Graphical method of finding solutions.

Substituting the origin ( (0,0) ) gives left side (0), not (6). Check whether a line passes through origin by substitution.

For the point \((2,5)\), substituting \(x=2\) and \(y=5\) gives \(5x-2y=5(2)-2(5)=10-10=0\). Therefore, the point lies on the line. For example, substituting \((2,3)\) gives \(10-6=4\), so it does not lie on the line. In an exam, quickly test a point by substituting its coordinates into the equation.

Direct answer: option C, (2,2). A point is on both lines only if substitution gives the stated right side in both equations. For (2,2), the first expression is x+2y=2+2(2)=2+4=6. The second is 2x+y=2(2)+2=4+2=6. Therefore (2,2) is the common point and hence the intersection. Option A, (1,4), gives 1+8=9 in the first equation and 6 in the second, so it fails the first. Option B, (4,1), gives 4+2=6 in the first but 8+1=9 in the second, so it fails the second. Option C passes both checks. Option D, (3,3), gives 9 in both expressions, not 6. Thus C is the only valid answer. Always test both equations, not just one.

Direct answer: option A, (5/2,3/2). A decimal with one digit after the decimal point can be changed to a fraction with denominator 10. Thus 2.5=25/10; divide numerator and denominator by 5 to get 5/2. Similarly, 1.5=15/10; divide by 5 to get 3/2. Keep the coordinate order unchanged: the first coordinate remains 5/2 and the second remains 3/2. Option A is correct. Option B swaps the coordinates, so it represents (1.5,2.5), not the given point. Option C has 25/10=2.5 but 15/100=0.15, not 1.5. Option D gives 2/5=0.4 and 1/5=0.2, so both values are wrong. Memory cue: decimal tenths become /10, then simplify; never reverse x and y.

A point lies on a line when its coordinates satisfy the equation of that line. Therefore, for two lines to meet at the origin, the coordinates \((0,0)\) must satisfy both equations in the pair. This is a direct substitution test and does not require solving the whole system.

For option B, \(2x+y=0\) gives \(2(0)+0=0\), and \(x-y=0\) gives \(0-0=0\). Thus both lines pass through the origin, so their intersection is at the origin because they are distinct lines. In the other options, at least one equation has a nonzero constant that fails when \(x=y=0\). Hence option B is correct.

The second equation is exactly three times the first: multiplying \(x+2y=7\) by 3 gives \(3x+6y=21\). Hence both equations represent the same graph, so the lines are coincident. For parallel and distinct lines, the ratios of the coefficients of \(x\) and \(y\) are equal but the ratio of the constants is different. Exam tip: if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the two lines are coincident.

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

The governing idea is that an intersection point must satisfy both linear equations simultaneously. Add the equations to eliminate y: (x + y) + (x − y) = 9 + 1, giving 2x = 10 and therefore x = 5. Substitute x = 5 into x + y = 9: 5 + y = 9, so y = 4. Hence the lines meet at (5, 4), and option B is correct. A reverses the coordinates; although 4 + 5 equals 9, the second equation would give 4 − 5 = −1 rather than 1. For C, 6 + 3 is not 9, and D fails the second equation as well. Substitution into both original equations confirms the answer uniquely.

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, compare the coefficients: a₁/a₂ = 1/2, whereas b₁/b₂ = 2/1 = 2. Since these ratios are unequal, the lines have different slopes and therefore meet once. Option A represents the same line because the first equation is twice the second, so it has infinitely many solutions. In option B, the x- and y-coefficients are proportional but the constants are not: the lines are parallel and there is no solution. In option D, the first equation is three times the second, so the lines coincide. Hence C is the only correct choice.

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.

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

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, option A is correct; option B gives incorrect values, while C and D misidentify or reverse the intercept points. Exam tip: set \(y=0\) for the x-intercept and \(x=0\) for the y-intercept.