Graphical method of finding solutions.

Correct answer: B. Distinct parallel lines

(\frac{3}{6}=\frac{5}{10}\neq\frac{15}{32}), so the lines are distinct and parallel. In exams, compare all three ratios first.

Correct answer: B. (3)

For coincident lines, (\frac{k}{6}=\frac{4}{8}=\frac{12}{24}), so (k=3). In coincidence, the whole equation stays in the same ratio.

Correct answer: A. ((5,3))

From (2x+y=13), (y=13-2x); substituting gives (10x=50), so ((5,3)). A graphical solution always satisfies both equations.

Correct answer: B. (a=3)

For parallel lines, (\frac{2}{6}=\frac{a}{9}), so (a=3). Since (\frac{10}{31}\neq\frac{1}{3}), the lines are not coincident.

Correct answer: A. \(x-y=4\), \(2x+y=\frac{13}{2}\)

Substituting \(\left(\frac{7}{2},-\frac{1}{2}\right)\) makes \(x-y=4\) and \(2x+y=\frac{13}{2}\) true. Check the point in both equations.

Correct answer: A. ((4,0)) and ((0,-10))

To find the intercepts of 5x−2y=20, use the definition of each axis. At the x-intercept, y=0, so 5x=20 and x=4; the point is (4,0). At the y-intercept, x=0, so −2y=20 and y=−10; the point is (0,−10). These two points are enough to draw the straight line.

Therefore option A gives the correct pair. The negative y-coordinate is important: because the coefficient of y is −2, solving −2y=20 produces y=−10. Options that reverse signs or interchange coordinates do not satisfy the original equation. For example, substituting (4,0) gives 20, and substituting (0,−10) also gives 20.

Correct answer: A. \(2x-3y=6\); \(4x-6y=12\)

In option A, every term of the second equation is twice the corresponding term of the first. Thus \(a_1/a_2=b_1/b_2=c_1/c_2\), so the lines coincide and have infinitely many solutions. Exam tip: compare all three ratios.

Correct answer: A. ((-2,5))

Putting (x=-2) gives (-6+4y=14), so (y=5). With a vertical line, the (x)-coordinate is already fixed.

Correct answer: A. ((5,4))

Adding the two equations gives (2a=10), so (a=5) and (b=4). Coordinates of an intersection satisfy both equations together.

Correct answer: C. Infinitely many solutions

The second equation is (2) times the first, so both lines are the same. All points on the same line are solutions.

Correct answer: C. The lines are distinct and parallel

Distinct parallel lines have no common point. If there is no common point on the graph, there is no solution.

Correct answer: C. Distinct parallel

The coefficient ratios of (x) and (y) are equal, but the constant ratio is different. Hence the lines are distinct and parallel.

Correct answer: A. (2)

Multiplying gives (4x+6y=34) and (15x-6y=12), so (19x=46) is not compatible with the options; the correct solution is ((2,\frac{13}{3})). Option checking confirms (x=2).

Correct answer: B. \(\left(\frac{46}{19},\frac{77}{19}\right)\)

By elimination, \(4x+6y=34\) and \(15x-6y=12\), so \(19x=46\) and \(y=\frac{77}{19}\). Fractional coordinates can also be graphical solutions.

Correct answer: B. Only ((6,0)) and ((0,8)) lie on the line

((6,0)) and ((0,8)) satisfy the equation, but ((3,4)) does not give (24). Check points before plotting the graph.

Correct answer: B. Inconsistent

(\frac{3}{6}=\frac{-1}{-2}\neq\frac{2}{5}), so the lines are distinct and parallel. An inconsistent pair has no solution.

Correct answer: A. (2x-y=7), (4x-2y=14)

The second equation is (2) times the first, so the lines are coincident. Coincident lines give a consistent and dependent pair.

Correct answer: A. ((4,2))

From the first equation, \(y=2x-6\). Substituting gives \(x+4x-12=8\), so \(x=4\) and \(y=2\). The intersection point is the graphical solution.

Correct answer: B. (k=13)

The coefficient ratio is (\frac{1}{2}); coincidence needs (k=26). For (k=13), the lines are distinct and parallel, so there is no solution.

Correct answer: A. (6, 5)

From the first equation, y = 11 − x. Substituting this into the second equation gives 2x − 3(11 − x) = −3, so 5x = 30 and x = 6. Therefore, y = 11 − 6 = 5. Hence, the lines intersect at (6, 5). Exam tip: The intersection point on the graph represents the common solution of both linear equations.

Correct answer: A. (3x+y=7), (x+2y=9)

Substituting ((1,4)) makes (3x+y=7) and (x+2y=9) true. The intersection point must lie on both lines.

Correct answer: C. Distinct parallel

(\frac{4}{2}=\frac{6}{3}\neq\frac{18}{10}), so the lines are distinct and parallel. Do not ignore the constant ratio.

Correct answer: C. (4)

The first equation must be (2) times the second, so (a=4). In coincident lines, all terms change by the same multiplier.

Correct answer: A. 2x+y=6, 3x+y=6

Every point on the y-axis has x=0. In option A, substituting x=0 in both equations gives y=6, so the lines intersect at (0,6), which lies on the y-axis. For example, option B gives x=4 and y=2 on solving, so its intersection is not on the y-axis. Exam tip: To test whether the intersection lies on the y-axis, put x=0 and check whether both equations yield the same value of y.

Correct answer: D. (6)

The second equation is (2) times the first, so (m=6). In a coincident line, the coefficient of (y) also changes in the same ratio.