(\frac{3}{6}=\frac{1}{2}\) and \(\frac{10}{25}=\frac{2}{5}\), so the constant-term ratio is different. Therefore, set \(\frac{4}{a}=\frac{3}{6}=\frac{1}{2}\), which gives \(a=8\). Hence, option C is correct. Exam tip: First equate the ratios of the coefficients of the variables, then verify that the ratio of 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, \(\frac{6}{2}=3\) and \(\frac{42}{14}=3\), so \(\frac{k}{5}=3\), giving \(k=15\). Therefore, option B is correct. With option C, \(k=18\), the ratio \(\frac{k}{5}\) is not 3, so the equations do not represent the same line. Exam tip: For infinitely many solutions, check that all three corresponding ratios are equal.

For two linear equations to have no solution, the ratios of the coefficients of x and y must be equal, while the ratio of the constant terms must be different. Therefore, l/8 = 9/18, so l/8 = 1/2 and l = 4. Also, 27/55 is not equal to 1/2, confirming that the two lines are parallel and distinct. Exam tip: For ‘no solution’, remember a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

The second equation is exactly twice the first: multiplying 9x+2y=31 by 2 gives 18x+4y=62. Hence, both equations represent the same line and the pair has infinitely many solutions. Exam tip: when the ratios of the corresponding coefficients and constant terms are all equal, the two lines are coincident.

Here, a₁/a₂ = 12/3 = 4 and b₁/b₂ = (−8)/(−2) = 4, whereas c₁/c₂ = 20/6 = 10/3. Thus, a₁/a₂ = b₁/b₂ ≠ c₁/c₂. The two lines are parallel and distinct, so they have no common point or solution. Exam tip: If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, a pair of linear equations has no solution.

The determinant of the coefficients is \\(15 \times 2 - 7 \times 4 = 2\\), which is non-zero. Therefore, the two lines intersect at exactly one point, so the pair has a unique solution. Infinitely many solutions would require the corresponding ratios of the coefficients and constant terms to be equal. Exam tip: If \\(a_1b_2-a_2b_1 \ne 0\\), a pair of linear equations has exactly one unique solution.

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, \(\frac{3}{12}=\frac{24}{96}=\frac{1}{4}\). Therefore, \(\frac{m}{28}=\frac{1}{4}\), giving \(m=7\). Hence, option C is correct. Exam tip: For infinitely many solutions, the ratios of the coefficients of \(x\), \(y\), and the constants must all be equal.

The ratios of the coefficients of x and y are equal: 11/22 = (-4)/(-8) = 1/2. However, the ratio of the constants on the right side is 26/53, which is not equal to 1/2. Thus, the two lines have the same slope but different intercepts, so they are parallel and distinct. They have no common point and hence no solution. Infinitely many solutions would require all three corresponding ratios to be equal. Exam tip: if a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair has no solution.

Here (5/9 \ne 7/13), so the lines will intersect at one point. If the first two ratios differ, one unique solution is obtained.

For two equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\), if the ratios of the coefficients of \(x\) and \(y\) are equal but the ratio of the constants is different, the lines are distinct and parallel. They never meet, so the system is inconsistent and has no solution. This condition must be checked carefully rather than judging only from one coefficient.

Here, \(4/8=3/6=1/2\), but \(125/260=25/52\), which is not \(1/2\). Thus the variable coefficients are proportional while the constants are not. The equations represent two different parallel lines, so no pair of prices satisfies both equations simultaneously. Hence the system is inconsistent, and option C is correct.

The second equation must be (3) times the first so (k=48). For infinitely many solutions all ratios must be equal.

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, \(\frac{a}{9}=\frac{4}{12}=\frac{1}{3}\), giving \(a=3\). Also, \(\frac{20}{61}\ne\frac{1}{3}\), so the two lines are distinct and parallel and therefore have no solution. In an exam, first equate the ratios of the variable coefficients to find the parameter, then verify the ratio of the constants.

The first two ratios are equal so the constant ratio must differ for inconsistency. Hence (m \ne 26).

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