Graphical method of finding solutions.

The y-intercept is the point where a line crosses the y-axis. Every point on the y-axis has \\(x=0\\). Substitute this value into \\(5x-3y=15\\): \\(5(0)-3y=15\\), so \\(-3y=15\\). Dividing by \\(-3\\) gives \\(y=-5\\). Therefore the y-intercept is \\((0,-5)\\), which is option B. The point \\((3,0)\\) is the x-intercept instead, since it is found by setting y equal to zero.

The result can also be checked by putting \\((0,-5)\\) into the equation: \\(5(0)-3(-5)=15\\), so the equation is satisfied. Because the y-coordinate is negative, the point lies five units below the origin on the y-axis. A positive point \\((0,5)\\) would give \\(-15\\), not 15, so it cannot be correct. Hence the supplied answer B is fully supported.

Direct answer: option C, coincident lines. Two equations represent the same line when one equation is a nonzero multiple of the other. Start with 2x+4y=16 and divide every term by 2: x+2y=8. This is exactly the second equation, so both graphs lie on top of each other. Every point on that one line satisfies both equations; therefore there are infinitely many common solutions. Option A, parallel and distinct, is wrong because distinct parallel lines have no common point, whereas these equations are identical after simplification. Option B, intersecting at one point, is wrong because identical lines do not cross at only one point. Option D, perpendicular, is wrong because perpendicular lines meet at a right angle and are not identical. Hence C is correct. Memory cue: proportional all three coefficients means coincident lines.

Multiplying the first equation by (2) gives (2x-6y=10), different from the second. Hence the lines are parallel and have no solution.

This condition represents distinct parallel lines. Therefore there is no common point and the pair is inconsistent.

When all three ratios are equal, both equations represent the same line. In this case, infinitely many solutions are obtained.

Since the first line is \(x=6\), the x-coordinate of the intersection must be 6. Substituting this in the second equation gives \(6+y=10\), so \(y=4\). Therefore, the intersection point is \((6,4)\). In \((6,10)\), the value of y has been incorrectly taken as 10. Exam tip: Always substitute the proposed point into both equations to verify it.

On the first line, \(y=-1\). Substituting this in the second equation gives \(2x-(-1)=9\), or \(2x+1=9\), so \(x=4\). Therefore, the point of intersection is \((4,-1)\). In option B the coordinates are reversed, while options C and D have an incorrect \(x\)-value or \(y\)-value, respectively. Exam tip: For a horizontal line of the form \(y=\text{constant}\), the y-coordinate of the intersection is known immediately.

Substituting the origin ( (0,0) ) gives left side (0), not (18). Check passing through origin by direct substitution.

At ( (4,2) ), (3(4)+2(2)=16), but (4+2=6). To be a solution of both lines, both equations must be true.

\(3.5=\frac{7}{2}\) and \(2.5=\frac{5}{2}\). Write decimal coordinates read from a graph as simplified fractions.

Distinct parallel lines never meet, so there is no common point and hence no solution. Check \(a_1/a_2=b_1/b_2\ne c_1/c_2\). Exam tip: equal slopes with different intercepts indicate parallel lines.

At (x=2), (6-y=4) gives (y=2), and at (x=3), (9-y=4) gives (y=5). Every table point must satisfy the equation.

Dividing the second equation by (2) gives (x+2y=8). Same left side with different constants gives parallel lines.

Here (\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{2}). Therefore both lines are coincident.

(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{1}{2}), but (\frac{c_1}{c_2}=\frac{9}{25}). Hence the lines are parallel and inconsistent.

The direct answer is B: treating coincident lines as having one solution. The second equation is exactly twice the first: 2(3x+2y=12) gives 6x+4y=24. Therefore both equations represent the same line, not two different lines. Every point on this common line satisfies both equations, so the pair has infinitely many solutions. A is wrong because the lines are not distinct parallel lines; distinct parallel lines have no solution, while coincident lines overlap. B correctly identifies the student’s mistake: a coincident pair was incorrectly assigned one solution. C is wrong because intersecting lines normally have exactly one solution, so calling them inconsistent is incorrect. D is irrelevant and false; reversing axis names is not the issue and would not explain the number of solutions. A useful check is to compare all ratios: 6/3=2, 4/2=2, and 24/12=2. Equal ratios indicate coincident lines. Memory cue: one intersection means one solution; complete overlap means infinitely many.

Adding the equations gives (2x=16), so (x=8) and (y=4). In a real problem, the meeting point is the graphical solution.

The governing concept is that the common point of two lines is the simultaneous solution of their equations. Subtract the second equation from the first: (2x + 3y) − (x + 3y) = 21 − 15. The y-terms cancel, giving x = 6. Substitute x = 6 into x + 3y = 15: 6 + 3y = 15, so 3y = 9 and y = 3. Therefore the lines meet at (6, 3), making option B correct. Verification gives 2(6) + 3(3) = 21 and 6 + 3(3) = 15. Option A fails the first equation, option C reverses the coordinates, and option D gives 18 rather than 21 in the first equation. Thus only B satisfies both equations.

An intersection point must satisfy both equations simultaneously. Subtract the first equation from the second: (2x + 4y) − (x + 4y) = 20 − 18, so x = 2. Substitute this value into x + 4y = 18: 2 + 4y = 18, which gives 4y = 16 and hence y = 4. Therefore the lines intersect at (2, 4), so option A is correct. A direct check confirms that 2 + 4(4) = 18 and 2(2) + 4(4) = 20. Option B swaps the coordinates and fails the equations, option C does not satisfy the first equation, and option D gives 22 in the first equation rather than 18. The elimination step is valid because subtracting the equations removes the common 4y term.

In \( \left(-3,2\right) \), the first coordinate is (x) and the second is (y). Do not change order with negative coordinates.

The intersection point must satisfy both linear equations simultaneously. For option A, 2(2) - (-1) = 4 + 1 = 5 and 2 + 2(-1) = 0, so both equations hold. The remaining options fail at least one equation when substituted. Hence the two graphs meet at (2, -1), making option A the unique answer.

Substituting the point \\(1,5\\) in option A gives \\(2(1)+5=7\\), and substituting \\(3,1\\) gives \\(2(3)+1=7\\). Thus, both points satisfy \\(2x+y=7\\), so this equation represents their line. For example, option C works for the first point because \\(1+5=6\\), but it fails for the second point since \\(3+1=4\\neq6\\). Exam tip: The quickest method is to substitute both given points into each option.

On the line x = 0, the value of x is zero. Substituting x = 0 in the first equation gives 3y = 12, so y = 4. Therefore, the intersection point is (0, 4). The point (4, 0) is the x-intercept, obtained by setting y = 0. In exams, remember that x = 0 represents the y-axis, while y = 0 represents the x-axis.

The line y = 0 represents the x-axis. To find the intersection, substitute y = 0 in 4x + y = 20: 4x = 20, so x = 5. Therefore, the intersection point is (5, 0). In (0, 5), the coordinates are interchanged, while (4, 0) does not satisfy the equation. Exam tip: y = 0 represents the x-axis, whereas x = 0 represents the y-axis.

The direct answer is C: infinitely many solutions. If two graphs overlap completely, they are coincident lines, meaning they are actually the same line. Every point on that line belongs to both graphs and makes both equations true. Since a straight line contains infinitely many points, the pair has infinitely many common solutions. A is wrong because no solution describes distinct parallel lines that never meet. B is wrong because exactly one solution describes two distinct intersecting lines meeting at one point. C is correct because complete overlap gives every point as a common solution. D is wrong because perpendicular lines meet at a right angle and do not overlap completely. For example, if one point on the common line is a solution, there are endlessly many other points on the same line as well. Do not count the visible drawing as one line and assume one solution; the important fact is that all its points satisfy both equations. Memory cue: overlap means identical equations geometrically, so the answer is infinitely many, not one.