Graphical method of finding solutions.

(\frac{5}{10}=\frac{-2}{-4}\neq\frac{9}{23}), so the lines are distinct and parallel. An inconsistent pair has no solution.

The governing concept is that the graphical solution is the intersection point, which must satisfy both equations. Add the equations to eliminate y: (3x − y) + (2x + y) = 10 + 15. This gives 5x = 25, so x = 5. Substitute x = 5 into 3x − y = 10: 15 − y = 10, hence y = 5. Therefore the lines intersect at (5, 5), so option A is correct. Verification in the second equation gives 2(5) + 5 = 15, confirming the point. Options B, C, and D do not satisfy both equations simultaneously, even if one may appear close on an inaccurately drawn graph.

The coefficient ratio is (\frac{1}{2}); coincidence needs (k=34). For (k=30), the lines will be distinct and parallel.

The intersection point is the common solution of both linear equations. From the first equation, \(y=16-2x\). Substituting this into the second equation gives \(x-2(16-2x)=-8\), so \(5x=24\) and \(x=\frac{24}{5}\). Therefore, \(y=16-2\left(\frac{24}{5}\right)=\frac{32}{5}\). Hence, the intersection point is \(\left(\frac{24}{5},\frac{32}{5}\right)\). Exam tip: verify the ordered pair in both original equations; \((4,8)\) satisfies the first equation but not the second.

Substituting ((2,-5)) makes (3x+y=1) and (x-y=7) both true. The intersection point must lie on both lines.

(\frac{5}{1}=\frac{10}{2}\neq\frac{25}{6}), so the lines are distinct and parallel. Check the constant term ratio also.

The first equation must be (3) times the second, so (a=12). In coincident lines, all terms change by the same multiplier.

The governing concept is verification of a point on a line by substitution. A point (x, y) lies on 3x + 8y = 24 only when its coordinates make the left side equal 24. For (8, 0), the value is 3(8) + 8(0) = 24. For (0, 3), it is 0 + 24 = 24. For (8/3, 2), it is 3(8/3) + 16 = 8 + 16 = 24. However, for (4, 2), the value is 3(4) + 8(2) = 12 + 16 = 28, not 24. Therefore (4, 2) does not lie on the line, making option D correct. The check also shows why plotting an incorrect point would produce a wrong graph.

The second equation is (2) times the first, so (m=10). In a coincident line, the coefficient of (y) also changes in the same ratio.

From the second equation, (y=3x-10). Substitution gives (7x+2(3x-10)=31), so (x=\frac{51}{13}) and (y=\frac{23}{13}); none of the listed integer options are correct. Matching calculation with options is necessary.

The equations are (x+y=58) and (x-y=12), giving (x=35), (y=23). In word problems, first define the variables clearly.

Simplifying the equations gives \(x+y=36\) and \(x-y=6\). Adding them yields \(2x=42\), so \(x=21\). Substituting this into \(x+y=36\) gives \(y=15\). Therefore, the graphs intersect at \((21,15)\). Option B incorrectly reverses the values of x and y, so it is not the solution. Exam tip: In the graphical method, the point of intersection represents the common solution of both equations.

For (5x-2y=11) and (x+y=7), (\frac{5}{1}\neq\frac{-2}{1}), so the lines intersect. Different coefficient ratios give a unique solution.

The second equation is (3) times the first, so (3k=12) and (k=4). For infinite solutions, the lines must be coincident.

From the first equation, (y=27-5x). Substituting gives (2x-3(27-5x)=-6), so (x=5), (y=2). This is the graph intersection.

Intersecting lines have only one common point. That point is the unique solution of both equations.

Substituting ((3,5)) gives \(3x+y=14\) but \(x-2y=-7\), so it is not correct; the true common point is \(\left(\frac{23}{7},\frac{29}{7}\right)\). Detecting option errors is also important.

The second equation is (2) times the first, so both are the same line. Only (1) distinct line will be visible on the graph.

Substituting ((-2,-3)) makes (x+y=-5) and (2x-y=-1) 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.