Conditions for solvability

The second equation is exactly three times the first: multiplying 6x+7y=41 by 3 gives 18x+21y=123. Thus both equations represent the same line, so the pair has infinitely many solutions. It does not represent two distinct parallel lines, which would give no solution; the lines are also not perpendicular because their slopes are equal. Exam tip: for equations in standard form, if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the pair has infinitely many solutions.

For the two equations in standard form, \(a_1/a_2=12/3=4\) and \(b_1/b_2=16/4=4\), whereas \(c_1/c_2=36/11\), which is not 4. Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the lines are distinct and parallel and the pair has no solution. Exam tip: when the first two coefficient ratios are equal but the constant ratio is different, the pair has no solution.

The coefficient-ratio test determines how two linear equations are related. If the ratios of the x and y coefficients are different, the lines have different slopes. Different slopes guarantee one intersection point, which means the pair is consistent and independent and has a unique solution. The constant terms do not change this conclusion once the first two ratios are unequal.

For these equations, \\(11/5=2.2\\), but \\(6/3=2\\). Hence \\(11/5\\ne6/3\\). The lines therefore have different slopes and intersect at exactly one point. They cannot be coincident, because coincident lines would have equal ratios for all corresponding coefficients. Thus option C is correct.

The relevant principle is that equations whose all corresponding terms have the same multiplier describe the same line. Multiply 2x+3y=11 by 8. The result is 16x+24y=88, exactly the first equation. In ratio form, 16/2=24/3=88/11=8, so the coefficients of x and y and the constants are all proportional. The two equations therefore do not impose two different conditions; they are duplicate descriptions of one line. Every point satisfying one equation satisfies the other, giving infinitely many solutions. Such a pair is called consistent and dependent, so option A is correct. It is not inconsistent, because the lines are not distinct parallel lines. It is not independent, because there is not one unique intersection point; and “unsolvable” is not appropriate when infinitely many solutions exist.

The coefficients of the first equation are 9 times those of the second: \(18/2=27/3=9\). However, the ratio of the constants is \(63/8\), which is not 9. Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the two lines are parallel but distinct and the pair is inconsistent; it has no solution. For coincident lines, all three ratios must be equal. Exam tip: always compare the constant-term ratio along with the coefficient ratios.

For a pair of linear equations, if \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\), the pair has a unique solution and is called consistent and independent. Here, \(\frac{13}{6} \ne \frac{4}{2}\), so the two lines intersect at exactly one point. Therefore, option C is correct. Option B would require infinitely many solutions, which occurs only when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). Exam tip: first compare \(a_1/a_2\) and \(b_1/b_2\).

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

Here, \(a_1/a_2=5/10=1/2\) and \(b_1/b_2=4/8=1/2\), but \(c_1/c_2=180/370=18/37\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the two lines are distinct and parallel and do not intersect. Therefore, the system is inconsistent and has no solution. Exam tip: Compare the ratios of the coefficients of \(x\) and \(y\) first, and then compare the ratio of the constants.

The determinant of the coefficients is \\(4(-3)-1(1)=-13\\ne0\\). Therefore, the two lines intersect at exactly one point, so the pair has one unique solution. Option B would apply only when the lines are parallel, with equal ratios of the corresponding coefficients. Exam tip: if \\(a_1b_2-a_2b_1\\ne0\\), a pair of linear equations has a unique solution.

Here, a₁/a₂ = 5/15 = 1/3, b₁/b₂ = 6/18 = 1/3, and c₁/c₂ = 32/96 = 1/3. Hence, all three ratios are equal, so the two equations represent coincident lines and have infinitely many solutions. In an exam, compare a₁/a₂, b₁/b₂, and c₁/c₂ in that order for equations written as ax + by + c = 0.

In option A, the second equation is exactly twice the first: multiplying \(2x+3y=6\) by 2 gives \(4x+6y=12\). Thus, both represent the same line and have infinitely many solutions. In B, the constant ratio differs, so the lines are parallel. Exam tip: compare all three ratios.

For the pair, \(a_1/a_2=7/14=1/2\) and \(b_1/b_2=8/15\), and these ratios are unequal. Thus, \(a_1/a_2\ne b_1/b_2\), so the two lines intersect at one point and the equations have a unique solution. Options A and C incorrectly treat the ratios as equal; coincident lines require all corresponding ratios to be equal. Exam tip: first compare \(a_1/a_2\) and \(b_1/b_2\).

For a pair of linear equations to be consistent and dependent, all corresponding coefficients and constants must be proportional. The coefficients in the second equation are twice those in the first: \(8=2\times4\) and \(10=2\times5\). Therefore, the constant must also be doubled, so \(s=2\times29=58\). Hence, option C is correct. Values such as 56 or 57 would not make the two equations represent the same line. Exam tip: for dependent equations, check whether \(a_1/a_2=b_1/b_2=c_1/c_2\).

The first two ratios are equal so the constant ratio must be different for inconsistency. Hence (r \ne 62).

When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), both equations represent the same line, so infinitely many common points exist. Option B represents distinct parallel lines. Exam tip: compare all three ratios.

For two linear equations to have infinitely many solutions, the ratios of their corresponding coefficients and constant terms must be equal: 8/2 = k/7 = 72/18. Since 8/2 and 72/18 are both 4, k/7 = 4, giving k = 28. Therefore, option C is correct. In an exam, check the ratio of the constant terms as well as the variable coefficients.

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, \(\frac{l}{10}=\frac{11}{22}=\frac{1}{2}\), so \(l=5\). Also, \(\frac{44}{91}\neq\frac{1}{2}\), confirming that the two lines are parallel and distinct. Exam tip: For ‘no solution’, first equate the ratios of the coefficients and then check that the ratio of constants is different.

The second equation is exactly twice the first: 24x+10y=2(12x+5y) and 82=2×41. Therefore, both equations represent the same line and the pair has infinitely many solutions. Option A would be correct if the ratios of the coefficients of x and y were equal but the ratio of the constants were different. In an exam, compare a₁/a₂, b₁/b₂, and c₁/c₂ to determine the position of the lines.

For the coefficients, \(\frac{14}{7}=\frac{-10}{-5}=2\), but the ratio of the constant terms is \(\frac{36}{19}\), which is not 2. Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the two lines are parallel and distinct and have no common solution. Exam tip: when the first two ratios are equal but the constant-term ratio is different, choose ‘no solution’.

For two linear equations, a unique solution exists when \\(a_1/a_2 \\ne b_1/b_2\\). Here, \\(17/8 \\ne 6/3\\); equivalently, the determinant is \\(17 \\times 3 - 8 \\times 6 = 3 \\ne 0\\). Therefore, the two lines intersect at exactly one point, so the system has one unique solution. Exam tip: Compare corresponding coefficients in the same order in both equations.