Conditions for solvability

Two linear equations have infinitely many solutions when the ratios of their corresponding coefficients are equal. Multiplying \(3x-4y=12\) by 2 gives \(6x-8y=24\). Therefore, the first equation matches it only when \(p=6\). For instance, if \(p=4\), the ratio of the coefficients of \(x\) is not equal to the other ratios. Exam tip: for infinitely many solutions, check \(a_1/a_2=b_1/b_2=c_1/c_2\).

For infinitely many solutions, both linear equations must represent the same line. Hence the ratios of corresponding coefficients must be equal: \(\frac{a}{2}=\frac{6}{3}=\frac{14}{7}=2\). Thus, \(\frac{a}{2}=2\) gives \(a=4\). If \(a=2\), the ratio of the coefficients of \(x\) does not match the other ratios. Exam tip: for infinitely many solutions, check \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

For no solution, the two lines must be parallel but distinct. Thus, the ratios of the coefficients of x and y must be equal, while the ratio of the constants must be different. Here, \(\frac{12}{2}=6\), so \(\frac{6}{k}=6\) gives \(k=1\). The equations then become \(x+2y=5\) and \(x+2y=8\), which represent distinct parallel lines. For \(k=2\) or other listed values, the coefficient ratios are not equal. Exam tip: first equate the coefficient ratios, then check that the constant ratio differs.

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, \(\frac{-10}{-5}=\frac{30}{15}=2\), so \(\frac{p}{3}=2\), giving \(p=6\). Therefore, option C is correct. Exam tip: For infinitely many solutions, all three corresponding coefficient and constant ratios must be equal; matching only two ratios is not sufficient.

For two linear equations to have no solution, they must represent distinct parallel lines. First simplify 12x + 18y = 54 by dividing every term by 6, which gives 2x + 3y = 9. The second equation already has the same left-hand side, namely 2x + 3y, but its right-hand side is c. If c = 9, both equations are identical and there are infinitely many solutions. If c is different from 9, they demand that the same expression equal two different constants, which is impossible; geometrically, the lines are parallel and distinct. Among the choices, c = 10 gives no solution. Therefore option C is correct.

Two linear equations have infinitely many solutions when the ratios of their corresponding coefficients and constants are equal. Multiplying the second equation by 3 gives \(6x+9y=27\), which must coincide with the first equation \(ax+9y=27\). Therefore, \(a=6\). Exam tip: for infinitely many solutions, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Multiply the first equation by 2: 2(4x+7y=31) gives 8x+14y=62, exactly the second equation. Hence the two equations are dependent and represent the same straight line. Since every point on that line satisfies both equations, the pair has infinitely many ordered-pair solutions. Therefore option D is correct, not a unique or inconsistent solution.

When coefficient ratios are different, the lines meet at one point. In exams, check the ratios of (a) and (b) first.

In this case, the lines are parallel and do not meet. In exams, if the (c) ratio differs, write no solution.

For the pair a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the condition a₁/a₂ = b₁/b₂ = c₁/c₂ means that every coefficient, including the constant term, is in the same ratio. Consequently, one equation is a scalar multiple of the other, so both equations represent the same line rather than two distinct lines. Every point on that common line satisfies both equations. Therefore, the pair is dependent and consistent and has infinitely many solutions. A unique solution would require unequal coefficient ratios, while no solution occurs when the first two ratios agree but the constant-term ratio differs.

Here (2/4=3/6=(-5)/(-10)), so the lines are coincident. In exams, if all ratios match, write infinitely many solutions.

Here (1/2=1/2) but ((-4)/(-9)) is different, so there is no solution. Parallel lines form an inconsistent pair.

Here (3/6 \ne 2/5), so the lines intersect at one point. Different (a) and (b) ratios give a unique solution.

The intersection point satisfies both equations. In a graph, one intersection means a unique solution.

Parallel lines never meet, so there is no common point. In exams, connect parallel lines with no solution.

Every point on the same line satisfies both equations. Therefore, such a pair has infinitely many solutions.

Here (5/10=(-1)/(-2)) but (2/3) is different, so no solution exists. Equal first two ratios and a different third ratio give parallel lines.

All three ratios are equal, so both equations represent the same line. This is called a consistent dependent pair.

Here (2/3 \ne (-5)/1), so there is a unique solution. If the first two ratios differ, the third ratio need not be checked.

When all three ratios are equal, both lines lie on each other. These are called coincident lines.