Conditions for solvability

Two linear equations \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\) have a unique solution when \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\). Here, \(a_1=q, b_1=6, a_2=5, b_2=3\), so the required condition is \(\frac{q}{5}\ne\frac{6}{3}\), or \(q\ne10\). Option A gives equal ratios, so it does not represent a unique solution. Exam tip: for a unique solution, the ratios of the coefficients of the two variables must be unequal.

For two linear equations to have infinitely many solutions, the ratios of corresponding coefficients and constants must be equal: a₁/a₂ = b₁/b₂ = c₁/c₂. Here, 2/4 = 9/18 = 1/2, so 3/k = 1/2, which gives k = 6. Therefore, option C is correct. Exam tip: For infinitely many solutions, always check a₁/a₂ = b₁/b₂ = c₁/c₂.

The first equation is 3x+2y=11. The second equation has coefficients 6 and 4, which are both twice the corresponding coefficients in the first equation. Therefore its left side is twice the first left side. If the equations represented the same line, its right side would also have to be twice 11, namely 22.

Thus t=22 makes the second equation 2(3x+2y)=22, exactly the same equation, giving infinitely many solutions. For every other listed value, the coefficient ratios remain equal but the constant ratio is different, so the lines are distinct and parallel and there is no solution. Therefore t must not be 22, and option C is correct.

Notice \(3x+6y=18\) divided by \(3\) gives \(x+2y=6\). So both equations are the same. Therefore, the graph is one line, meaning infinitely many solutions.

Multiplying the second equation by 2 gives 2x+4y=14, whereas the first equation is 2x+4y=10. Thus, the ratios of the coefficients of x and y are equal, but the ratio of the constant terms is different. Therefore, the two lines are distinct and parallel. Option A is incorrect because coincident lines require all three corresponding ratios to be equal. Exam tip: if a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel and distinct.

The ratios of the coefficients of x and y are not equal: 5/3 ≠ 8/4. Therefore, the lines are neither coincident nor parallel; they intersect at one unique point. Solving the equations gives the intersection point (3, 0). Exam tip: If a₁/a₂ ≠ b₁/b₂, the pair of linear equations has a unique solution, so the lines intersect at one point.

The direct answer is D. A consistent and independent pair has exactly one solution, so its two lines must intersect at one point. For equations a1x+b1y=c1 and a2x+b2y=c2, this happens when a1/a2 is not equal to b1/b2. In A, 1/2=1/2, so the equations are the same line and have infinitely many solutions. In B, 2/4=3/6 but 7/15 is different, so the lines are parallel and inconsistent; there is no solution. In C, 3/6=1/2 and 1/2=1/2, so the equations are the same line and have infinitely many solutions. In D, 1/2 is not equal to 3/5, so the lines intersect once and the pair is consistent and independent. Memory cue: unequal coefficient ratios mean one unique solution.

In option A, \(a_1/a_2=1/2\) and \(b_1/b_2=4/8=1/2\), but \(c_1/c_2=6/13\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), so the two lines are distinct and parallel and the pair has no solution. Hence, it is inconsistent. In option C, all three ratios are equal, so the pair is dependent and consistent. Exam tip: For an inconsistent pair, check whether \(a_1/a_2=b_1/b_2\ne c_1/c_2\).

In the second option the second equation is (3) times the first. Therefore both are the same line and the pair is dependent.

All terms are multiplied by (2) so the equations give the same line. In such questions identifying the multiplier is the fastest method.

In standard form, the coefficients are \(a_1=3,b_1=6,c_1=12\) and \(a_2=1,b_2=2,c_2=5\). Thus, \(\frac{a_1}{a_2}=3\) and \(\frac{b_1}{b_2}=3\), but \(\frac{c_1}{c_2}=\frac{12}{5}\). Therefore, the first two ratios are equal while the third is different. The lines are distinct and parallel, so the pair has no solution. Exam tip: If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the pair is inconsistent.

For the equations 4x+y=9 and 7x+2y=13, compare the ratios of the coefficients of x and y. We obtain a₁/a₂=4/7 and b₁/b₂=1/2. Since 4/7 is not equal to 1/2, the two coefficient ratios are unequal. This is the condition for two non-parallel lines, so their graphs intersect at exactly one point and the pair has one unique solution. The constant-term ratio 9/13 need not be compared to establish this conclusion once the first two ratios differ. Option A contains a false equality, while C and D would describe coincident lines. Therefore B is correct.

Two lines with different slopes intersect at exactly one point. Hence, the pair of linear equations has one unique solution and is called consistent and independent. Lines with the same slope but different intercepts are inconsistent, while equations representing the same line have infinitely many solutions. Exam tip: different slopes always imply a unique solution.

