Graphical method of finding solutions.

At ( (2,5) ), (2(2)+3(5)=19), but (2+5=7). To be a solution of both lines, both equations must be true.

A straight line is determined by its equation. Drawing 2x + y = 8 a second time does not change any coefficient or constant, so the second graph has exactly the same set of points as the first graph. The two representations therefore coincide and appear as one line, although algebraically they have been drawn twice. Every point on this line is common to both equations, which corresponds to infinitely many common solutions. Hence option C is correct. Option A would require parallel but distinct equations, option B is not possible because two lines cannot have two separate intersections, and option D is false because the equation does produce a valid graph.

The intersection point must satisfy both equations simultaneously. Add the equations: (x + y) + (x − y) = 9 + 1, so 2x = 10 and x = 5. Substitute x = 5 into x + y = 9: 5 + y = 9, giving y = 4. Thus the lines meet at (5, 4), so option B is correct. A reverses the coordinates and would give 4 + 5 = 9 but 4 − 5 = −1, not 1. Options C and D also fail at least one of the original equations. Graphically, this ordered pair is the single common point.

To find the \\(x\\)-intercept, put \\(y=0\\): \\(3x=24\\), so \\(x=8\\) and the point is \\((8,0)\\). To find the \\(y\\)-intercept, put \\(x=0\\): \\(4y=24\\), so \\(y=6\\) and the point is \\((0,6)\\). Therefore, the ordered pair in the stated sequence is \\((8,0)\\), \\((0,6)\\). Exam tip: set \\(y=0\\) for the x-intercept and \\(x=0\\) for the y-intercept.

The common point of two lines must satisfy both equations at the same time. The equations are \(x+4y=14\) and \(2x+4y=16\). Subtract the first equation from the second. The terms containing \(4y\) cancel, leaving \(x=2\). Substitute this value into the first equation: \(2+4y=14\), so \(4y=12\) and \(y=3\).

Therefore the lines meet at \((2,3)\), which is option A. Verification is easy: in the first equation, \(2+12=14\), and in the second, \(4+12=16\). Since the same point satisfies both equations, it is their intersection on the graph. Other listed points fail at least one equation.

The intersection point lies on both lines. Substituting (3, 2) gives 2(3) − 2 = 4 and 3 + 2(2) = 7, so it satisfies both equations. The other options do not satisfy both equations simultaneously. Exam tip: verify a graphical intersection point by substituting its coordinates into both equations.

In the point \( \left(-2,5\right) \), the first coordinate is (x) and the second is (y). Do not change the order while reading negative coordinates.

A pair has a unique graphical solution when its two lines intersect at exactly one point. For option C, the coefficient ratios are a₁/a₂ = 1/2 and b₁/b₂ = 2/1 = 2, which are unequal. Therefore the lines have different slopes and must meet once. In option A, the second equation is half the first, so the lines coincide. In option B, the left sides are proportional but the constants are not, giving distinct parallel lines and no solution. In option D, the first equation is three times the second, so the lines coincide. Hence option C is correct.

Multiplying the first equation by (2) gives (2x-2y=6), while the second is (2x-2y=10). Hence the lines are parallel and have no solution.

A pair is consistent and dependent when both equations represent the same line, so it has infinitely many solutions. In option A, multiplying 2x + y = 6 by 2 gives 4x + 2y = 12, exactly the second equation. Thus the two equations are dependent and every point on their common line is a solution. Option B represents parallel distinct lines because the left sides are identical but the constants differ, so it has no solution. Option C contains two non-proportional equations and generally has one solution. Option D fixes x and y separately and does not form a dependent pair of identical line equations. Therefore A is correct.

At (x=2), (2y=6) gives (y=3), and at (x=6), (2y=2) gives (y=1). Simple table values make the graph clear.

At (x=2), (4-y=3) gives (y=1), and at (x=3), (6-y=3) gives (y=3). Table points must satisfy the equation.

Dividing the second equation by (2) gives (x+2y=6), which is parallel to the first. Same left side with different constants gives parallel lines.

Here (\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{2}). Therefore the lines are coincident and have infinitely many solutions.

(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{1}{2}), but (\frac{c_1}{c_2}=\frac{5}{7}). Hence the lines are parallel and inconsistent.

The second equation is (2) times the first, so the lines are coincident. Coincident lines have infinitely many solutions, not only (1).

Adding both equations gives (2x=12), so (x=6) and (y=4). In a real situation, the meeting point is the graphical solution.

Substituting ( (4,2) ) makes both equations true. In graphical method, the common point of both lines is the solution.

The graphical solution of two linear equations is their common point, or the point where the two corresponding lines intersect. We can verify each candidate by substitution. For (3, 5), the first equation gives 2(3) + 5 = 6 + 5 = 11, and the second gives 3 + 5 = 8. Therefore (3, 5) lies on both lines and is their intersection point. Option B is correct. For (5, 3), the first equation gives 13; for (4, 4), it gives 12; and for (2, 6), it gives 10, so those points fail the first equation even though some may satisfy the second. Since a graphical solution must satisfy both equations simultaneously, only option B is valid.

( (-1,4) ) satisfies both equations. With negative coordinates, pay attention to signs while checking.