Graphical method of finding solutions.

From the first equation, (y=38-6x). Substituting gives (3x-2(38-6x)=-1), so (x=5), (y=8). This is the graph intersection.

Two distinct parallel lines never intersect, so they have no common point. Therefore, the pair has no solution and is called an inconsistent pair. Option B is incorrect because an intersection point exists only when the lines meet. Exam tip: distinct parallel lines represent zero solutions.

Substituting ((4,3)) does not give \(2x+5y=29\), so it is not correct; the true solution is \(\left(\frac{64}{17},\frac{73}{17}\right)\). Check a point in both equations.

Substituting ((4,3)) gives \(2x+5y=31\) but not \(3x-y=7\); the true common point is \(\left(\frac{66}{17},\frac{79}{17}\right)\). Verify in both equations before choosing.

The governing concept is identifying a pair of lines by comparing the ratios of their coefficients. Rewrite the first equation by dividing every term by 5: 15x + 20y = 60 becomes 3x + 4y = 12. This is exactly the second equation, not merely a proportional equation with a different constant. Equivalently, 15/3 = 20/4 = 60/12 = 5. Equal ratios for all three corresponding coefficients mean that both equations describe the same geometric line. Such lines are called coincident lines, and every point on one is also on the other, giving infinitely many common solutions. Therefore option C is correct. Intersecting lines have unequal slope ratios, distinct parallel lines have proportional x and y coefficients but a different constant ratio, and perpendicular lines require slopes whose product is −1.

Substituting ((-4,1)) makes (x+y=-3) and (2x-y=-9) both true. Substituting the intersection point in both equations is the fastest check.

The second equation is (2) times the first, so both are the same line. The solution is all points on that line, not the whole plane.

From the given equation, \(5y=-10x+25\), so \(y=-2x+5\). Comparing this with \(y=mx+c\), the coefficient of \(x\) gives the slope \(m=-2\); therefore, option B is correct. Option A misses the negative sign. In an exam, isolate \(y\) first to identify the slope directly.

The second equation is (3) times the first, so every point on (3x+4y=36) is a solution. ((4,6)) lies on this line.

(\frac{3}{6}=\frac{-8}{-16}\neq\frac{11}{25}), so the lines are distinct and parallel. Such a pair has no solution.

The equations are (x+y=154) and (x-y=22). Solving gives (x=88), (y=66), which is the graph intersection.

The second equation is (3) times the first, so (3a=18) and (a=6). Include the constant term in ratio checking.

The coefficient ratio is (\frac{1}{2}), so coincidence needs (k=90). For (k=88), the lines are distinct and parallel.

In the graphical method, the intersection point of the two lines represents the common solution of both equations. From \(x+y=5\), we get \(y=5-x\). Substituting this in \(5x-y=19\) gives \(5x-(5-x)=19\), so \(6x=24\) and \(x=4\). Consequently, \(y=1\), hence \((p,q)=(4,1)\) and \(p-q=4-1=3\). Therefore, option C is correct. Exam tip: the coordinates of the intersection of two lines give the simultaneous solution of their equations.

For parallel lines, (\frac{9}{27}=\frac{2}{m}), so (m=6). Since (\frac{18}{55}\neq\frac{1}{3}), the lines are not coincident.

Elimination gives \(19x=112\), so \(x=\frac{112}{19}\) and \(y=\frac{41}{19}\). A graphical solution may also have fractional coordinates.

Substituting ((2,8)) gives \(13\cdot2+4\cdot8=58\), not (52). Check every point in the equation before drawing the graph.

The second equation must be (3) times the first, so (k=126). In coincident lines, the constant term also changes in the same ratio.

Dividing the first equation by 5 and the second by 6 gives \(x+y=28\) and \(x-y=4\). Adding these equations yields \(2x=32\), so \(x=16\), and consequently \(y=12\). Therefore, the intersection point of the two lines is \((16,12)\). In option B, the coordinates are reversed; it satisfies \(x+y=28\) but gives \(x-y=-4\), so it is incorrect. Exam tip: In the graphical method, the common point of the two lines represents the solution of the pair of equations.

The coefficient ratios \\(2/5\\) and \\(9/11\\) are unequal. Therefore, the two lines have different slopes and intersect at exactly one point, giving a unique solution. Option B would be correct only if the lines were distinct and parallel. Exam tip: for \\(a_1x+b_1y+c_1=0\\) and \\(a_2x+b_2y+c_2=0\\), if \\(a_1/a_2 \\ne b_1/b_2\\), the pair has a unique solution.