Graphical method of finding solutions.

In (x=7), (x) is fixed and (y) can vary. Hence, it is a vertical line parallel to the (y)-axis.

The governing coordinate-geometry idea is that an equation y = a fixes the vertical coordinate while allowing x to take any real value. For y = 6, points such as (0,6), (2,6), and (−3,6) all lie on the graph. Since their y-coordinate is constant and their x-coordinates vary, they form a horizontal line. Every horizontal line is parallel to the x-axis, so option A is correct. It is not parallel to the y-axis, because a vertical line has the form x = a. It does not pass through the origin because the origin has y = 0, and it is not a single point because infinitely many x-values are possible.

Every point on the line \(x=9\) has x-coordinate 9, and every point on the line \(y=3\) has y-coordinate 3. Therefore, their common point is \((9,3)\). In option A, the coordinates are reversed. Exam tip: in \((x,y)\), the first value is the x-coordinate and the second value is the y-coordinate.

Putting \(y=0\) gives \(5x+0=25\), so \(x=5\). Therefore, the point is \((5,0)\), the x-intercept of the line. Option A is the y-intercept because it is obtained by putting \(x=0\), which gives \(y=25\). Exam tip: set \(y=0\) to find the x-intercept and \(x=0\) to find the y-intercept.

Putting x = 0 gives 0 + 5y = 25, so y = 5. Therefore, the point is (0, 5), which is the y-intercept of the line. Remember: to find the x-intercept, put y = 0; this gives (25, 0).

Adding the two equations gives \(2x=16\), so \(x=8\). Substituting this into \(x+y=12\) gives \(y=4\). Therefore, the point of intersection is \((8,4)\). In option B, \(x-y=4-8=-4\), so it does not satisfy the second equation. Exam tip: Verify a common point by substituting its coordinates into both equations.

Substituting ( (4,3) ) gives (4+3(3)=13) and (4+3=7). If both equations are satisfied, that is the intersection point.

Subtracting the second equation from the first gives \(x=3\). Substituting this into \(x+y=8\) gives \(y=5\). Therefore, the intersection point is \((3,5)\). Check: \(2(3)+5=11\) and \(3+5=8\). In the graphical method, the common point of the two lines represents the solution of the pair. Exam tip: verify both coordinates in both equations.

Two correct points are sufficient to draw one straight line. A third point may be taken for checking.

Two distinct parallel lines have no common point. Since a common point on the graph represents a solution of the pair, such a pair has no solution and is called an inconsistent pair. Exam tip: intersecting lines give one solution, coincident lines give infinitely many solutions, and distinct parallel lines give no solution.

Two linear equations are represented by two straight lines on a graph. If the lines meet at exactly one point, that point satisfies both equations. Therefore, the pair has one and only one ordered pair as its solution. Such a pair is called consistent because at least one solution exists, and independent because the solution is unique.

Hence the correct description is “consistent and independent,” option C. Parallel distinct lines have no common point and represent an inconsistent pair. Coincident lines have infinitely many common points and represent a consistent dependent pair. Since the question states that the lines intersect at one point, neither of those alternatives applies.

Dividing both sides of the first equation, \(2x+2y=18\), by \(2\) gives \(x+y=9\), which is exactly the second equation. Hence, both equations represent the same line and have infinitely many solutions. Exam tip: equations represent the same line when one is a non-zero multiple of the other.

Both have the same coefficients of (x) and (y), but different constants. Therefore, they are distinct parallel lines.

The governing idea is that a point lies on a line when its coordinates satisfy the equation. Here y is given as 2, so substitute y=2 in x−y=5: x−2=5. Adding 2 to both sides gives x=7. Therefore the ordered pair is (x,y)=(7,2), which is option B. Option A reverses the coordinate order, while option C incorrectly treats 5 as the x-coordinate without solving. Option D would give 3−2=1, not 5, so it does not lie on the line. The graph of the line must therefore contain (7,2).

A point lies on a line if its coordinates satisfy the equation of the line. For option C, \(2(3)-5=6-5=1\), so \((3,5)\) lies on the line. For example, option D gives \(2(4)-6=2\), not 1, so it is incorrect. Exam tip: Substitute the given \(x\)- and \(y\)-coordinates directly into the equation to test a point.

In the point ( (0,8) ), (x=0), so it lies on the (y)-axis. Identifying points on axes helps in graph reading.

In the point ( (9,0) ), (y=0), so it lies on the (x)-axis. This identification is very useful for intercepts.

To find the x-intercept, put y=0 in the equation, giving x=14 and the point (14, 0). To find the y-intercept, put x=0, giving y=14 and the point (0, 14). Therefore, option B is correct. Exam tip: set the other coordinate equal to zero to find an intercept.

For a straight line, any two distinct points satisfying the equation can be used. Put y = 0: 2x = 16, so x = 8, giving (8, 0). Put x = 0: 4y = 16, so y = 4, giving (0, 4). Thus option C gives the two intercepts. The other pairs contain at least one point that does not satisfy the equation.

The governing concept is the geometric form of a linear equation. An equation x=c represents a vertical line containing all points whose x-coordinate is c. Thus x=2 and x=8 are both vertical lines, but they are at different horizontal positions. Since one point cannot have x-coordinate 2 and 8 at the same time, the two lines have no common point. Distinct vertical lines never meet, so they are parallel; option D is correct. They are not intersecting because there is no solution to x=2 and x=8 together. They do not coincide because their constants differ, and neither is the x-axis, whose equation is y=0.