Graphical method of finding solutions.

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

The direct answer is option A, (72,54), where x is the cost of the first item and y is the cost of the second item. The total cost gives x+y=126. Since the first item is 18 costlier, x-y=18. Add the equations: 2x=144, so x=72. Substitute into x+y=126: 72+y=126, hence y=54. Therefore the two graph lines intersect at (72,54). Option A satisfies both conditions: 72+54=126 and 72-54=18. Option B, (54,72), makes the first item 18 cheaper, not costlier. Option C, (70,56), has total 126 but difference 14. Option D, (74,52), has total 126 but difference 22. The order matters because the question specifically calls the first item more expensive. Memory cue: for “first is more than second,” write x-y=the extra amount, not y-x.

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.

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.

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

A consistent and independent pair has one unique solution. On a graph, it appears as one intersection point of two lines.

If all three ratios are equal, both equations represent the same line. Hence the lines are coincident and give infinitely many solutions.

From the second equation, (y=4x-9). Substituting gives (6x+5(4x-9)=39), so (x=3). The graph intersection gives this (x)-coordinate.

Putting \(y=7x-20\) in \(x+3y=12\) gives \(22x=72\), so \(x=\frac{36}{11}\) and \(y=\frac{32}{11}\). Fractional coordinates can also be correct graphical solutions.

((10,0)) and ((0,12)) satisfy the equation, but ((5,6)) does not give (60). Check points before drawing the graph.

(\frac{7}{14}=\frac{-3}{-6}\neq\frac{18}{41}), so the lines are distinct and parallel. An inconsistent pair has no solution.

From the first equation, (y=4x-13). Substituting gives (x+2(4x-13)=14), so (x=4) and (y=3). The intersection point is the graphical solution.

The coefficient ratio is (\frac{1}{2}); coincidence needs (k=54). For (k=50), the lines will be distinct and parallel.