Graphical method of finding solutions.

The second equation is (2) times the first, so both lines are the same. All points on the same line are solutions.

A consistent and independent pair has one unique solution. On a graph, it appears as one intersection point of two lines.

If all three ratios are equal, both equations represent the same line. Hence the lines are coincident and give infinitely many solutions.

From the second equation, (y=4x-9). Substituting gives (6x+5(4x-9)=39), so (x=3). The graph intersection gives this (x)-coordinate.

Putting \(y=7x-20\) in \(x+3y=12\) gives \(22x=72\), so \(x=\frac{36}{11}\) and \(y=\frac{32}{11}\). Fractional coordinates can also be correct graphical solutions.

((10,0)) and ((0,12)) satisfy the equation, but ((5,6)) does not give (60). Check points before drawing the graph.

The governing concept is the graphical condition for a pair of linear equations. Two non-identical straight lines having equal slopes are parallel. Here m₁ = m₂ = −7/4, while the y-intercepts are different, so the lines cannot be the same line; they remain distinct parallel lines. Distinct parallel lines never meet at any point in the Cartesian plane. Since a common solution would be represented by an intersection point, the pair has no common solution. Option B is therefore correct. Option A would describe lines with unequal slopes that intersect once, option C would describe coincident lines with the same slope and the same intercept, and option D is incorrect because perpendicular lines have slopes whose product is −1, not equal slopes.

(\frac{7}{14}=\frac{-3}{-6}\neq\frac{18}{41}), so the lines are distinct and parallel. An inconsistent pair has no solution.

From the first equation, (y=4x-13). Substituting gives (x+2(4x-13)=14), so (x=4) and (y=3). The intersection point is the graphical solution.

The coefficient ratio is (\frac{1}{2}); coincidence needs (k=54). For (k=50), the lines will be distinct and parallel.

From the second equation, (x=3y-11). Substituting gives (9y-33+2y=25), so (y=5). Then (x=5), so the intersection is ((5,5)).

Substituting ((-1,6)) makes (2x+y=4) and (x-y=-7) both true. The intersection point must lie on both lines.

(\frac{9}{3}=\frac{12}{4}\neq\frac{45}{20}), so the lines are distinct and parallel. Check the constant term ratio also.

The first equation must be (4) times the second, so (a=20). In coincident lines, all terms change by the same multiplier.

The governing concept is testing whether a point lies on a line by substitution. A point (x, y) lies on 5x + 11y = 55 only when its coordinates make the left-hand side equal to 55. For (11, 0), 5(11) + 11(0) = 55, so it lies on the line. For (0, 5), 5(0) + 11(5) = 55, so it also lies on the line. For (22/5, 3), 5(22/5) + 11(3) = 22 + 33 = 55, so this point is valid as well. For (4, 3), the value is 5(4) + 11(3) = 20 + 33 = 53, not 55. Therefore option D is the only point that does not lie on the line.

The second equation is (3) times the first, so (m=21). In a coincident line, the coefficient of (y) also changes in the same ratio.

From the first equation, (y=43-8x). Substituting gives (2x-3(43-8x)=-5), so (x=5) and (y=3). Hence the (y)-coordinate is (3).

The equations are (x+y=74) and (x-y=16), giving (x=45), (y=29). In word problems, first define the variables clearly.

Dividing the first equation by 3 and the second by 4 gives x + y = 33 and x − y = 7. Adding these equations gives 2x = 40, so x = 20 and y = 13. Therefore, the intersection point of the two lines, and hence the graphical solution, is (20, 13). Option B reverses the coordinates; for (13, 20), x − y = −7, so it is incorrect. Exam tip: the point where the two graphs intersect represents the solution of the pair of linear equations.

The second equation is (4) times the first, so (4k=20) and (k=5). For infinite solutions, the lines must be coincident.