Graphical method of finding solutions.

The point where both lines meet gives the pair (x,y) satisfying both equations. In exams, always treat the intersection point as the solution.

One intersection point means the equations have exactly one solution. Remember, intersecting lines are consistent and independent.

Distinct parallel lines never intersect, so there is no common point. In exams, such lines are inconsistent.

Every point on coincident lines satisfies both equations. In exams, call this a consistent and dependent case.

For ( (2,3) ), (2+3=5), so it lies on the line. While checking options, substitute (x) and (y) in the equation.

In ( (3,2) ), (3-2=1), so it lies on the line. The easy method is to test each option in the equation.

Substituting x = 0 in 2x + y = 6 gives 2(0) + y = 6, so y = 6. Thus, the y-intercept of the line is 6 and the corresponding point is (0, 6). Option B is incorrect because when x = 0, the term 2x becomes 0. Exam tip: To find the y-intercept, put x = 0.

Substituting \(y=0\) in the equation gives \(x+2(0)=8\), so \(x=8\). Therefore, the x-intercept of the line is the point \((8,0)\). The value 4 may result from confusing the coefficient 2 with the required calculation. Exam tip: To find the x-intercept, always put \(y=0\).

The point ( (3,1) ) satisfies both (3+1=4) and (3-1=2). On the graph, this will be the intersection point.

The point of intersection must satisfy both equations. For (3, 3), \(3+3=6\) and \(3-3=0\), so it is the graphical solution. Option B satisfies the first equation but \(6-0\neq0\), so it does not lie on the second line. Exam tip: Substitute the coordinates into both equations to verify the intersection point.

For a pair of linear equations represented by two intersecting lines, the common solution is the coordinate pair of their point of intersection. At the point (2, 4), the horizontal coordinate gives x = 2 and the vertical coordinate gives y = 4. Therefore, the ordered solution is (x, y) = (2, 4), so option A is correct. Option B swaps the coordinates, which changes the point to (4, 2). Options C and D assign the same value to both variables and therefore do not represent the stated intersection. A unique intersection indicates a consistent, independent pair with one solution.

For (x=2), (y=2(2)=4), so ( (2,4) ) is correct. To check a point on a line, substitute its values in the equation.

Putting (x=4) gives (y=4+1=5), so ( (4,5) ) lies on the line. In such questions, check the options directly.

In the equation \(x=3\), the \(x\)-coordinate of every point is 3, while the \(y\)-coordinate can have any value. Hence, all such points lie on a vertical line parallel to the \(y\)-axis. Option B would be correct for \(y=3\). Exam tip: a constant \(x\)-value gives a vertical line, whereas a constant \(y\)-value gives a horizontal line.

In the equation \(y=2\), the y-coordinate is always 2, while x can take any value. Thus, all its points are of the form \((x,2)\), forming a horizontal line parallel to the x-axis. A line parallel to the y-axis has an equation of the form \(x=\text{constant}\). Exam tip: \(y=c\) represents a horizontal line, whereas \(x=c\) represents a vertical line.

The line (x=4) has (x=4) for all its points and (y=5) has (y=5) for all its points. Their common point is ( (4,5) ).

Substituting \(x=2\) in the equation gives \(2+y=7\). Therefore, \(y=7-2=5\), so option A is correct. The value 7 is the right-hand side of the equation, not the value of \(y\). For the graphical method, the point \((2,5)\) can be plotted on the line.

Substituting x = 2 gives 3(2) + y = 9, so 6 + y = 9. Therefore, y = 9 − 6 = 3, and the corresponding point is (2, 3). The value 6 is only the value of 3x, not y. Exam tip: When preparing a table for graphing, substitute the chosen x-value first and then calculate y.

At least (2) points are enough to draw a straight line. In exams, a third point may be used for checking.

The y-intercept is the point where a graph crosses the y-axis. Every point on the y-axis has x-coordinate 0, so substitute x = 0 into x + y = 3. This gives 0 + y = 3, hence y = 3. Therefore, the y-intercept is (0, 3), making option A correct. Option B is the x-intercept because it has y = 0. The points (1, 1) and (0, 0) do not satisfy the equation: 1 + 1 equals 2, and 0 + 0 equals 0, not 3. Thus they cannot be intercepts of this line.