Conditions for solvability

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.

For two linear equations to have infinitely many solutions, the ratios of corresponding coefficients must be equal: a₁/a₂ = b₁/b₂ = c₁/c₂. Here, 3/9 = 12/36 = 1/3. Therefore, k/15 = 1/3, giving k = 5. Hence, option C is correct. Exam tip: Find the common ratio using the coefficients of x and the constants, then use it to determine the coefficient of y.

Two linear equations have no solution when \,\(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). Here, \,\(\frac{a}{6}=\frac{4}{8}=\frac{1}{2}\), so \,\(a=3\). Also, \,\(\frac{10}{25}=\frac{2}{5}\ne\frac{1}{2}\), so the two lines are distinct and parallel, giving no solution. Exam tip: first equate the ratios of the x- and y-coefficients, then check the constant-term ratio.

The first two ratios are equal, so the constant ratio must be different for inconsistency. Hence, (m \ne 34).

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.

The coefficients of the first equation are proportional to the corresponding coefficients of the second equation: \\(\frac{7}{14}=\frac{4}{8}=\frac{1}{2}\\). However, the ratio of the constants is \\(\frac{19}{41}\\), which is not equal to \\(\frac{1}{2}\\). Thus, \\(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\\), so the two lines are distinct parallel lines and do not intersect. Hence, there is no solution. Exam tip: For infinitely many solutions, all three ratios must be equal.

Here (4/9 \ne 7/15), so the lines intersect at one point. If the first two ratios are different, one unique solution is obtained.

If all three ratios are equal, both equations form the same line. Therefore, the pair is consistent and dependent.

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

For infinitely many solutions, (p/10=6/15=18/45) must hold. Since the common ratio is (2/5), (p=4) would be required.

For two linear equations to have infinitely many solutions, a_1/a_2=b_1/b_2=c_1/c_2 must hold. Here, a_1/a_2=7/14=1/2 and c_1/c_2=29/58=1/2 . Therefore, q/10=1/2 , giving q=5 . For instance, if q=4, the coefficient ratio becomes 4/10=2/5, so the equations are not dependent. Exam tip: For infinitely many solutions, check that all three ratios are equal.

Two linear equations have a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(a_1=6, a_2=3, b_1=k, b_2=5\), so \(\frac{6}{3}\ne\frac{k}{5}\), or \(2\ne\frac{k}{5}\). This gives \(k\ne10\). If \(k=10\), the ratios of the coefficients of \(x\) and \(y\) would be equal, so the equations would 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 corresponding coefficients and constants must be equal: \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). Here, \(\frac{5}{15}=\frac{35}{105}=\frac{1}{3}\), so \(\frac{a}{18}=\frac{1}{3}\), giving \(a=6\). Therefore, option C is correct. Exam tip: infinitely many solutions require all three corresponding ratios to be equal, not just one pair of ratios.

A pair of linear equations has no solution when the ratios of the coefficients of \(x\) and \(y\) are equal, but the ratio of the constants is different. Thus, \(\frac{b}{12}=\frac{8}{24}=\frac{1}{3}\), which gives \(b=4\). Also, \(\frac{32}{71}\ne\frac{1}{3}\), so the two lines are distinct and parallel, giving no solution. Exam tip: First equate the coefficient ratios, then check that the constant ratio is different.

For the two equations, the ratios of the coefficients of x and y are equal: 14/2 = 21/3 = 7. However, the ratio of the constant terms is 49/8, which is not equal to 7. Thus, the two lines have the same slope but different positions, so they are distinct parallel lines. Coincident lines would require all three ratios to be equal. Exam tip: If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair represents distinct parallel lines.