Conditions for solvability

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.

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.

The direct answer is option C: consistent and independent. For a pair a1x+b1y=c1 and a2x+b2y=c2, if a1/a2 is not equal to b1/b2, the two lines have different slopes. Different slopes force the lines to intersect at exactly one point. A common point means the pair is consistent; exactly one common point means it is independent. Option A, inconsistent, applies when the lines are distinct and parallel, typically when a1/a2=b1/b2 but c1/c2 differs. Option B, consistent and dependent, applies to coincident lines with infinitely many solutions, when all three ratios are equal. Option D, no solution, is another description of inconsistency and contradicts the given unequal first two ratios. Memory cue: unequal first two ratios means one intersection, so choose consistent 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.

The direct answer is option C: infinitely many solutions. A line written as y=mx+c is determined by its slope m and y-intercept c. If two lines have the same slope and the same y-intercept, their equations are identical, so they are the very same line. Every point on that line satisfies both equations, giving infinitely many common solutions. Option A, no solution, applies to distinct parallel lines: they have the same slope but different intercepts. Option B, one unique solution, applies to lines with different slopes that cross once. Option C is correct because identical lines overlap at every point. Option D, exactly two solutions, is not a possible result for a pair of ordinary straight-line equations. Remember: same slope plus same intercept means coincident lines, not merely parallel lines.

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.