Conditions for solvability

A pair of linear equations has infinitely many solutions when the ratios of the coefficients of x, the coefficients of y, and the constant terms are all equal. For the given equations, 12/20 = 3/5 and 18/30 = 3/5. Therefore, the middle ratio s/10 must also be 3/5. Hence s = 10 × 3/5 = 6. Option C is correct. If s were 4, 5, or 8, the y-coefficient ratio would differ from the other two ratios. The equations would then represent distinct intersecting or parallel lines rather than coincident lines, so they could not have infinitely many common solutions.

Equating coefficient ratios, (\frac{u}{4}=\frac{9}{6}) gives (u=6). The constant ratio is different, so there is no solution.

A pair of linear equations has a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(\frac{13}{26}=\frac12\) and \(\frac{-t}{-10}=\frac{t}{10}\). Therefore, a unique solution requires \(\frac12\ne\frac{t}{10}\), which gives \(t\ne5\). At \(t=5\), the coefficients on the left-hand sides are proportional, so the solution cannot be unique. Exam tip: include the negative signs while comparing the coefficients of \(y\).

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, \(\frac{-3}{-5}=\frac{6}{10}=\frac{3}{5}\). Therefore, \(\frac{v+2}{10}=\frac{3}{5}\), giving \(v+2=6\) and hence \(v=4\). Option \(v=3\) is a close distractor, but it makes the first ratio \(\frac{1}{2}\), not \(\frac{3}{5}\). Exam tip: for infinitely many solutions, compare all three coefficient ratios.

The coefficient ratio is (\frac{1}{7}). For no solution, (\frac{d}{28}\neq\frac{1}{7}), so (d\neq4).

The determinant of the coefficient matrix is \(D=a(3b)-(2a)b=ab\). A pair of linear equations has a unique solution only when \(D\neq 0\). Hence, the required condition is \(ab\neq 0\), meaning that neither \(a\) nor \(b\) can be zero. Conditions such as \(a+b=0\) or \(a=b\) do not by themselves guarantee a unique solution. Exam tip: for a unique solution, check whether \(a_1b_2-a_2b_1\neq0\).

Equating coefficient ratios, (\frac{l}{9}=\frac{6}{18}) gives (l=3). The constant ratio is different.

For infinitely many solutions, (\frac{6}{15}=\frac{h+1}{20}=\frac{12}{30}) must hold. This gives (h=7).

For a pair of linear equations, compare the corresponding coefficients and constants. Multiplying x+2y=3 by 4 gives 4x+8y=12, exactly the second equation. Thus both equations represent the same straight line, and every point on that line satisfies both equations. Therefore infinitely many ordered pairs are solutions. A unique solution would require intersecting lines, while parallel distinct lines would give no solution.

Write the equations in the standard form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. Their coefficient ratios are a1/a2 = 3/6 = 1/2 and b1/b2 = (−1)/(−2) = 1/2, but the constant ratio is c1/c2 = (−4)/(−9) = 4/9, which is not 1/2. Thus the coefficients are proportional while the constants are not. Geometrically, the two lines have the same slope but different intercepts, so they are distinct parallel lines. Distinct parallel lines never meet; therefore the pair has no solution, making option C correct.

Compare the ratios of the coefficients: \(\frac{a_1}{a_2}=\frac{4}{8}=\frac{1}{2}\), whereas \(\frac{b_1}{b_2}=\frac{5}{9}\). Since \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), the two lines intersect at exactly one point, so the pair has a unique solution. Infinitely many solutions would require all three ratios to be equal. Exam tip: compare \(a_1/a_2\) and \(b_1/b_2\) first.

No solution occurs when lines are parallel and distinct. Its ratio form is (\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}).

For a pair of linear equations, if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the two lines have the same slope but different intercepts. Hence, they are distinct parallel lines and never intersect, so the pair has no solution and is inconsistent. Exam tip: equal coefficient ratios with an unequal constant-term ratio indicate “no solution” immediately.

A pair of linear equations in two variables represents two lines. If the ratios of the coefficients of x and y are unequal, the lines do not have the same slope. Lines with different slopes meet at exactly one point, so the pair has one and only one ordered pair satisfying both equations. This is called a unique solution, which is option C.

The condition is \(a_1/a_2\ne b_1/b_2\). Because the coefficient ratios differ, the lines cannot be parallel or coincident. The constants may affect the exact location of the intersection, but they do not change this conclusion when the two coefficient ratios are unequal. Therefore, “no solution” and “infinitely many solutions” are not suitable choices here. The correct conclusion is an unique, or अद्वितीय, solution.

Coincident lines represent the same line. Hence all three ratios are equal and infinitely many solutions occur.

For a pair of linear equations, let D be the determinant of the coefficients of x and y, while Dx and Dy are the auxiliary determinants. Cramer's rule gives a unique solution only when D is non-zero. If D = 0, the coefficient rows are dependent, so the represented lines are either coincident or parallel. When at least one auxiliary determinant is non-zero, the corresponding ratios are not all equal; hence the equations are inconsistent. Geometrically, the lines are distinct and parallel. They have no common point, so the pair has no solution. Therefore option C is correct, whereas infinitely many solutions require D = Dx = Dy = 0.

Two lines with the same slope are either coincident or parallel. Different (y)-intercepts show that the lines are not coincident, so they are distinct parallel lines and never intersect. Hence, the pair of equations has no solution. Exam tip: equal slopes with different intercepts indicate no solution, whereas equal slopes with equal intercepts indicate infinitely many solutions.

The slope tells us how steep a line is. If two lines have different slopes, they cannot remain parallel and they cannot be the same line. Their directions are different, so they meet at one and only one point. Therefore, the pair of linear equations is consistent and independent, and it has a unique solution. The correct choice is C, “Unique solution.”

In the graph, the common point gives the values of both variables simultaneously. Choice A would apply to distinct parallel lines, which have equal slopes and never meet. Choice B and choice D describe coincident, or dependent, lines; such lines have the same slope and share infinitely many points. Since the slopes here are different, exactly one intersection is possible, so C follows directly.

For non-unique solutions, the determinant must be zero. From ((k-3)k-8=0), (k=\frac{3\pm\sqrt{41}}{2}).

Equating coefficient ratios gives (k=0). The constant ratio (\frac{3}{8}) is different, so there is no solution.