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.

The direct answer is option A, the point \((2,4)\). A point \((x,y)\) lies on the line \(y=2x\) exactly when its coordinates satisfy that equation. For option A, substitute \(x=2\): the right side is \(2(2)=4\), which equals \(y=4\). Therefore \((2,4)\) is on the line. Option B, \((3,5)\), fails because \(2(3)=6\), not 5. Option C, \((1,3)\), fails because \(2(1)=2\), not 3. Option D, \((4,2)\), fails because \(2(4)=8\), not 2. The safest method is substitution: take the x-coordinate, multiply it by 2, and compare with the y-coordinate. The line has slope 2, so y must always be twice x. Memory cue: for \(y=2x\), every valid point has the pattern \((x,2x)\).

The direct answer is A: \((4,5)\). A point \((x,y)\) lies on \(y=x+1\) when its coordinates satisfy the equation. For A, substitute \(x=4\): the required value is \(y=4+1=5\), exactly the given second coordinate. So A works. For B, \((5,4)\), the equation requires \(y=5+1=6\), not 4. For C, \((2,4)\), the required value is \(2+1=3\), not 4. For D, \((0,2)\), the required value is \(0+1=1\), not 2. The safest method is to substitute each point, or calculate \(y=x+1\) from its x-coordinate. A line contains precisely the points that make its equation true. Do not interchange x and y: the first coordinate is x and the second is y. Memory cue: check whether the second coordinate equals the first coordinate plus one.

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 governing concept is finding an axis intercept from a linear equation. The y-intercept is the point where the graph meets the y-axis, and every point on the y-axis has x-coordinate 0. Substitute x = 0 into x + y = 3: 0 + y = 3, so y = 3. Therefore the y-intercept is (0, 3), making option A correct. Option B has y = 0 and is therefore the x-intercept of this line. The point (1, 1) does not lie on the line because 1 + 1 = 2, not 3. Similarly, (0, 0) gives a sum of 0, not 3. Thus an intercept must satisfy both the equation and the coordinate condition for the relevant axis.

The governing concept is the coordinate condition for an x-intercept. A line meets the x-axis at a point whose y-coordinate is 0. Substitute y = 0 into x + y = 9: x + 0 = 9, so x = 9. Hence the x-intercept is (9, 0), and option A is correct. Option B is the y-intercept because its x-coordinate is 0. The points (1, 8) and (4, 5) do satisfy the equation, since their coordinate sums are 9, but neither lies on the x-axis: their y-coordinates are 8 and 5 rather than 0. This illustrates that merely satisfying the equation is not enough; an x-intercept must also lie on the x-axis.

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

To compare the two lines, simplify the first equation by dividing every term by 2: 2x + 2y = 10 becomes x + y = 5. This is exactly the second equation, not merely an equation with a similar slope. Therefore both equations represent the same geometric line, called coincident lines. Every point on that line satisfies both equations, so the pair has infinitely many solutions. Option A is correct. Distinct parallel lines would have the same slope but different intercepts, while intersecting lines would have different slopes and one common point. Options B, C, and D therefore do not describe this pair.

The two equations give different lines with unequal slopes. Hence, they will have one unique intersection point.

If two lines have no common point, they are distinct parallel lines, so the pair of equations has no solution. Such a pair is called inconsistent. In contrast, a dependent pair represents coincident lines and has infinitely many solutions. Exam tip: No point of intersection on the graph means an inconsistent pair.