Graphical method of finding solutions.

Correct answer: C. (4)

The second equation must be (2) times the first, so (kx=4x) and (k=4). In the coincident case, the whole line remains the same.

Correct answer: A. ((2,3))

Putting (x=2) in (2x+3y=13) gives (4+3y=13), so (y=3). With a vertical line, (x) is already fixed.

Correct answer: A. (3, -1)

The line y = -1 is horizontal, so the y-coordinate of the intersection must be -1. Substituting y = -1 in the second equation gives 3x - 2(-1) = 11, or 3x + 2 = 11, hence x = 3. Therefore, the intersection point is (3, -1). In option B, the coordinates are reversed, while option D has the correct y-coordinate but 3(5) - 2(-1) = 17, not 11. Exam tip: always substitute the obtained point into both equations to verify it.

Correct answer: A. ((40,60))

The equations are (x+y=100) and (y-x=20), giving (x=40), (y=60). In word problems, first form the two correct linear equations.

Correct answer: A. ((11,7))

The equations are (x+y=18) and (x-y=4), so (x=11), (y=7). On the graph, the intersection is this point.

Correct answer: B. Parallel distinct

(\frac{2}{6}=\frac{1}{3}\neq\frac{7}{20}), so the lines are distinct and parallel. This case gives no solution.

Correct answer: B. The entire same line

The second equation is (3) times the first, so both lines fully overlap. The solution is all points on the same line.

Correct answer: B. No solution

(\frac{2}{4}=\frac{-3}{-6}\neq\frac{6}{13}), so the lines are parallel. Distinct parallel lines have no intersection.

Correct answer: A. ((2,-3)) satisfies both equations

The intersection point always lies on both lines, so it satisfies both equations. A graphical solution can always be checked in both equations.

Correct answer: B. 10

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.