Graphical method of finding solutions.

The direct answer is A: (25,17), where x is the first type and y is the second type. Total plants give x+y=42. The first type is 8 more, so x−y=8. Add the equations: 2x=50, hence x=25. Substitute into x+y=42: 25+y=42, so y=17. Therefore the graph intersection is (25,17). Option A is correct. Option B reverses the types: its total is 42, but the first coordinate is not 8 more than the second. Option C gives total 42 but difference 6, not 8. Option D gives total 42 but difference 10. The variable order matters in a word problem. Memory cue: define first type as x, second as y; “first is 8 more” becomes x−y=8.

Dividing the equations by 3 and 2 respectively gives x + y = 20 and x − y = 6. Adding these equations gives 2x = 26, so x = 13 and y = 7. Thus, the intersection point of the two lines is (13, 7). Option B has the correct sum but gives a difference of −6, not 6. Exam tip: In the graphical method, the point where the two lines intersect represents the solution of the pair of equations.

For (3x-y=6) and (x+y=4), (\frac{3}{1}\neq\frac{-1}{1}), so the lines intersect. Different coefficient ratios give a unique solution.

The second equation is (2) times the first, so (2k=6) and (k=3). For infinite solutions, the lines must be coincident.

From the first equation, (y=19-4x). Substituting gives (x-2(19-4x)=-7), so (x=3), (y=7). This is the graph intersection.

Coincident lines represent the same line. Therefore every point on that line satisfies both equations.

The direct answer is A: (4,1). A point lies on both lines only when it satisfies both equations. Test (4,1): in 2x+y=9, 2(4)+1=8+1=9; in x−y=3, 4−1=3. Both statements are true, so (4,1) is the common point and graphical solution. Option B, (3,0), gives 2x+y=6, not 9. Option C, (5,−1), gives 2x+y=9 but x−y=6, not 3, so it lies on only one line. Option D, (2,5), gives 2x+y=9 but x−y=−3, not 3. Thus checking both equations is essential. Memory cue: a common point must pass two separate substitutions, not just one.

Both equations are the same line, so only one distinct line will be visible on the graph. This is the case of infinitely many solutions.

Substituting ((-3,2)) makes (x+y=-1) and (2x-y=-8) 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, \(2y=-6x+18\). Dividing by 2 gives \(y=-3x+9\). Comparing this with \(y=mx+c\), the coefficient of \(x\) is \(m=-3\). Option 3 has the wrong sign. In an exam, identify the slope as the coefficient of \(x\) after isolating \(y\).

The direct answer is option A, (1,3). The second equation, 3x+6y=21, is exactly three times the first equation x+2y=7. Therefore both equations represent the same line, and every point on x+2y=7 is a solution. Test option A: 1+2(3)=1+6=7, so it lies on the common line. Option B, (3,1), gives 3+2=5, not 7. Option C, (7,1), gives 7+2=9, not 7. Option D, (2,2), gives 2+4=6, not 7. Although the equations have infinitely many common points, among the listed choices only option A is on the line. A graphical solution means a common point, not necessarily a unique point. Memory cue: if one equation is a multiple of the other, check the point in just one equation, but it must satisfy it exactly.

(\frac{1}{2}=\frac{-3}{-6}\neq\frac{2}{9}), so the lines are distinct and parallel. Such a pair has no solution.

The direct answer is A: (52,38), with x as the first item’s cost and y as the second item’s cost. The total gives x+y=90. The first is 14 costlier, so x−y=14. Add the equations: 2x=104, hence x=52. Substitute into x+y=90: 52+y=90, so y=38. Thus the graph intersection is (52,38). Option A is correct. Option B reverses the item order and makes the first cheaper by 14. Option C has total 90, but its difference is 10, not 14. Option D also totals 90, but its difference is 18. The words “first” and “second” determine coordinate order. Memory cue: total plus difference gives twice the larger value: x=(90+14)/2=52.

The second equation is (2) times the first, so (2a=10) and (a=5). Include the constant term in ratio checking.

The coefficient ratio is (\frac{1}{2}), so coincidence needs (k=24). For (k=20), the lines are distinct and parallel.

From the first equation, (s=10-2r). Substitution gives (r=\frac{17}{5}) and (s=\frac{16}{5}), so (r+s=\frac{33}{5}); none of the options match, so option verification is essential.

The second equation directly gives (p-q=2). The intersection point satisfies both equations, so full solving is not always necessary.

(\frac{2}{4}=\frac{7}{14}\neq\frac{14}{31}), so the lines are distinct and parallel. Such a graph has no common point.

The second equation is (3) times the first, so (3k=6) and (k=2). In coincident lines, all terms are in the same ratio.

For parallel lines, (\frac{4}{8}=\frac{a}{10}), so (a=5). Since (\frac{16}{35}\neq\frac{1}{2}), they will not be coincident.

Substituting \(\left(-\frac{3}{2},4\right)\) makes both \(2x+y=1\) and \(x+2y=\frac{13}{2}\) true. The intersection point should be checked in both equations.

Putting (y=0) gives (x=3), and putting (x=0) gives (y=-7). Intercepts give simple and useful points for graphing.

Putting (x=-4) gives (-8+3y=7), so (y=5). With a vertical line, the (x)-coordinate is already fixed.

Adding the two equations gives (2p=18), so (p=9) and (q=4). Coordinates of the intersection satisfy both equations together.