Graphical method of finding solutions.

(\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.

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.

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.

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.

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

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

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

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

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

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).

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

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

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

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

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.

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.

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.

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

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

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

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.

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.