Graphical method of finding solutions.

(\frac{4}{2}=\frac{6}{3}\neq\frac{18}{10}), so the lines are distinct and parallel. Do not ignore the constant ratio.

The first equation must be (2) times the second, so (a=4). In coincident lines, all terms change by the same multiplier.

Every point on the y-axis has x=0. In option A, substituting x=0 in both equations gives y=6, so the lines intersect at (0,6), which lies on the y-axis. For example, option B gives x=4 and y=2 on solving, so its intersection is not on the y-axis. Exam tip: To test whether the intersection lies on the y-axis, put x=0 and check whether both equations yield the same value of y.

The governing method is point verification: a point lies on a line exactly when its coordinates satisfy the line equation. For (7, 0), 2(7) + 7(0) = 14, so it lies on the line. For (0, 2), 0 + 14 = 14, so it also lies on the line. For (7/2, 1), 2(7/2) + 7(1) = 7 + 7 = 14, so it lies on the line as well. For (1, 2), 2(1) + 7(2) = 2 + 14 = 16, not 14. Therefore option D is the only point that does not lie on the line. The distinction comes from substitution, not from visual estimation of a graph.

The second equation is (2) times the first, so (m=6). In a coincident line, the coefficient of (y) also changes in the same ratio.

The governing concept is that the intersection coordinates satisfy both equations. Solve the pair by elimination. Multiply the second equation by 2: 10x − 4y = 14. Add this to the first equation, 3x + 4y = 25, to obtain 13x = 39, so x = 3. Substitute x = 3 into 5x − 2y = 7: 15 − 2y = 7, hence −2y = −8 and y = 4. Therefore the y-coordinate is 4, making option C correct. The other choices can result from reading the x-coordinate as the requested value, using only one equation incorrectly, or making an arithmetic error during elimination. Substitution into both original equations confirms the point (3, 4).

The equations are (x+y=42) and (x-y=8), giving (x=25), (y=17). In word problems, first define the variables clearly.

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.

Substituting ((4,1)) makes both (2x+y=9) and (x-y=3) true. Such a common point is the graphical solution.

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 governing concept is the slope-intercept form y = mx + c, where m is the slope and c is the y-intercept. In option A, both equations have m = 2, so their slopes are equal, while their intercepts are c = 5 and c = −3, which are different. Therefore the lines never meet and are distinct parallel lines. Option B has opposite slopes, so it does not satisfy the condition. In option C the slopes are 1 and 2, so they are unequal and the lines intersect. Option D has equal slopes and equal intercepts, meaning it describes the same line rather than two distinct lines. Thus option A is the only correct choice.

The second equation is (3) times the first, so every point on (x+2y=7) is a solution. ((1,3)) lies on this line.

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