Graphical method of finding solutions.

Correct answer: C. Infinitely many solutions

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

Correct answer: C. The lines are distinct and parallel

An inconsistent pair has no common point, so the lines are distinct and parallel. This is the no-solution case.

Correct answer: A. Intersecting

The coefficient ratios are not equal, so the lines meet at one point. This is the case of a unique solution.

Correct answer: A. \(\left(\frac{43}{13},\frac{37}{13}\right)\)

Putting \(y=6x-17\) in \(x+2y=9\) gives \(13x=43\), so \(x=\frac{43}{13}\) and \(y=\frac{37}{13}\). Fractional coordinates can also be correct graphical solutions.

Correct answer: A. All three points lie on the line

Substituting all three points makes (5x+4y=40) true. In a graph, three correct points should lie on the same straight line.

Correct answer: B. Inconsistent

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

Correct answer: B. (k=30)

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

Correct answer: A. \(\left(\frac{24}{5},\frac{32}{5}\right)\)

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.

Correct answer: A. (3x+y=1), (x-y=7)

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

Correct answer: C. Distinct parallel

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

Correct answer: C. (12)

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

Correct answer: C. (10)

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.

Correct answer: A. (1)

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.

Correct answer: A. ((35,23))

The direct answer is option A, (35,23), where x is the number of first-type books and y is the number of second-type books. “Total 58” gives x+y=58. “The first type is 12 more” gives x-y=12. Add the equations: 2x=70, so x=35. Substitute into x+y=58: 35+y=58, hence y=23. Thus the graphical intersection is (35,23). Option A works: 35+23=58 and 35-23=12. Option B, (23,35), reverses the types; its difference is -12. Option C, (34,24), has the correct total 58 but difference 10, not 12. Option D, (36,22), also totals 58 but difference 14. A graph would show the two lines x+y=58 and x-y=12 meeting at the answer point. Memory cue: define the larger quantity as x, then use x-y=the stated difference.

Correct answer: A. \((21,15)\)

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.

Correct answer: B. (5x-2y=11), (x+y=7)

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.

Correct answer: C. (4)

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

Correct answer: A. ((5,2))

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

Correct answer: B. Only the intersection point is the solution

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

Correct answer: A. ((3,5))

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.

Correct answer: B. (1)

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

Correct answer: A. (x+y=-5), (2x-y=-1)

Substituting ((-2,-3)) makes (x+y=-5) and (2x-y=-1) both true. Substituting the intersection point in both equations is the fastest check.

Correct answer: C. The solution set is the line (5x+6y=30)

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.

Correct answer: A. 2

Starting with \\(8x-4y=16\\), rearrange to get \\(-4y=16-8x\\), and then divide by \\(-4\\): \\(y=2x-4\\). Comparing this with \\(y=mx+c\\), the coefficient of \\(x\\) is the slope, so \\(m=2\\). Therefore, option A is correct. Option B results from a sign error while isolating \\(y\\). Exam tip: always express the equation in \\(y=mx+c\\) form before reading the slope.

Correct answer: A. ((3,4))

The second equation is (3) times the first, so every point on (2x+3y=18) is a solution. ((3,4)) lies on this line.