Graphical method of finding solutions.

Substituting ( (0,0) ) gives (2(0)+2(0)=0), so the line passes through the origin. Checking the origin is easy.

A point lies on a line when its coordinates satisfy the equation. Testing option B gives 3(4) - 4(3) = 12 - 12 = 0, so (4, 3) is on the line. The other choices give 9 - 16 = -7, 3 - 8 = -5, and 6 - 4 = 2 respectively, so they do not satisfy the equation. Therefore B is unambiguous.

Exactly one solution is obtained when both lines intersect at one point. That point is the common solution of both equations.

In the graphical method, a solution is represented by a common point of the two lines. Distinct parallel lines never intersect, so they have no common point and the pair has no solution. Lines intersecting at one point give one solution, whereas coincident lines have infinitely many solutions. Exam tip: distinct parallel lines represent an inconsistent pair.

In the graphical method, each linear equation is drawn as a straight line, and common solutions are the points shared by the two lines. If the lines are distinct and parallel, they never meet, so there is no common solution. If they cross at one point, there is exactly one solution.

Infinitely many solutions occur only when both equations represent the very same line. Then every point on that line is common to both graphs, and each such point satisfies both equations. Therefore, option B is correct. Cutting the coordinate axes does not by itself create infinitely many common solutions, and a single given point cannot represent all the solutions of a pair of coincident lines.

The coordinates of the intersection point must satisfy both equations. Subtracting the second equation from the first gives \((x+3y)-(x+y)=15-7\), so \(2y=8\) and \(y=4\). Using \(x+y=7\), we get \(x=3\). Therefore, the intersection point is \((3,4)\). For example, the close distractor \((4,3)\) satisfies the second equation but not the first. Exam tip: The graphical solution is the common point of both lines.

The point of intersection must satisfy both linear equations simultaneously. Substituting \((6,4)\) gives \(2(6)+4=16\) and \(6+4=10\), so it is the graphical solution. For example, \((5,5)\) satisfies the second equation but not the first. Exam tip: verify a proposed intersection point by substituting its coordinates into both equations.

( (3,2) ) satisfies both equations. In graphical method this intersection point is the solution.

In the graphical method, the intersection point of the two lines is the solution of the pair of equations. Adding the equations gives 4x = 12, so x = 3. Substituting this in x + y = 5 gives y = 2. Therefore, the intersection point is (3, 2). The nearby option (2, 3) is incorrect because 3(2) − 3 = 3, not 7. Exam tip: Verify a coordinate pair in both equations before selecting the answer.

Distinct parallel lines never intersect, so there is no common point satisfying both equations. Hence, the pair has no solution. Exam tip: distinct parallel lines represent an inconsistent pair of equations.

The governing concept is comparing two linear equations after reducing them to a common standard form. Divide every term of the first equation 4x + 2y = 10 by 2. This gives 2x + y = 5, which is exactly the second equation. Therefore both equations describe the same set of points and represent one line drawn twice. Such lines are coincident and have infinitely many common solutions, so option D is correct. Parallel lines would have proportional x and y coefficients but different constants, intersecting lines would have different slopes, and perpendicularity is not indicated here. Scalar simplification proves coincidence directly.

Multiplying the first equation by (2) gives (4x+10y=30), different from the second. Same coefficient ratio with different constant gives parallel lines.

Putting ( (2,-1) ) gives (2-2(-1)=4) and (2(2)-(-1)=5). Watch signs carefully with negative coordinates.

To find the x-intercept, put y = 0: 3x = 12, so x = 4 and the point is (4, 0). To find the y-intercept, put x = 0: 2y = 12, so y = 6 and the point is (0, 6). Therefore, option B is correct. Exam tip: at the x-intercept y is zero, while at the y-intercept x is zero; option C uses the wrong values for both intercepts.

Substituting \(x=0\) gives \(4(0)-3y=12\), so \(-3y=12\) and \(y=-4\). Therefore, the point is \((0,-4)\). The option \((0,4)\) results from missing the negative sign. Exam tip: setting \(x=0\) gives the point where the line intersects the y-axis.

The governing concept is graphical solution of a pair of linear equations: the coordinates of the intersection point must satisfy both equations simultaneously. Substitute x=5 and y=1 into each option. For option B, x+y=5+1=6 and x-y=5-1=4, so both equations are true. Therefore (5,1) lies on both lines represented by option B, making it their common graphical solution. In option A, the two required values are interchanged; option C uses 5 and 1 as separate results rather than the sum and difference; and option D gives 7 and 3, which do not match the point. Hence B is the only unambiguous answer.

Checking ( (4,3) ) gives (2(4)+3(3)=17), not (18). Correct calculation gives (x=3) and (y=2).

Putting ( (3,2) ) gives (2(3)+3(2)=12), so it is not correct. The correct solution is ( (\frac{21}{5},\frac{16}{5}) ), so recalculation is needed in such options.

The governing concept is that the graphical solution of two linear equations is their common intersection point. The coordinates must satisfy both equations. Test option A: for (4,3), the first equation gives x+2y=4+2(3)=10, and the second gives 2x−y=2(4)−3=5. Thus (4,3) lies on both lines. The other options fail this simultaneous check: (3,4) gives 11 and 2; (5,2) gives 9 and 8; and (2,5) gives 12 and −1. Since only one point satisfies both equations exactly, the lines meet at (4,3). Algebraically, solving the pair also confirms this intersection, so option A is unambiguous.

( (5,2) ) satisfies both equations. In exams, quickly check options by substituting them in both equations.