Graphical method of finding solutions.

Putting \(x=-4\) gives \(3\left(-4\right)-2y=10\), so \(y=-11\). In a vertical line, (x) is already fixed.

The point of intersection must satisfy both line equations. From the first line, \(y=5\). Substituting this into the second equation gives \(4x-3(5)=17\), so \(4x=32\) and \(x=8\). Hence, the intersection point is \((8,5)\). Remember that \(y=5\) is a horizontal line, so its points all have y-coordinate 5.

When all three ratios are equal, both equations represent the same line. Therefore there are infinitely many common points.

When coefficient ratios are unequal, the lines intersect at one point. Therefore the pair is consistent and independent.

When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the two lines have the same slope but are not the same line. Hence, they are distinct parallel lines and have no common point. Coincident lines require all three ratios to be equal, while intersecting lines require the first two ratios to be unequal. Exam tip: Compare the ratios in the order \(a\), \(b\), and \(c\) to identify the graphical condition.

The intersection point is the common solution of both linear equations. Adding the equations gives \(4x=16\), so \(x=4\). Substituting this into \(x+2y=11\) gives \(4+2y=11\), hence \(y=\frac{7}{2}\). Therefore, the intersection point is \(\left(4,\frac{7}{2}\right)\). In option B, the values of \(x\) and \(y\) are interchanged. Exam tip: Always substitute the ordered pair into both equations to verify the intersection point.

\(\left(3,\frac{5}{2}\right)\) satisfies both equations. Read fraction coordinates carefully using scale on the graph.

At \(\left(1,5\right)\), \(2\left(1\right)+5\left(5\right)=27\), but \(1+5=6\). To be a common solution, both equations must be true.

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

The first coordinate of a point is (x) and the second is (y). Do not change order while reading negative fraction coordinates.

Multiplying the first equation by (2) gives (6x-4y=12), while the second is (6x-4y=18). Hence the lines are parallel and distinct.

The line \(y=0\) represents the \(x\)-axis, so the point of intersection must have \(y=0\). