Graphical method of finding solutions.

The governing concept is the y-intercept of a graph. The y-intercept is the point where a line crosses the y-axis, and every point on the y-axis has x-coordinate 0. Substituting x = 0 into 3x + y = 18 gives 3(0) + y = 18, so y = 18. Therefore the y-intercept is (0,18), making option C correct. Option A is the x-intercept, obtained by setting y = 0. Option B incorrectly treats the coefficient 3 as an intercept, while option D has the correct zero x-coordinate but an incorrect y-value. The distinction between x = 0 for the y-axis and y = 0 for the x-axis 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.

The direct answer is option B: 2x + y = 5. To simplify an equation without changing its solution set, divide every term on both sides by the same nonzero number. Starting with 8x + 4y = 20, divide all terms by 4: 8x/4 + 4y/4 = 20/4, giving 2x + y = 5. Option A, 4x + y = 20, divides only some terms and leaves the constant unchanged, so it is not equivalent. Option B divides every term correctly. Option C, x + 4y = 5, changes the coefficients incorrectly. Option D, 2x + 4y = 20, changes only one coefficient and is not the result of common division. Memory cue: when simplifying an equation, apply the same operation to every term on both sides.

The direct answer is C: coincident lines. In the first equation, every term is multiplied by 4: 4(2x+y=5) gives 8x+4y=20, exactly the second equation. Thus both equations describe the same set of points, not two different lines. Step by step, compare coefficients: 8/2=4, 4/1=4, and 20/5=4, so the ratios are equal. A is wrong because parallel distinct lines have no common point, whereas these lines have all points in common. B is wrong because intersecting lines have one common point, not infinitely many. C is correct because the graphs overlap completely and every point on the line is a solution of both equations. D is wrong because perpendicular lines meet at a right angle and these are the same line. Memory cue: if all three coefficients, including the constant, have the same multiplier, the lines are coincident and there are infinitely many solutions.

The governing concept is point verification by substitution. An ordered pair (x, y) satisfies an equation when replacing x and y by its coordinates makes the statement true. For (3, 7), option A gives x + y = 3 + 7 = 10, exactly matching its right-hand side. Option B fails because the same sum is 10, not 12. In option C, x − y = 3 − 7 = −4, not 10. In option D, 2x + y = 2(3) + 7 = 6 + 7 = 13, not 16. Thus the point lies on the line represented by option A and on none of the other listed lines. The direct substitution check is sufficient and gives one unambiguous answer, so A is correct.

The governing concept is checking whether an ordered pair belongs to the graph of a linear equation. A point (x, y) lies on y = x + 5 only when its y-coordinate equals the value obtained from x + 5. For option B, x = 2, so x + 5 = 2 + 5 = 7, exactly the given y-coordinate. Therefore (2, 7) lies on the line and option B is correct. The distractors fail under direct substitution: for (5, 5), the equation requires y = 10; for (7, 2), it requires y = 12; and for (0, 4), it requires y = 5. None of those coordinate pairs satisfies the equation. This verification also means that the graph passes through B, making the answer unique.

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.

The direct answer is A: a line parallel to the x-axis. In the equation y=8, the y-coordinate is fixed at 8, while x is free to take any real value. Thus points such as (0,8), (1,8), and (−3,8) all lie on the graph. These points form a horizontal line, and every horizontal line is parallel to the x-axis. A is correct. B is wrong because a line parallel to the y-axis has a fixed x-coordinate, such as x=8, not a fixed y-coordinate. C is wrong because a line through the origin must contain (0,0), but y=8 contains (0,8), not the origin. D is wrong because one equation in this form allows infinitely many points, not only one point. Remember: y=constant gives a horizontal line; x=constant gives a vertical line. The number 8 tells the height, not a single isolated point.

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.

Dividing every term of the first equation, \(3x+3y=24\), by 3 gives \(x+y=8\), which is the second equation. Hence, both equations represent the same line and have infinitely many common solutions. Exam tip: If one linear equation is a non-zero constant multiple of the other, their graphs are coincident, or the same line.

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

The governing concept is obtaining a coordinate pair from a linear equation by substitution. The equation is x − y = 8 and y is given as 3. Substitute y = 3: x − 3 = 8. Adding 3 to both sides gives x = 11. Since an ordered pair is written as (x,y), the required point is (11,3), so option B is correct. Option A reverses the coordinates. Option C copies the constant 8 as x without solving. Option D results from the incorrect operation 8 − 3; when isolating x in x − 3 = 8, 3 must be added. Verification is immediate: 11 − 3 = 8.

For the point (3, 7), substituting x = 3 and y = 7 gives 3x − y = 3(3) − 7 = 9 − 7 = 2. Therefore, the point lies on the line. The other points do not satisfy the equation; for example, (1, 3) gives 3(1) − 3 = 0. In an exam, test a point by directly substituting its coordinates into the equation.

Direct answer: option D, the y-axis. A point is written as (x,y), where the first number tells horizontal position and the second tells vertical position. Every point on the y-axis has x-coordinate zero, because it is neither to the left nor to the right of the vertical axis. Here the point is (0,12), so x=0 and y=12; therefore it lies 12 units above the origin on the y-axis. Option A, the x-axis, is wrong because points on that axis have y=0. Option B, both axes, is wrong because only the origin (0,0) belongs to both axes. Option C, no axis, is wrong because x=0 is exactly the test for the y-axis. Option D is correct. Memory cue: x=0 means y-axis; y=0 means x-axis.