Graphical method of finding solutions.

Dividing the first equation by (5) gives (x-2y=4). Therefore the two lines have infinitely many common points.

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 first equation is (2) times the second, so the lines are coincident. Coincident lines have 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.