Conditions for solvability

Check the three ratios: \(a_1/a_2=8/24=1/3\), \(b_1/b_2=13/39=1/3\), and \(c_1/c_2=(-59)/(-177)=1/3\). Thus, \(a_1/a_2=b_1/b_2=c_1/c_2\); the two lines are coincident and the pair has infinitely many solutions. Options A and D incorrectly state that one ratio is unequal. Exam tip: if all three ratios are equal, there are infinitely many solutions; if only the first two are equal and the third differs, there is no solution.

Here, \(a_1=12, b_1=-5, c_1=37\) and \(a_2=24, b_2=-10, c_2=79\). Thus, \(\frac{a_1}{a_2}=\frac{12}{24}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}\), whereas \(\frac{c_1}{c_2}=\frac{37}{79}\ne\frac{1}{2}\). Therefore, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the two lines are parallel and distinct, giving no solution. Exam tip: When the first two coefficient ratios are equal but the constant-term ratio is different, the pair has no solution.

Here (13/26 \ne 6/13) so the lines intersect at one point. In this case, one unique solution is obtained.

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, \(\frac{16}{42}=\frac{8}{21}\), so \(\frac{c}{21}=\frac{8}{21}\), giving \(c=8\). Also, \(\frac{64}{130}=\frac{32}{65}\), which is not equal to \(\frac{8}{21}\); therefore, the equations are inconsistent. The other values make the coefficients' ratios unequal and lead to a unique solution. Exam tip: equate the ratios of the variable coefficients first, then check that the constant-term ratio 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, \(\frac{6}{18}=\frac{54}{162}=\frac{1}{3}\), so \(\frac{d}{30}=\frac{1}{3}\), giving \(d=10\). If d were 9 or 11, the coefficient ratios would not remain \(1:3\). Exam tip: first compare the ratios of the known coefficients and constants, then use that ratio to find the unknown coefficient.

Two linear equations have a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, the condition is \(\frac{15}{7}\ne\frac{p}{2}\), which gives \(30\ne7p\), or \(p\ne\frac{30}{7}\). Hence, option B is correct. In option A, the two coefficient ratios become equal, so the system does not have a unique solution. Exam tip: For a unique solution, check that the ratios of corresponding coefficients are unequal.

For two linear equations to have infinitely many solutions, the ratios of their corresponding coefficients and constants must be equal. Here, \(27/9=3\) and \(15/5=3\), so the second equation must be 3 times the first one. Hence, \(n=3\times47=141\), making option C correct. Exam tip: For infinitely many solutions, check \(a_1/a_2=b_1/b_2=c_1/c_2\).

When \(a_1/a_2=b_1/b_2\), the two lines have the same slope. Since \(c_1/c_2\) is different, they cannot be the same line; hence they are distinct parallel lines with no solution. Exam tip: equal first two ratios and an unequal third ratio indicate an inconsistent pair.

Every term in the second equation is three times the corresponding term in the first: 27=3×9, 42=3×14, and 213=3×71. Thus, both equations represent the same line, so every point on that line satisfies both equations. Therefore, the pair has infinitely many solutions. A no-solution or distinct-parallel-lines conclusion would require the coefficients of x and y to be proportional but the constants not to be proportional; here all three ratios are equal. Exam tip: If a₁/a₂ = b₁/b₂ = c₁/c₂, the pair has infinitely many solutions.

Here, \(a_1/a_2=21/3=7\) and \(b_1/b_2=28/4=7\), whereas \(c_1/c_2=84/15=28/5\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), which represents inconsistent equations whose lines are parallel. Therefore, the pair has no solution. Exam tip: When the first two coefficient ratios are equal but the constant-term ratio is different, the pair has no solution.

The relationship between two linear equations can be classified by comparing corresponding coefficient ratios. Unequal ratios for the x and y coefficients show that the lines have unequal slopes. Such lines cannot be parallel or identical; they intersect at one point and therefore produce one unique solution.

For this pair, \\(17/8=2.125\\), whereas \\(10/5=2\\). Thus \\(17/8\\ne10/5\\). Because the first two ratios are unequal, the equations represent two lines with different directions. They meet at exactly one point, regardless of the constant-term ratio. Therefore the correct conclusion is one unique solution, given in option C.

A pair of linear equations is dependent when both equations represent exactly the same line. Such a pair is consistent because it has solutions, but it does not have just one solution; every point on the common line satisfies both equations. To identify this situation, compare all corresponding coefficients and constants after multiplying or dividing an equation by a non-zero number.

Here, multiplying the second equation by \(12\) gives \(12(2x+3y=14)\), or \(24x+36y=168\). This is exactly the first equation. Therefore, the two equations are the same equation and their graphs coincide. They have infinitely many common solutions, so the pair is consistent and dependent. Choice A is correct; “independent” would mean that the lines meet at only one point.

Compare the corresponding coefficients: \(\frac{22}{2}=11\) and \(\frac{33}{3}=11\), but \(\frac{99}{11}=9\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\), which is the condition for an inconsistent pair. The two lines are distinct and parallel, so the equations have no solution. Exam tip: compare \(a_1/a_2\), \(b_1/b_2\), and \(c_1/c_2\); the pair is dependent only when all three ratios are equal.

For the given equations, \(a_1/a_2=16/8=2\) and \(b_1/b_2=7/4\). Since \(2\ne 7/4\), the two lines intersect at one point, so the pair has a unique solution. Therefore, it is consistent and independent. A consistent dependent pair would require \(a_1/a_2=b_1/b_2=c_1/c_2\). Exam tip: First compare the ratios of the coefficients of \(x\) and \(y\).

The second equation is (3) times the first. Therefore, both conditions give the same information and have infinitely many solutions.

The ratios of the coefficients are \(\frac{8}{16}=\frac{3}{6}=\frac{1}{2}\), but the ratio of the constant terms is \(\frac{280}{575}=\frac{56}{115}\neq\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\), so the two lines are parallel and have no common solution; the system is inconsistent. Option B would be correct only if all three ratios were equal, giving infinitely many solutions. Exam tip: Compare the three coefficient and constant ratios directly to identify the type of system quickly.

For the given equations, a₁/a₂ = 6/3 = 2 and b₁/b₂ = 5/(-2) = -5/2. Since a₁/a₂ ≠ b₁/b₂, the two lines intersect at exactly one point, so the pair has a unique solution. Infinitely many solutions would require a₁/a₂ = b₁/b₂ = c₁/c₂, which is not true here. Exam tip: first compare the ratios of the coefficients of x and y.

The ratios of the corresponding coefficients and constants are \(\frac{9}{27}=\frac{16}{48}=\frac{74}{222}=\frac{1}{3}\). Hence, all three ratios are equal. When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the two linear equations represent the same line and have infinitely many solutions. Exam tip: Always compare the ratio of the constant terms as well; checking only the coefficients of x and y is not sufficient.

Here, \(a_1=14, b_1=21, c_1=98\) and \(a_2=2, b_2=3, c_2=17\). We get \(\frac{a_1}{a_2}=\frac{14}{2}=7\) and \(\frac{b_1}{b_2}=\frac{21}{3}=7\), but \(\frac{c_1}{c_2}=\frac{98}{17}\) is not 7. Hence, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the two lines are parallel and the pair has no solution. Option B is incorrect because \(\frac{98}{17}\ne7\). Exam tip: compare all three ratios; equality of only the first two ratios means no solution.

Here, (10/20) \ne (11/21), since 10/20 = 1/2 while 11/21 has a different value. Therefore, the two lines intersect at exactly one point, so the pair has a unique solution. Option A would apply when the first two ratios are equal but the constant-term ratio is different; option C requires all three ratios to be equal. Exam tip: for two linear equations, a₁/a₂ \ne b₁/b₂ always indicates a unique solution.