Conditions for solvability

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.

The first two ratios are equal so the constant ratio must be different for inconsistency. Hence (t \ne 66) is correct.

For two linear equations to have infinitely many solutions, the ratios of the corresponding coefficients and constants must be equal: 4/12 = a/21 = 36/108. Since 4/12 and 36/108 are both 1/3, we get a/21 = 1/3, so a = 7. Therefore, option C is correct. Exam tip: For infinitely many solutions, all three corresponding ratios must be equal; checking only two ratios is not sufficient.

For two linear equations to have a unique solution, the condition is \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{5}{15}\ne\frac{2}{k}\); cross-multiplication gives \(5k\ne30\), so \(k\ne6\). Therefore, option B is correct. In option A, \(k=6\) makes the corresponding coefficients proportional, so the equations do not have a unique solution. Exam tip: a unique solution requires unequal ratios of the corresponding coefficients.

Here, \(a_1/a_2=6/18=1/3\) and \(b_1/b_2=(-7)/(-21)=1/3\), whereas \(c_1/c_2=13/40\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), which represents an inconsistent pair of equations. The two lines are distinct and parallel, so they have no solution. Exam tip: Compare the ratios of the coefficients of \(x\), \(y\), and the constants in this order to identify the solution status quickly.

The solution status of two linear equations can be determined by comparing a₁/a₂ and b₁/b₂. For 2x + 7y = 160 and 5x + 16y = 370, the first coefficient ratio is 2/5, while the second is 7/16. These are unequal because 2 × 16 = 32 whereas 7 × 5 = 35. Thus the two equations represent lines with different slopes. Lines with different slopes intersect at exactly one point, so the pair is consistent and independent and has one unique solution. The constant terms are not needed once the first two ratios are unequal. “No solution” would require equal coefficient ratios but a different constant ratio; infinitely many solutions would require all three ratios to be equal. Therefore option A is correct.

The first equation is 7x+9y=43. Multiplying every term by 2 gives 14x+18y=86, which is exactly the second equation. Multiplying an equation by a nonzero number does not change its set of solutions; it only writes the same relationship in a different form.

Therefore both equations represent one and the same straight line on the graph. Every point on that line satisfies both equations, so the pair has infinitely many solutions. Option B is correct. Distinct parallel lines would have proportional x- and y-coefficients but non-proportional constants; here the constant is also doubled.

Multiply the first equation, 5x − 3y = 11, by 2. The result is 10x − 6y = 22, exactly the second equation. Equivalently, the ratios of the corresponding coefficients and constants are all equal: 10/5 = (−6)/(−3) = 22/11 = 2. Thus the equations describe coincident lines. They share every point on that line, so infinitely many ordered pairs satisfy them; a unique solution is not possible.

Multiplying every term of 6x − 5y = 17 by 3 gives 18x − 15y = 51, which is the second equation. Hence the two equations are not independent; they are different algebraic forms of the same line. In ratio form, 18/6 = (−15)/(−5) = 51/17 = 3, confirming coincidence. A coincident pair has every point of the common line as a solution, so the number of solutions is infinite, not one or zero.

For the pair 7x + 10y = 39 and 5x + 8y = 31, compare the ratios of the corresponding coefficients. The x-coefficient ratio is 7/5 and the y-coefficient ratio is 10/8 = 5/4. These are not equal: cross multiplication gives 7 × 8 = 56, while 10 × 5 = 50. Therefore the two lines have different slopes and must intersect at exactly one point. The pair is consequently consistent and independent, with one unique solution. Comparing the constant ratio is unnecessary once the first two ratios differ. No solution would require equal x- and y-coefficient ratios but a different constant ratio; infinitely many solutions would require all three ratios to be equal. Thus option C states the correct conclusion.

The first equation is 9x − 2y = 31. Multiplying it by 3 produces 27x − 6y = 93, exactly the second equation. Therefore, the pair represents two coincident lines, not two intersecting or distinct parallel lines. The coefficient ratios also confirm this: 27/9 = (−6)/(−2) = 93/31 = 3. Every point satisfying the first equation automatically satisfies the second, so infinitely many solutions exist.

Compare the ratios of the corresponding x- and y-coefficients in the equations 6x + 17y = 71 and 11x + 31y = 130. The ratios are 6/11 and 17/31. They are unequal, since 6 × 31 = 186 whereas 17 × 11 = 187. Therefore the two equations represent lines with different slopes. Two lines with different slopes intersect at one and only one point, so the pair is consistent and independent and has a unique solution. It is not necessary to compare the constant ratio after finding the first two ratios unequal. If the first two ratios were equal but the constant ratio differed, the pair would be inconsistent; if all three were equal, it would have infinitely many solutions. Hence option C is correct.

Multiply the first equation, 18x + 5y = 79, by 2. It becomes 36x + 10y = 158, exactly the second equation. Thus both equations have the same graph and represent one coincident line. The phrase “the lines are the same” is therefore the most precise statement; such a pair has infinitely many solutions. Distinct parallel lines would have proportional x and y coefficients but a different constant ratio, while intersecting lines would not have proportional coefficients.

The second equation is obtained by multiplying every term of the first equation by 3: 3(2x − 5y = 1) gives 6x − 15y = 3. Therefore, both equations represent the same line. In ratio form, 6/2 = (−15)/(−5) = 3/1 = 3, confirming that all three ratios are equal. Coincident lines have infinitely many common points, so the pair has infinitely many solutions. Two linear equations cannot have exactly two isolated intersection points.

The governing concept is the consistency condition for a pair of linear equations in two variables. Write the equations in the form a₁x+b₁y+c₁=0 or compare the corresponding coefficients directly. For the x-coefficients, 4/12=1/3, and for the y-coefficients, (-7)/(-21)=1/3. Thus the two left-hand sides are proportional. However, the constants have ratio 5/17, which is not equal to 1/3. Therefore the two equations represent distinct parallel lines, so they do not intersect and have no common solution. Hence option C is correct. Infinitely many solutions would require the constant ratio to match as well; a unique solution would require unequal coefficient ratios.

Multiplying 2x + 3y = 5 by 2 gives 4x + 6y = 10, exactly the second equation. Thus the two equations are equivalent and represent coincident lines. The ratio test confirms this: 4/2 = 6/3 = 10/5 = 2. Every point on the common line satisfies both equations, so the pair has infinitely many solutions. The value x = 0 is not the only possibility; for example, x = 1 gives y = 1, which satisfies both equations.

The governing concept is the no-solution condition for two linear equations: a₁/a₂=b₁/b₂ but c₁/c₂ must be different. Here the coefficient ratio for y is 4/8=1/2. To make the x-coefficient ratio equal to it, require (k+1)/6=1/2. Multiplying by 6 gives k+1=3, so k=2. For this value, the equations have proportional x- and y-coefficients, but their constants have ratio 9/15=3/5, which is not 1/2. Hence the corresponding lines are distinct and parallel, producing no solution. Therefore option B is correct. The other choices do not make the two variable-coefficient ratios equal, so they do not satisfy the required condition.

A pair of linear equations has infinitely many solutions when the corresponding coefficients, including the constants, are proportional: a₁/a₂ = b₁/b₂ = c₁/c₂. For the given equations, 7/14 = 1/2 and 13/26 = 1/2. Therefore, the y-coefficient ratio must also be 1/2, so (m − 4)/6 = 1/2. Multiplying by 6 gives m − 4 = 3, and hence m = 7. With this value, the first equation becomes exactly one-half of the second, so both represent the same line and every point on that line is a common solution. Thus option B is correct. The other values destroy proportionality and cannot give infinitely many solutions.