Conditions for solvability

Compare the ratios of the corresponding coefficients: \(8/16=3/6=1/2\), whereas the ratio of the constant terms is \(19/45\), which is not \(1/2\). Thus, the two lines are parallel and distinct, so they have no common point and the pair has no solution. Infinitely many solutions would require all three ratios to be equal. Exam tip: if \(a_1/a_2=b_1/b_2\ne c_1/c_2\), the pair has no solution.

Here (5/9 \ne 11/20), so the lines intersect at one point. If the first two ratios are different, one unique solution is obtained.

If the first two ratios are different, the lines intersect at one point. Therefore, the pair is consistent and independent.

If the first two ratios are equal and the third differs, the lines are parallel and distinct. Therefore, there is no common solution.

For two linear equations to have infinitely many solutions, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) must hold. Here, \(\frac{10}{15}=\frac{30}{45}=\frac{2}{3}\), so \(\frac{k}{12}=\frac{2}{3}\), giving \(k=8\). Hence, option B is correct. Exam tip: for infinitely many solutions, verify equality of all three coefficient and constant-term ratios, not just two of them.

A pair of linear equations has infinitely many solutions when the ratios of the corresponding coefficients and constant terms are equal. Here, the second equation must be twice the first: 18x = 2(9x) and 72 = 2(36). Therefore, 14 = 2q, giving q = 7. Hence, option C is correct. Exam tip: For infinitely many solutions, check the condition a₁/a₂ = b₁/b₂ = c₁/c₂.

Two linear equations have a unique solution when \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\). Here, \(\frac{8}{2} \ne \frac{p}{3}\), which gives \(4 \ne \frac{p}{3}\), so \(p \ne 12\). If \(p=12\), the ratios of the coefficients become equal, but the ratio of the constant terms does not; hence the equations have no solution rather than a unique solution. In an exam, compare the coefficient ratios first to identify the condition for a 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, \(\frac{7}{21}=\frac{49}{147}=\frac{1}{3}\). Therefore, \(\frac{a}{18}=\frac{1}{3}\), giving \(a=6\). Hence, option C is correct. Exam tip: For infinitely many solutions, all three corresponding coefficient and constant ratios must be equal; equality of only two ratios is not sufficient.

For two linear equations to have no solution, the condition is \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\). Here, \(\frac{b}{10}=\frac{12}{30}=\frac{2}{5}\), so \(b=4\). Also, \(\frac{48}{101}\neq\frac{2}{5}\), so the two lines are parallel and inconsistent. Exam tip: first equate the ratios of the coefficients of x and y, then verify that the ratio of the constant terms is different.

To classify two linear equations graphically, compare the ratios of corresponding coefficients. If the ratios of the coefficients of \(x\) and \(y\) are equal but the ratio of the constants is different, the equations represent distinct parallel lines. Such lines have the same direction but lie at different positions, so they never meet.

For these equations, \(18/3=6\) and \(24/4=6\), so the variable coefficients are proportional. However, \(54/11\ne6\). Therefore the constant terms do not have the same proportion. The lines are parallel but not coincident, and the pair has no common solution. Thus the correct graph description is option C, distinct parallel lines.

Every term of the first equation is 5 times the corresponding term of the second: 20=5×4, 15=5×3, and 85=5×17. Hence, both equations represent the same line and have infinitely many common solutions. Option B is incorrect because distinct parallel lines require the coefficient ratios to be equal but different from the constant-term ratio. Exam tip: if a₁/a₂ = b₁/b₂ = c₁/c₂, the two lines are coincident.

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, so the pair has a unique solution. Here, \(\frac{3}{8} \ne \frac{13}{29}\); therefore, the lines intersect at one point. Distinct parallel lines would require \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), which is not true here. Exam tip: compare the ratios of the coefficients of \(x\) and \(y\) first.

Same slope and same intercept mean both lines are identical. Therefore, the pair has infinitely many solutions.

The slope tells the direction of a line, while the y-intercept tells where it crosses the y-axis. Lines with the same slope move in the same direction. If their y-intercepts are different, they start at different heights and therefore cannot be the same line. They remain separate and parallel.

Because distinct parallel lines never meet, there is no ordered pair satisfying both linear equations. In the language of systems, the pair is inconsistent and has zero solutions. It is not dependent or infinitely solvable; those descriptions apply when both equations represent the same line. Therefore the correct choice is option B, inconsistent.

Two lines with different slopes cannot be parallel or coincident, so they intersect at exactly one point. Hence, the pair of linear equations is consistent and independent and has one unique solution. Options B and D describe infinitely many solutions, which occur for coincident lines having the same slope. Exam tip: different slopes imply one unique solution; equal slopes with different intercepts imply no solution; and equal slopes with equal intercepts imply infinitely many solutions.

Here, \(\frac{6}{18}=\frac{1}{3}\), \(\frac{11}{33}=\frac{1}{3}\), and \(\frac{-47}{-141}=\frac{1}{3}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), so the two lines are coincident and the pair has infinitely many solutions. In an exam, retain the signs of the constant terms while forming the ratios.

Here, \(a_1=10, b_1=-7, c_1=31\) and \(a_2=20, b_2=-14, c_2=65\). We get \(\frac{a_1}{a_2}=\frac{10}{20}=\frac{1}{2}\) and \(\frac{b_1}{b_2}=\frac{-7}{-14}=\frac{1}{2}\), but \(\frac{c_1}{c_2}=\frac{31}{65}\ne\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the pair is inconsistent and has no solution because the corresponding lines are parallel. Exam tip: Compare the ratios of the coefficients of both variables first, and then compare them with the ratio of the constants.

Here (11/22 \ne 4/9), so the lines intersect at one point. In this case, one unique solution is obtained.

For two linear equations to have no solution, the ratios of the coefficients of the variables must be equal, while the ratio of the constant terms must be different. Thus, \(\frac{c}{15}=\frac{14}{35}=\frac{2}{5}\), giving \(c=6\). Also, \(\frac{56}{121}\neq\frac{2}{5}\), so the two lines are distinct and parallel. Hence, option (6) is correct. Exam tip: For no solution, use \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\).

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, \(\frac{5}{20}=\frac{40}{160}=\frac{1}{4}\). Thus, \(\frac{d}{28}=\frac{1}{4}\), which gives \(d=7\). Therefore, option C is correct. Remember that the ratios of both variable coefficients and the constants must be equal; matching only one ratio is insufficient.