A line written as y=mx+c is determined by its slope m and y-intercept c. If two lines have the same value of m and the same value of c, their equations are identical: y=m₁x+c₁ and y=m₁x+c₁. Therefore they are not two separate parallel lines; they lie exactly on top of each other and are called coincident lines. Every point on this common line satisfies both equations, so the corresponding pair has infinitely many solutions and is consistent and dependent. Distinct parallel lines have the same slope but different intercepts. Intersecting lines have different slopes, and perpendicular lines have slopes whose product is -1 when defined. Hence option C is correct.

For the two equations, 13/26 = 5/10 = 1/2, but 22/45 is not equal to 1/2. Thus, a₁/a₂ = b₁/b₂, while c₁/c₂ is different. Therefore, the two lines are parallel and distinct, so they have no common solution and the pair is inconsistent. Exam tip: When a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair is inconsistent.

For the two equations, \(a_1=4, b_1=-7, a_2=8, b_2=-13\). Here, \(\frac{a_1}{a_2}=\frac{4}{8}=\frac{1}{2}\), whereas \(\frac{b_1}{b_2}=\frac{-7}{-13}=\frac{7}{13}\); the two ratios are unequal. Therefore, the lines intersect at exactly one point, so the system has one unique solution. Exam tip: If \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), a pair of linear equations has a unique solution.

Every term of the first equation is 5 times the corresponding term of the second: \(15=5\times3\), \(10=5\times2\), and \(5=5\times1\). Thus, both equations represent the same line and have infinitely many common solutions. Therefore, option C is correct. Option A would be correct only if the two lines were distinct and parallel. Exam tip: If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), a pair of linear equations has infinitely many solutions.

Dividing the first equation by 8 gives 2x-y=3, whereas the second equation is 2x-y=4. The ratios of the coefficients of x and y are equal, 16/2 = (-8)/(-1), but the ratio of the constant terms is different. Thus, the two lines are parallel and distinct, so they have no common solution. Exam tip: If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair of linear equations is inconsistent and has no solution.

Here, \(a_1=6, b_1=11\) and \(a_2=3, b_2=5\). We have \(a_1/a_2=6/3=2\), whereas \(b_1/b_2=11/5\); the two ratios are unequal. Therefore, the pair is consistent and independent and has a unique solution, so the lines intersect at one point. For parallel or coincident lines, the relevant coefficient ratios must be equal, while perpendicular lines require the product of their slopes to be \(-1\). Exam tip: whenever \(a_1/a_2 \ne b_1/b_2\), conclude that the pair has a unique solution.

For two equations a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0, if a₁/a₂=b₁/b₂ but this common ratio is not equal to c₁/c₂, the coefficients of x and y are proportional while the constants are not. Consequently, one equation cannot be a complete scalar multiple of the other. The graphs therefore have the same slope but different positions, so they are distinct parallel lines. Parallel distinct lines do not share any point, which means the pair has no solution and is called inconsistent. Coincident lines require all three ratios to be equal, whereas intersecting lines require the first two ratios to be unequal. Thus option C is correct.

Here, \(a_1/a_2=17/9\) and \(b_1/b_2=4/2=2\), so \(a_1/a_2\ne b_1/b_2\). Therefore, the two lines intersect at exactly one point, giving one unique solution. Option A would require all three ratios to be equal, while option B would require the coefficient ratios to be equal but the constant-term ratio to be different. Exam tip: first compare \(a_1/a_2\) and \(b_1/b_2\).

The first equation is (6) times the second so both lines are coincident. Coincident lines have infinitely many solutions.

Here (5/1=10/2) but (25/6) is different so the lines are parallel and distinct. Such lines have no common solution.

The governing concept is the classification of a pair of linear equations by comparing the ratios of corresponding coefficients and constants. Write the equations as 12x+18y-30=0 and 2x+3y-5=0. For the first equation, each coefficient is six times the corresponding coefficient in the second: 12=6(2), 18=6(3), and 30=6(5). Thus the first equation is exactly six times the second, not a different equation. Both equations represent the same straight line, so every point on that line satisfies both equations. The pair has infinitely many solutions and is therefore consistent and dependent. Hence option A is correct. An inconsistent pair has parallel distinct lines, while an independent consistent pair has intersecting lines and exactly one solution; neither situation applies here.

The governing idea is the ratio test for a pair of linear equations. Compare 14x+21y=70 with 2x+3y=10. Multiplying the second equation by 7 gives 7(2x+3y)=7(10), hence 14x+21y=70, which is exactly the first equation. Equivalently, the coefficient and constant ratios are 14/2=21/3=70/10=7. Since all three ratios are equal, the two equations represent coincident lines rather than two separate lines. Consequently, the system has infinitely many common solutions and is classified as consistent and dependent. Thus option A is correct. Option B would require parallel distinct lines, which would have equal coefficient ratios but an unequal constant ratio. Option C would require unequal coefficient ratios, and “unsolvable” is not the standard classification for this case.