Graphical method of finding solutions.

The equations are (x+y=90) and (x-y=14). Solving gives (x=52), (y=38), which is the graph intersection.

The second equation is (2) times the first, so (2a=10) and (a=5). Include the constant term in ratio checking.

The coefficient ratio is (\frac{1}{2}), so coincidence needs (k=24). For (k=20), the lines are distinct and parallel.

From the first equation, (s=10-2r). Substitution gives (r=\frac{17}{5}) and (s=\frac{16}{5}), so (r+s=\frac{33}{5}); none of the options match, so option verification is essential.

The second equation directly gives (p-q=2). The intersection point satisfies both equations, so full solving is not always necessary.

(\frac{2}{4}=\frac{7}{14}\neq\frac{14}{31}), so the lines are distinct and parallel. Such a graph has no common point.

The second equation is (3) times the first, so (3k=6) and (k=2). In coincident lines, all terms are in the same ratio.

The governing concept is that the intersection point is the ordered pair satisfying both linear equations simultaneously. From x − y = 1, we obtain y = x − 1. Substitute this into the first equation: 3x + 2(x − 1) = 18. Simplifying gives 3x + 2x − 2 = 18, so 5x = 20 and x = 4. Then y = 4 − 1 = 3. Thus the two graphs meet at (4, 3), making option A correct. Checking confirms it: 3(4) + 2(3) = 18 and 4 − 3 = 1. The other ordered pairs fail at least one equation.

For parallel lines, (\frac{4}{8}=\frac{a}{10}), so (a=5). Since (\frac{16}{35}\neq\frac{1}{2}), they will not be coincident.

Substituting \(\left(-\frac{3}{2},4\right)\) makes both \(2x+y=1\) and \(x+2y=\frac{13}{2}\) true. The intersection point should be checked in both equations.

Putting (y=0) gives (x=3), and putting (x=0) gives (y=-7). Intercepts give simple and useful points for graphing.

Putting (x=-4) gives (-8+3y=7), so (y=5). With a vertical line, the (x)-coordinate is already fixed.

The governing idea is that the intersection must satisfy both equations. The equation y = −2 represents a horizontal line, so every point on it has y-coordinate −2. Substitute this fixed value into 4x + 3y = 10: 4x + 3(−2) = 10, giving 4x − 6 = 10. Hence 4x = 16 and x = 4. Therefore the unique common point is (4, −2), so option A is correct. A quick check gives 4(4) + 3(−2) = 16 − 6 = 10. Option D has the correct y-coordinate but x = 5 gives 20 − 6 = 14, while the remaining options do not lie on y = −2.

Adding the two equations gives (2p=18), so (p=9) and (q=4). 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.

An inconsistent pair has no common point, so the lines are distinct and parallel. This is the no-solution case.

The coefficient ratios are not equal, so the lines meet at one point. This is the case of a unique solution.

Putting \(y=6x-17\) in \(x+2y=9\) gives \(13x=43\), so \(x=\frac{43}{13}\) and \(y=\frac{37}{13}\). Fractional coordinates can also be correct graphical solutions.

Substituting all three points makes (5x+4y=40) true. In a graph, three correct points should lie on the same straight line.

The governing concept is the relationship between slopes and the number of intersections. Two non-identical lines with equal slopes have the same direction and therefore are parallel. Since the y-intercepts are different, the lines are not the same line; they are distinct parallel lines. Distinct parallel lines never meet, so their pair of equations has no common ordered pair and hence no solution. Therefore option B is correct. A unique solution would require different slopes so that the lines intersect once. Infinitely many solutions would occur only when both the slope and the intercept were equal, producing coincident lines. Perpendicular lines would require the product of their slopes to be −1, which is not true here.