Graphical method of finding solutions.

Subtracting the equations gives \(2x=14\), then \(x=7\) and \(7+5y=25\) gives \(y=\frac{18}{5}\). This is the graphical intersection.

Solving both equations gives \(y=\frac{22}{5}\) and \(x=\frac{26}{5}\). \(\frac{52}{10}\) can also be written as \(\frac{26}{5}\).

The intersection point \((2,4)\) must satisfy both line equations. Substituting \(x=2\) and \(y=4\) into the first equation gives \(2k+4(4)=22\), or \(2k+16=22\). Hence, \(2k=6\) and \(k=3\), so option B is correct. As a check, the point also satisfies the second equation because \(2+4=6\). Exam tip: when the intersection point is given, substitute its coordinates into the equation containing the unknown parameter.

In \(\left(7,-3\right)\), \(x=7\) and \(y=-3\). Reversing coordinates and changing sign makes the answer wrong.

Subtracting the equations gives \(3x=12\), so \(x=4\) and \(y=10\). Whatever the context, the intersection point is the solution.

The direct answer is option A, b=4. Two lines are parallel when their x- and y-coefficient ratios are equal, but they are distinct when their constant terms do not have the same ratio. The second equation is 3x+12y=30. Dividing it by 3 gives x+4y=10. The first equation is x+by=8, so b must be 4 to make the x and y coefficients proportional. Then the equations are x+4y=8 and x+4y=10: they have the same slope but different intercepts, so they are parallel and never meet. Option A works because b=4. Option B, b=3, gives x+3y=8, which has a different slope. Option C, b=12, also gives a different slope. Option D, b=1, gives x+y=8 and is not parallel to the second line. Memory cue: same coefficient ratio means parallel; a different constant ratio means distinct.

Substituting \(\left(0,-5\right)\) gives \(5\left(0\right)-6\left(-5\right)=30\). A negative (y)-intercept is plotted downward on the graph.

Here \(x+y=\frac{7}{2}+\frac{9}{2}=\frac{16}{2}=8\). Values of (x) and (y) are read directly from the intersection point.

For parallel lines, (\frac{k}{3}=\frac{-2}{-6}), so (k=1). The constants ratio is different, so the lines are distinct parallel lines.

From \(x-y=1\), \(x=y+1\), and substituting in the first equation gives \(9y=29\). Hence \(y=\frac{29}{9}\) and \(x=\frac{38}{9}\).

Multiplying the first equation by (2) gives (10x+4y=18), while the second is (10x+4y=25). Hence the lines are parallel and have no solution.

From \(x+4y=22\), \(x=22-4y\), and substituting in the first equation gives \(y=\frac{73}{16}\). Then \(x=\frac{15}{4}\).

Here \(x-y=-\frac{7}{3}-\frac{2}{3}=-3\). Values of (x) and (y) are read directly from the intersection point.

Subtracting the equations gives \(x=8\), then \(8+5y=26\) gives \(y=\frac{18}{5}\). This is the meeting point of both paths.

Here (\frac{2}{4}=\frac{3}{6}\neq\frac{12}{18}), so the lines are parallel. In exams, check ratios first.

The second equation is (2) times the first, so both lines are the same. If one line overlaps the other, there are infinitely many solutions.

The direct answer is option B: (4,3). To find the intersection, solve both equations together. Add x+y=7 and x-y=1: the y terms cancel, giving 2x=8. Divide by 2 to get x=4. Substitute into x+y=7: 4+y=7, so y=3. Therefore the common point is (4,3). B is correct because it satisfies both equations: 4+3=7 and 4-3=1. A, (3,4), satisfies the first equation, but 3-4=-1, not 1. C, (5,2), gives 5+2=7, but 5-2=3, not 1. D, (2,5), gives 2+5=7, but 2-5=-3, not 1. The intersection is the single point that lies on both lines, so checking both equations is a reliable method. Memory cue: add equations with opposite y signs, then substitute the found coordinate.

Both lines have the same slope (2), but different intercepts. Equal slope and different intercepts mean parallel lines.

The vertical line (x=5) and the horizontal line (y=-3) intersect at ((5,-3)). Remember the order ((x,y)).

The second equation is (2) times the first, so the lines are coincident. In coincident lines, every point is a solution.

Here (\frac{5}{10}=\frac{2}{4}\neq\frac{10}{25}), so the lines are parallel. Parallel lines have no common point.

The direct answer is option B, (4,1). A graphical solution is the point where both lines meet, so the point must satisfy both equations. Subtract x+y=5 from 2x+y=9: x=4. Substitute x=4 into x+y=5: 4+y=5, so y=1. Therefore the intersection is (4,1). Checking both equations gives 2(4)+1=9 and 4+1=5. Option A, (3,2), satisfies x+y=5 but gives 2x+y=8, not 9. Option B satisfies both, so it is correct. Option C, (2,3), satisfies x+y=5 but gives 7 in the first equation. Option D, (1,4), satisfies x+y=5 but gives 6 in the first equation. The graph would show both lines crossing at the same ordered pair. Memory cue: test both coordinates in both equations, not just one.

The direct answer is C: a1/a2=b1/b2=c1/c2. For equations a1x+b1y+c1=0 and a2x+b2y+c2=0, infinitely many solutions mean both graphs are the same line, called coincident lines. The coefficients of x, y, and the constant term must all have the same ratio, so all three ratios are equal. Option C is correct. Option A means the x and y coefficient ratios differ; the lines generally intersect once, giving one solution. Option B has equal x and y ratios but a different constant ratio; this gives distinct parallel lines and no solution. Option D only says one ratio is zero, which is neither the complete condition nor enough to identify coincident lines. Memory cue: equal all three ratios means same line and infinitely many solutions; equal first two but not the third means parallel and no solution.

The second equation is (3) times the first, so both lines are coincident. In such questions, check the multiplier of the whole equation.

Substituting ((-2,5)) gives (x+y=3) and (2x-y=-9), both true. Test the point in both equations.