Graphical method of finding solutions.

Both equations give the same line, so all points on that line are solutions. Note that not all ordered pairs, only points on the line, are solutions.

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.

The direct answer is C: the x-coordinate is 3. Add the equations: (3x+y)+(2x−y)=10+5. The y terms cancel, giving 5x=15. Dividing by 5 gives x=3. The corresponding y can be found as 1, so the intersection is (3,1), but only x is requested. Option A, 1, is the y-coordinate, not the x-coordinate. Option B, 2, does not satisfy the combined equation 5x=15. Option C is correct because x=3. Option D, 5, is the coefficient produced before division, not the solution. In a graph, the intersection satisfies both equations, so algebraically solving the pair gives its coordinates. Memory cue: when adding, opposite y terms cancel; then solve for the requested coordinate.

(\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.

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.

The direct answer is option C. Coincident lines are actually the same line, so one equation can be obtained by multiplying the other by one common non-zero number. For a1x+b1y+c1=0 and a2x+b2y+c2=0, this requires a1/a2=b1/b2=c1/c2, wherever the ratios are defined. Therefore option C is correct. Option A, unequal a- and b-ratios, describes different slopes and therefore generally intersecting or non-parallel lines, not coincident lines. Option B has equal coefficient ratios but an unequal constant ratio; this is the condition for parallel, distinct lines, so the two lines never meet. Option D says b1/b2=0, which is not the general condition and would also require special coefficient values; it does not express equality of the complete equations. The important distinction is: all three ratios equal means coincident, first two equal but third unequal means parallel and distinct. Memory cue: for coincident lines, remember “all three ratios 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.

The coefficient ratio is (\frac{3}{6}=\frac{2}{4}=\frac{1}{2}); for coincidence, (k=14) is needed. With (k=7), the lines are parallel and distinct.

The second line is (2) times the first, so both lines are the same. The same line meets at infinitely many points.

For intersection on the (y)-axis, (x=0) is needed. In the correct option, putting (x=0) gives (y=4) in both equations.

The direct answer is option B, (4,3). A graph intersection must satisfy both equations. From x-y=1, add y to both sides to get x=y+1, or y=x-1. Substitute y=x-1 into 3x+y=15: 3x+(x-1)=15, so 4x-1=15, 4x=16, and x=4. Then y=x-1=3. Thus the lines meet at (4,3). Option A, (3,4), gives 3(3)+4=13 rather than 15, although it gives 3-4=-1 rather than 1 in the second equation. Option B gives 3(4)+3=15 and 4-3=1, so it is correct. Option C, (5,0), satisfies the first equation but gives 5 instead of 1 in the second. Option D, (2,5), satisfies neither equation: 11 is not 15 and -3 is not 1. Memory cue: the intersection point must pass both substitution checks.

Substituting ((2,5)) gives (2\cdot2+5\cdot5=29), not (20). Every chosen plotting point must satisfy the equation.

(\frac{1}{2}=\frac{4}{8}\neq\frac{9}{21}), so the lines are distinct parallel lines. Such lines have no solution.