Graphical method of finding solutions.

Starting with \\(8x-4y=16\\), rearrange to get \\(-4y=16-8x\\), and then divide by \\(-4\\): \\(y=2x-4\\). Comparing this with \\(y=mx+c\\), the coefficient of \\(x\\) is the slope, so \\(m=2\\). Therefore, option A is correct. Option B results from a sign error while isolating \\(y\\). Exam tip: always express the equation in \\(y=mx+c\\) form before reading the slope.

The second equation is (3) times the first, so every point on (2x+3y=18) is a solution. ((3,4)) lies on this line.

(\frac{2}{4}=\frac{-5}{-10}\neq\frac{7}{16}), so the lines are distinct and parallel. Such a pair has no solution.

The equations are (x+y=126) and (x-y=18). Solving gives (x=72), (y=54), which is the graph intersection.

The second equation is (3) times the first, so (3a=12) and (a=4). Include the constant term in ratio checking.

The coefficient ratio is (\frac{1}{2}), so coincidence needs (k=42). For (k=40), the lines are distinct and parallel.

The intersection point of the two graphs satisfies both equations. Adding the equations gives \(5x=15\), so \(x=p=3\). Substituting this into \(x+y=4\) gives \(y=q=1\). Hence, \(p-q=3-1=2\), so option B is correct. Exam tip: The intersection of the graphs can be found efficiently by solving the two linear equations simultaneously.

For parallel lines, (\frac{7}{14}=\frac{b}{10}), so (b=5). Since (\frac{21}{50}\neq\frac{1}{2}), the lines are not coincident.

Putting \(x=3y-4\) gives \(5(3y-4)+2y=23\), so \(y=\frac{43}{17}\) and \(x=\frac{61}{17}\). Fractional coordinates can also be correct graphical solutions.

Substituting ((0,4)) gives (4\cdot0-9\cdot4=-36), not (36). Check every point in the equation before drawing the graph.

Solving both equations gives (x=14) and (y=9). On the graph, the intersection of the two lines will be ((14,9)).

(\frac{4}{8}=\frac{9}{18}\neq\frac{36}{75}), so the lines are distinct and parallel. Distinct parallel lines have no common point.

The second equation is (3) times the first, so (3k=15) and (k=5). In coincident lines, all terms are in the same ratio.

The intersection point is the ordered pair satisfying both linear equations simultaneously. From x - y = 3, rearrange to y = x - 3. Substitute this expression into 5x + 2y = 24: 5x + 2(x - 3) = 24, so 5x + 2x - 6 = 24, giving 7x = 30, or x = 30/7. Thus the exact intersection is (30/7, 9/7), not any listed option. Therefore the supplied MCQ has no valid correct option; option A, (6,3), gives 36 and 3 in the two equations, so it does not satisfy the first equation. The original key is mathematically incorrect and the item must be revised.

For parallel lines, (\frac{3}{9}=\frac{a}{12}), so (a=4). Since (\frac{15}{47}\neq\frac{1}{3}), they will not be coincident.

Substituting \(\left(\frac{5}{2},-\frac{3}{2}\right)\) makes both \(2x+y=\frac{7}{2}\) and \(x-2y=\frac{11}{2}\) true. Check the intersection point in both equations.

Putting (y=0) gives (x=4), and putting (x=0) gives (y=-9). Intercepts give convenient points for graphing.

Putting (x=-5) gives (-15+2y=9), so (y=12). With a vertical line, the (x)-coordinate is already fixed.

The intersection must satisfy both equations. The equation y = -4 represents a horizontal line, so the y-coordinate of every point on it is fixed at -4. Substitute y = -4 into 5x + 2y = 17: 5x + 2(-4) = 17, so 5x - 8 = 17. Hence 5x = 25 and x = 5. The intersection point is therefore (5, -4), making option A correct. Option B reverses the coordinate order. Option C keeps the correct y-value but gives an x-value that does not satisfy the second equation, while D fails both coordinate requirements. Verification gives 5(5) + 2(-4) = 25 - 8 = 17.

Adding the two equations gives (2p=24), so (p=12) and (q=5). Coordinates of the intersection satisfy both equations together.