Graphical method of finding solutions.

The coefficient ratio is (\frac{3}{6}=\frac{2}{4}=\frac{1}{2}); for coincidence, (k=14) is needed. With (k=7), the lines are parallel and distinct.

The second line is (2) times the first, so both lines are the same. The same line meets at infinitely many points.

For intersection on the (y)-axis, (x=0) is needed. In the correct option, putting (x=0) gives (y=4) in both equations.

The governing coordinate-plane fact is that every point on the x-axis has y-coordinate equal to zero. The x-coordinate may be any real number, so it is represented by a parameter such as a. Consequently, if two lines meet on the x-axis, their common point must have the form (a, 0), making option B correct. Option A describes a general point on the y-axis because its x-coordinate is zero. Option C has equal coordinates and lies on the line y = x, not necessarily on the x-axis. Option D also has a generally nonzero y-coordinate, so it cannot lie on the x-axis unless a is zero, which is not the required general form.

From (x-y=1), (y=x-1), and (3x+x-1=15) gives (x=4), (y=3). The lines meet at ((4,3)) on the graph.

Substituting ((2,5)) gives (2\cdot2+5\cdot5=29), not (20). Every chosen plotting point must satisfy the equation.

(\frac{1}{2}=\frac{4}{8}\neq\frac{9}{21}), so the lines are distinct parallel lines. Such lines have no solution.

The second equation must be (2) times the first, so (kx=4x) and (k=4). In the coincident case, the whole line remains the same.

Putting (x=2) in (2x+3y=13) gives (4+3y=13), so (y=3). With a vertical line, (x) is already fixed.

The line y = -1 is horizontal, so the y-coordinate of the intersection must be -1. Substituting y = -1 in the second equation gives 3x - 2(-1) = 11, or 3x + 2 = 11, hence x = 3. Therefore, the intersection point is (3, -1). In option B, the coordinates are reversed, while option D has the correct y-coordinate but 3(5) - 2(-1) = 17, not 11. Exam tip: always substitute the obtained point into both equations to verify it.

The equations are (x+y=100) and (y-x=20), giving (x=40), (y=60). In word problems, first form the two correct linear equations.

The equations are (x+y=18) and (x-y=4), so (x=11), (y=7). On the graph, the intersection is this point.

(\frac{2}{6}=\frac{1}{3}\neq\frac{7}{20}), so the lines are distinct and parallel. This case gives no solution.

The second equation is (3) times the first, so both lines fully overlap. The solution is all points on the same line.

(\frac{2}{4}=\frac{-3}{-6}\neq\frac{6}{13}), so the lines are parallel. Distinct parallel lines have no intersection.

The intersection point always lies on both lines, so it satisfies both equations. A graphical solution can always be checked in both equations.

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

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.

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

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.