Graphical method of finding solutions.

The governing concept is obtaining a coordinate point on a linear equation by assigning one coordinate and calculating the other. Substitute x = 4 into x + y = 11: 4 + y = 11. Subtracting 4 from both sides gives y = 7. Therefore the ordered pair is (x, y) = (4, 7), so option A is correct. Option B reverses the x- and y-coordinates and would give 7 + 4 = 11 only because the equation is symmetric, but it does not satisfy the stated condition x = 4. Options C and D use 11 as one coordinate without solving correctly. The point can be plotted as an ordinary point on the line.

Substituting \(x=1\) gives \(2(1)+y=10\), or \(2+y=10\). Therefore, \(y=8\). Choosing 6 would make the left side equal to 8, not 10, so it is incorrect. Exam tip: to find a point on a linear equation, substitute the known coordinate and solve for the other variable.

To find the y-intercept, put x=0 in the equation. Then \(3(0)+y=15\), so y=15 and the point is \((0,15)\). Option A is the x-intercept because it is obtained by putting y=0. Exam tip: every point on the y-axis has x-coordinate 0.

The \(x\)-intercept is the point where the line crosses the \(x\)-axis, so \(y=0\). Substituting this into the equation gives \(4x=20\), hence \(x=5\). Therefore, the \(x\)-intercept is \((5,0)\). Remember that \((0,20)\) is the \(y\)-intercept because \(x=0\) there.

Both have the same coefficients of (x) and (y), but different constants. Hence, the lines are parallel and have no solution.

The direct answer is option C, coincident lines. Multiply the first equation by 3: 3(2x+y=4) gives 6x+3y=12, which is exactly the second equation. Therefore both equations represent the same line, not two separate lines. Every point on one is on the other, so there are infinitely many solutions. Option A, parallel lines, is wrong because the equations are not distinct. Option B, intersecting at one point, is wrong because identical lines share infinitely many points. Option D, perpendicular lines, is wrong because no perpendicular relationship is produced by simply multiplying an entire equation by 3. Memory cue: if one complete linear equation is a nonzero multiple of the other, the lines are coincident.

In the graphical method, the point of intersection of two lines is the common solution of both equations. Since the intersection point is \((6,2)\), we have \(x=6\) and \(y=2\), so option D is correct. Option A incorrectly reverses the coordinates. Exam tip: In \((x,y)\), the first coordinate is always \(x\) and the second is \(y\).

Substituting ( (2,5) ) gives (2+5=7), so (x+y=7) is correct. To check a point, substitute it in the equation.

The direct answer is option B: (3,7). For a point (x,y) to lie on y = x + 4, substituting its x-coordinate must produce its y-coordinate. Option A, (4,4), gives x + 4 = 4 + 4 = 8, not 4, so it is wrong. Option B, (3,7), gives 3 + 4 = 7, so it satisfies the equation and is correct. Option C, (7,3), gives 7 + 4 = 11, not 3, so it is wrong. Option D, (0,3), gives 0 + 4 = 4, not 3, so it is wrong. The ordered pair matters: changing the positions of x and y changes the result. Memory cue: calculate x + 4 and compare it directly with y.

Substituting \(x=2\) gives \(y=3(2)-1=6-1=5\). Therefore, the point is \((x,y)=(2,5)\). In option A, the value of \(y\) is incorrectly taken as 3. In the exam, remember to write coordinates in the order \((x,y)\).

The direct answer is option D, a line parallel to the y-axis. In x=7, the x-coordinate is fixed at 7, while y may take any real value. Thus points such as (7,0), (7,2), and (7,-3) lie on the line. Joining them gives a vertical line, and vertical lines are parallel to the y-axis. Option A is wrong because the origin (0,0) does not satisfy x=7. Option B is wrong because a line parallel to the x-axis has constant y, not constant x. Option C is wrong because x=7 clearly represents many points and therefore a line. Option D correctly describes it. Exam cue: x fixed means vertical; y fixed means horizontal.

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 concept is substitution in the equation of a line and the correct order of coordinates in an ordered pair. The given equation is x − y = 5 and y = 2. Substituting y = 2 gives x − 2 = 5. Adding 2 to both sides produces x = 7. Therefore the point is written as (x,y) = (7,2), so option B is correct. Option A reverses the x- and y-coordinates. Option C uses the constant 5 as the x-coordinate without solving the equation. Option D fails direct verification because 3 − 2 = 1, not 5. Substitution confirms that 7 − 2 = 5.

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.