Graphical method of finding solutions.

The governing concept is the coordinate definition of an intercept. A y-intercept is the point where a graph crosses the y-axis, and every point on the y-axis has x-coordinate 0. Substitute x = 0 in the equation: 2(0) − 5y = 10, so −5y = 10 and y = −2. Therefore the required point is (0, −2), which is option A. Option B has y = 0 and is an x-intercept candidate, while option C results from ignoring the negative sign. Option D does not satisfy the equation because 2(5) − 5(0) = 10, although it is actually the x-intercept.

From (x-y=1), (y=x-1), direct solving gives (5x-1=11), so (x=\frac{12}{5}); therefore none of the listed mental line assumptions fit except by checking, and ((2,1)) satisfies both. In hard questions, verify options carefully.

Putting \(y=x-1\) gives \(4x+x-1=11\), so \(x=\frac{12}{5}\) and \(y=\frac{7}{5}\). Fractional coordinates can also be graphical solutions.

For parallel lines, (\frac{2}{4}=\frac{k}{6}), so (k=3). Also (\frac{8}{12}\neq\frac{1}{2}), so they are not coincident.

For coincident lines, (\frac{p}{4}=\frac{3}{6}=\frac{9}{18}), so (p=2). In the coincident case, all ratios must be equal.

Adding the two equations gives (5x=15), so (x=3). This is the (x)-coordinate of the graphical intersection.

(\frac{1}{2}=\frac{3}{6}\neq\frac{6}{15}), so the lines are parallel. Parallel lines never meet.

((0,0)) satisfies both (x+y=0) and (2x-y=0). In the other options, the intersection is not the origin or the lines are coincident.

Substituting ((6,0)) gives (2\cdot6+3\cdot0=12), so it lies on the line. Intercept points are convenient for graphing.

The second line is (2) times the first, so both are the same line. A same line has infinitely many solution points.

For intersecting lines, the coefficient ratios of (x) and (y) are not equal. This is the condition for a unique solution.

From (x-2y=4), (y=\frac{1}{2}x-2), so the slope is (\frac{1}{2}). To find slope, write the line as (y=mx+c).

The slopes are different, so the lines meet at one point. Two lines with different slopes have one unique solution.

The governing idea is that the x-intercept is found by setting y = 0, while the y-intercept is found by setting x = 0. For the x-intercept, 3x − 4(0) = 12, so 3x = 12 and x = 4; hence it is (4, 0). For the y-intercept, 3(0) − 4y = 12, giving −4y = 12 and y = −3; hence it is (0, −3). Thus option A is correct. Options B and C arise from sign or coefficient errors. Option D reverses the intercept order and also gives incorrect signs, so it cannot represent the requested pair.

For coincident lines, (\frac{2}{6}=\frac{a}{9}=\frac{10}{30}), so (a=3). All coefficients must be in the same ratio.

Subtracting the second equation from the first gives (4x=8), so (x=2) and (y=6). Thus the graph point has (y)-coordinate (6).

In the option, (\frac{1}{2}=\frac{2}{4}\neq\frac{5}{9}), so the lines are parallel. Equal coefficient ratio and different constant ratio mean no solution.

For coincident lines, both equations are in the same ratio. Therefore all three ratios are equal.

From the second equation, (y=5-x). Substituting gives (2x-(5-x)=4), so (x=3) and (y=2). The graphical intersection is this solution.

The second equation must be (2) times the first, so (k=12). For coincident lines, the constant term also changes in the same ratio.