Graphical method of finding solutions.

Direct answer: option A, the x-axis. In an ordered pair (x,y), the second coordinate gives the vertical position. A point lies on the x-axis exactly when its y-coordinate is zero, because the x-axis is horizontal and has no height above or below it. For (14,0), y=0 and x=14, so the point is 14 units to the right of the origin on the x-axis. Option A is correct. Option B, the y-axis, would require x=0, but x=14. Option C, both axes, is wrong because a point is on both axes only when x=0 and y=0, namely the origin. Option D, no axis, is wrong because y=0 directly proves that the point is on the x-axis. Exam cue: check the coordinate equal to zero before doing anything else.

To find the x-intercept, put \(y=0\), which gives \(x=18\); hence the point is \((18,0)\). To find the y-intercept, put \(x=0\), which gives \(y=18\); hence the point is \((0,18)\). Therefore, option B is correct. The points in option A satisfy \(x+y=9\), not the given equation. Exam tip: set y=0 for the x-intercept and x=0 for the y-intercept.

The governing idea is that two distinct points determine a straight line. For the x-intercept, put y = 0: 3x = 24, so x = 8, giving (8, 0). For the y-intercept, put x = 0: 6y = 24, so y = 4, giving (0, 4). Thus option C gives both correct points; the other pairs contain incorrect intercepts or points not satisfying the equation.

The governing concept is the graph of an equation in the form x = a. It represents a vertical line because x is fixed at a while y may have any value. Thus x = 5 and x = 12 are vertical lines with different horizontal positions. They cannot share a point, since a common point would need its x-coordinate to be both 5 and 12. Distinct vertical lines therefore are parallel, so option D is correct. They do not coincide because their constants differ, and they do not intersect because they have no common point. Neither is the x-axis; the x-axis has equation y = 0. The other options confuse orientation, position, or axis identification.

The governing concept is the relationship between the number of solutions and the common points of two graphs. An equation y = a fixes the y-coordinate and permits every x-value, so it represents a horizontal line. Therefore y = 3 and y = 11 are distinct horizontal lines. A common solution would have to satisfy y = 3 and y = 11 simultaneously, which is impossible because 3 and 11 are different. Hence the lines have no common point and the system has no solution, making option B correct. Coincident lines would have infinitely many solutions, while two intersecting non-parallel lines would have one solution. Neither applies here.

In an ordered pair (x, y), the first coordinate is the x-coordinate and the second coordinate is the y-coordinate. Therefore, in (7, 13), y = 13. The value 7 is the x-coordinate, not the y-coordinate. Exam tip: the second coordinate of (x, y) always gives the value of y.

In an ordered pair \\( (x,y) \\), the first coordinate represents \\(x\\) and the second represents \\(y\\). Therefore, for the intersection point \\( (15,2) \\), \\(x=15\\). The value 2 in option A is the value of \\(y\\), not \\(x\\). Exam tip: Always read the coordinates in the order \\(x,y\\).

The direct answer is A: infinitely many solutions. Multiplying every part of x+y=7 by 3 gives 3x+3y=21, which is the second equation. Therefore the two equations represent one identical line. Any point on that line satisfies both equations, so there are infinitely many common ordered pairs, such as (0,7), (1,6), and (7,0). A is correct because coincident lines share every point. B is wrong because no solution occurs when distinct parallel lines never meet. C is wrong because one solution occurs when two different lines intersect once. D is wrong because two lines cannot have only two common points; distinct straight lines meet at most once, while identical lines have infinitely many common points. On a graph, the two drawings would completely overlap. A useful cue is to compare ratios of corresponding coefficients and constants: 3/1=3, 3/1=3, and 21/7=3, so the lines are coincident.

Multiplying the first equation by (3) gives (3x-3y=21), which differs from (3x-3y=24). Therefore, the lines are parallel.

Substituting \(x=5\) into the equation \(x+5y=15\) gives \(5+5y=15\). Hence, \(5y=10\) and \(y=2\). Therefore, the point is \((x,y)=(5,2)\). Exam tip: verify a proposed point by substituting both coordinates into the original equation.

Substituting y = 1 in 5x + y = 26 gives 5x + 1 = 26, so 5x = 25 and x = 5. Therefore, the ordered pair is (5, 1). In option B, the x- and y-coordinates are interchanged. As an exam tip, verify a point by substituting both coordinates into the original equation.

Substituting the point \((4,2)\) into both equations gives \(4+2(2)=8\) and \(2(4)+2=10\). Therefore, \((4,2)\) is the common point and hence the graphical intersection of the two lines. The nearby option \((3,2)\) is incorrect because it gives \(3+2(2)=7\), not 8, in the first equation. Exam tip: verify an intersection point in both equations.

The intersection point must satisfy both equations. Subtracting the first equation from the second gives x = 5. Substituting this into the first equation, 5 + 4y = 17, so y = 3. Therefore, the graphical solution is (5, 3). Exam tip: Substitute each option into both equations for a quick check; (3, 5) does not satisfy the first equation.

The governing concept is accurate scale selection in graph construction. A scale tells how numerical coordinate units correspond to physical distances on the graph paper. If the scale is wrong or applied inconsistently, a coordinate such as (2, 5) may be marked at an incorrect horizontal or vertical distance. Consequently, the plotted points do not represent the given values, and the line or intersection may appear in the wrong position. Therefore option A is correct. The algebraic equations do not change merely because the scale is chosen poorly, and naming the line or reading the question is unrelated to the immediate graphical error. A correct scale should be selected before plotting.

The governing convention for an ordered pair is (x, y): the first coordinate gives horizontal displacement and the second gives vertical displacement. Thus (9, 4) means x = 9 and y = 4, whereas (4, 9) means a completely different point with x = 4 and y = 9. Writing one in place of the other reverses the coordinate order, so option B is correct. The error is not about naming axes, because the coordinates themselves have been interchanged. It is not about failing to draw a line, and option D describes a correct action rather than a mistake. Careful use of (x, y) prevents this error.

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.

The governing concept is the use of axis conditions to find intercepts. At the x-intercept, the point lies on the x-axis, so y = 0. Substituting gives 2x + 5(0) = 20, hence 2x = 20 and x = 10; the x-intercept is (10, 0). At the y-intercept, x = 0 because the point lies on the y-axis. Thus 2(0) + 5y = 20, so y = 4 and the y-intercept is (0, 4). Therefore option A is correct. Option B reverses the numerical coordinates, but (4, 0) does not satisfy the equation. Option C gives incorrect intercept values, and option D also lists the correct-looking values in the wrong order for the requested pair.

The governing concept is the relationship between the two equations. Multiplying x + 3y = 9 by 2 gives 2x + 6y = 18, which is exactly the second equation. Consequently, both equations represent the same straight line rather than two different lines. Every point lying on that line satisfies both equations, so there are infinitely many common solutions. Therefore option C is correct. Option A would apply to two lines that intersect at one point, option B to distinct parallel lines, and option D is not a possible number of intersections for two coincident or distinct straight lines in this context.

A straight line is determined by its equation. Drawing 2x + y = 8 a second time does not change any coefficient or constant, so the second graph has exactly the same set of points as the first graph. The two representations therefore coincide and appear as one line, although algebraically they have been drawn twice. Every point on this line is common to both equations, which corresponds to infinitely many common solutions. Hence option C is correct. Option A would require parallel but distinct equations, option B is not possible because two lines cannot have two separate intersections, and option D is false because the equation does produce a valid graph.