Graphical method of finding solutions.

To find the \(x\)-intercept, put \(y=0\) because the point lies on the \(x\)-axis. Thus, \(5x+2(0)=20\), so \(5x=20\) and \(x=4\). Therefore, the \(x\)-intercept is \((4,0)\). Option B is the \(y\)-intercept because it has \(x=0\). Exam tip: set \(y=0\) for the \(x\)-intercept and \(x=0\) for the \(y\)-intercept.

Substituting \(x=0\) gives \(2(0)-7y=14\), so \(-7y=14\) and \(y=-2\). Therefore, the point is \((0,-2)\). Option C results from setting \(y=0\), which gives the x-intercept rather than the required point. Exam tip: Set \(x=0\) to find the y-intercept.

Substituting \(y=0\) gives \(6x+0=18\), so \(x=3\). Therefore, the point is \((3,0)\), which represents the x-intercept of the line on the x-axis. In option B, the coordinates are interchanged, while 18 cannot be directly taken as a coordinate. Exam tip: Set \(y=0\) to find the x-intercept.

Dividing every term of the first equation by 3 gives \(x+2y=8\), which is exactly the second equation. Hence both equations represent the same line on the graph, so the lines are coincident and have infinitely many solutions. Parallel distinct lines require the constant terms not to be in the same ratio as the coefficients. Exam tip: compare \(a_1/a_2\), \(b_1/b_2\), and \(c_1/c_2\); if all three ratios are equal, the lines are coincident.

Dividing the first equation by (2) gives (x+2y=5), while the second is (x+2y=9). Hence the lines are parallel.

The direct answer is C: the graphs are coincident. The first equation is \(5x-10y=20\). Divide every term by 5, which is nonzero: \(x-2y=4\). This is exactly the second equation, so both equations describe the same straight line. Every point on that line satisfies both equations, giving infinitely many common solutions. Option A, intersecting at one point, would describe two different non-parallel lines, but these equations are identical after simplification. Option B, parallel and distinct, would require different lines with the same direction, not the same equation. Option C is correct because the two graphs lie exactly on top of each other. Option D, like axes, is not a standard conclusion and is not supported by these equations; the lines are not generally the coordinate axes. Exam cue: simplify both equations first. If one becomes exactly the other, the lines are coincident and there are infinitely many solutions.

Multiplying the first equation by (2) gives (2x-8y=6), different from (2x-8y=14). Hence the lines are parallel.

When all three ratios are equal, both equations represent the same line. In this case there are infinitely many solutions.

Unequal coefficient ratios mean the lines intersect at one point. Therefore the pair is consistent and independent.

For the pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the graphical classification uses the coefficient ratios. If a₁/a₂ = b₁/b₂ but this common ratio is different from c₁/c₂, the x- and y-coefficients are proportional while the constant terms are not. The equations therefore have the same slope but different intercepts. Their graphs are distinct parallel lines and have no common solution, so option B is correct. Coincident lines require all three ratios to be equal, while intersecting lines require the first two ratios to be unequal. Meeting at the origin is not implied by this ratio condition.

(\frac{a_1}{a_2}=\frac{b_1}{b_2}=3), but (\frac{c_1}{c_2}=\frac{7}{5}). Hence the lines are parallel and inconsistent.

Putting \(x=5\) gives \(2\left(5\right)+y=17\), so \(y=7\). In a vertical line, (x) is already fixed.

Putting \(y=-3\) gives \(4x-\left(-3\right)=19\), so \(x=4\). In a horizontal line, (y) is fixed.

The line \(x=0\) represents the y-axis. For the point of intersection, substitute \(x=0\) in \(3x+2y=12\), giving \(2y=12\) and hence \(y=6\). Therefore, the intersection point is \((0,6)\). Option C is incorrect because \((0,4)\) gives \(3x+2y=8\), not 12. Exam tip: Always substitute both coordinates of the obtained point back into the original equations to verify the intersection.

To find the intersection, solve the two equations simultaneously. Since the first line has \(y=0\), substitute this in the second equation: \(5x+2(0)=25\), giving \(x=5\). Therefore, the intersection point is \((5,0)\). Option C incorrectly takes \(x=25\), ignoring the coefficient 5. Exam tip: when one line is \(y=0\), substitute \(y=0\) directly into the other equation.

At \(\left(3,6\right)\), \(2\left(3\right)+3\left(6\right)=24\), but \(3+6=9\). For a common solution, both equations must be true.

In the point \(\left(-4,3\right)\), the first coordinate is (x) and the second is (y). Do not change order with negative coordinates.

\(4.5=\frac{9}{2}\) and \(1.5=\frac{3}{2}\). Write decimal coordinates read from a graph as simplified fractions.

\(\left(0,0\right)\) satisfies both \(2x-y=0\) and \(x+3y=0\). To check origin, put \(x=0,\ y=0\).

At (x=2), (6-y=5) gives (y=1), and at (x=4), (12-y=5) gives (y=7). Every table point must satisfy the equation.

Both given points must satisfy the equation of the line. For \((2,4)\), \(x+y=2+4=6\), and for \((5,1)\), \(x+y=5+1=6\). Therefore, the correct equation is \(x+y=6\). Option B satisfies only the first point, not the second. Exam tip: Substitute every given point into an equation before selecting it.

The direct answer is B: the mistake is treating coincident lines as having only one solution. A pair of linear equations represents two lines. Here, multiplying x+2y=8 by 2 gives 2x+4y=16, exactly the first equation. Therefore both equations describe the same line, not two different lines. Every point on that line satisfies both equations, so there are infinitely many solutions. Option A is wrong because parallel distinct lines have no solution, whereas these lines are the same. Option B is correct because coincident lines have infinitely many common points. Option C is wrong because intersecting lines have exactly one solution and are consistent. Option D is unrelated; naming axes does not decide the number of solutions. Memory cue: same ratios in all coefficients mean coincident lines and infinitely many solutions.

Dividing the second equation by (2) gives (x+2y=7). Same left side with different constants gives parallel lines.

Dividing (2x+6y=30) by (2) gives (x+3y=15). Same left side with different constants gives distinct parallel lines.

Subtracting the equations gives \(x=4\), then \(4+y=8\) gives \(y=4\). In a real situation, the meeting point is the intersection.