Graphical method of finding solutions.

The governing concept is the intercept of a graph. A y-intercept is the point where a line meets the y-axis, and every point on the y-axis has x = 0. Substitute x = 0 into 3x + y = 18: 3(0) + y = 18, so y = 18. Hence the intercept is the ordered pair (0, 18), making option C correct. Option A is the x-intercept because it results from setting y = 0. Options B and D confuse the coefficient 3 with an intercept or do not satisfy the required axis condition. The distinction between x = 0 and y = 0 is essential.

To find the \(x\)-intercept, put \(y=0\), because every point on the \(x\)-axis has a zero \(y\)-coordinate. Thus, \(5x+0=30\), giving \(x=6\); hence the intercept is \((6,0)\). Option B is incorrect because substituting \((5,0)\) does not satisfy the equation. Exam tip: set \(y=0\) for the \(x\)-intercept and \(x=0\) for the \(y\)-intercept.

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

Dividing the whole equation by (4) gives (2x+y=5). The simplified form makes line comparison easier.

The second equation is (4) times the first, so both are the same line. Such lines have infinitely many solutions.

The governing concept is testing whether an ordered pair satisfies an equation. Substitute x = 3 and y = 7 into each suitable expression. For option A, x + y = 3 + 7 = 10, so the equation is true. Option B would require 3 + 7 to equal 12, which is false. For option C, x - y = 3 - 7 = -4, not 10. For option D, 2x + y = 2(3) + 7 = 13, not 16. Therefore the point lies on the line represented by option A, and no other option is satisfied.

The governing concept is membership of a point on a line. An ordered pair (x, y) lies on y = x + 5 only when its coordinates satisfy the equation. For option B, x = 2 and y = 7; substituting gives x + 5 = 2 + 5 = 7, exactly the stated y-value. Option A gives 5 + 5 = 10, not 5. Option C gives 7 + 5 = 12, not 2, and option D gives 0 + 5 = 5, not 4. Hence (2, 7) is the only point on the line, so option B is correct.

Substituting \(x=3\) into the equation gives \(y=4(3)-2=12-2=10\). Therefore, the corresponding ordered pair is \((x,y)=(3,10)\). Options B and D interchange the \(x\)- and \(y\)-coordinates, while option A results from an arithmetic error. Exam tip: in an ordered pair, the first coordinate is \(x\) and the second is \(y\).

In (x=11), (x) is fixed and (y) can vary. Hence, it is a vertical line parallel to the (y)-axis.

In (y=8), (y) is fixed and (x) can vary. Hence, it is a horizontal line parallel to the (x)-axis.

The equation x=4 describes a vertical line. Every point on it has horizontal coordinate 4, while its vertical coordinate can be any real number. Similarly, y=9 describes a horizontal line. Every point on it has vertical coordinate 9, while its horizontal coordinate can be any real number.

Their common point must satisfy both conditions simultaneously: its x-coordinate must be 4 and its y-coordinate must be 9. Therefore the intersection is (4,9), so option B is correct. The order matters in an ordered pair: (9,4) would have x=9 and y=4, so it does not satisfy either required combination. Points (4,0) and (0,9) lie on only one of the two lines.

Substituting y = 0 gives 6x + 0 = 36, so x = 6. Therefore, the point is (6, 0), which is the x-intercept of the line. The point (36, 0) is incorrect because 36 is not the value of x after solving the equation. In exams, remember that setting y = 0 gives the x-intercept.

Substituting \(x=0\) gives \(0+6y=36\), so \(y=6\). Therefore, the point is \((0,6)\), which is the y-intercept of the line. Option C incorrectly takes \(y=36\) without dividing by 6. In an exam, remember: set \(y=0\) to find the x-intercept and \(x=0\) to find the y-intercept.

Adding the two equations gives \(2x=22\), so \(x=11\). Substituting this into \(x+y=16\) gives \(y=5\). Therefore, the intersection point is \((11,5)\). Option B has the coordinates reversed; although \(5+11=16\), \(5-11\neq6\). Exam tip: Always substitute a proposed point into both equations.

The common solution must satisfy both equations simultaneously. A useful algebraic method is to compare the two equations and remove the common term x. The equations are x+2y=11 and x+y=7. Subtracting the second equation from the first gives y=4. Substituting this value into x+y=7 gives x+4=7, so x=3.

Hence the ordered pair is (3,4), which is option A. A direct check confirms it: in the first equation, 3+2(4)=3+8=11; in the second, 3+4=7. Thus the point (3,4) lies on both lines and is their intersection. The order matters: (4,3) would give 4+2(3)=10, not 11, so it cannot be the solution.

Subtracting the second equation from the first gives x = 4. Substituting this in x + y = 9 gives 4 + y = 9, so y = 5. Therefore, the point of intersection is (4, 5). The distractor (3, 6) satisfies the second equation, but 2x + y = 12, not 13. In an exam, verify the ordered pair in both equations.

Two correct points are sufficient to draw a straight line. A third point may be used only for checking.

Distinct parallel lines never intersect, so they have no common point and the pair of linear equations has no solution. In contrast, intersecting lines give one solution, while coincident lines give infinitely many solutions. Exam tip: distinct parallel lines represent an inconsistent pair of equations.

A solution of two linear equations is a point that lies on both corresponding lines. When the graphs are exactly the same line, every point drawn on that line belongs to both graphs. Hence every one of those points satisfies both equations, rather than just one specially selected point.

Because a straight line contains infinitely many points, the equations have infinitely many common solutions. Therefore, option B is correct. One point of intersection would indicate a unique solution, which occurs for non-parallel lines that meet once. Distinct parallel lines have no common point and therefore no solution. The possibilities “only two” and “one” do not describe coincident lines.

When two lines intersect at exactly one point, the pair has one unique solution. Therefore, it is called a consistent and independent pair. In contrast, parallel lines form an inconsistent pair, while coincident lines form a consistent and dependent pair. Exam tip: one intersection point means one unique solution and hence a consistent, independent pair.