Graphical method of finding solutions.

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.

The governing idea is that the intersection point satisfies both equations simultaneously. Since the first line is y = 3, the y-coordinate of the intersection must be 3. Substitute this value into the second equation: 2x − 5(3) = 1, so 2x − 15 = 1, hence 2x = 16 and x = 8. Therefore the intersection is (8, 3), which is option A. Option B reverses the coordinates. Option C keeps y = 3 but does not satisfy the second equation, since 2(5) − 15 = −5. Options D has the coordinates in an unsuitable order and also fails the fixed y = 3 condition. The horizontal line y = 3 makes the coordinate check especially direct.

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.

The governing graphical criterion is the relationship between slopes and intercepts. Two distinct nonvertical lines with equal slopes have the same direction and therefore are parallel. Because their y-intercepts are different, they are not the same line; distinct parallel lines never meet at any point. Hence the pair has no solution, so option B is correct. A unique solution would require the lines to intersect, which normally occurs when their slopes differ. Infinitely many solutions would occur only if both equations represented the very same line, requiring matching intercepts as well. Perpendicular lines have slopes whose product is −1, but here the product is (−4/3)(−4/3) = 16/9, not −1.

(\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 governing condition for infinitely many solutions is that the two linear equations represent the same, coincident line. The left side of the second equation is exactly twice the left side of the first: 2(3x + 2y) = 6x + 4y. Therefore its constant term must also be twice the first constant: 2p = 20. Dividing by 2 gives p = 10, so option B is correct. If p were 5, the first equation multiplied by 2 would have constant 10, not 20. If p were 15 or 20, the coefficient ratios would no longer match the constant ratio, producing parallel distinct lines or a unique intersection rather than infinitely many solutions. Thus equality of all corresponding ratios is essential.

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.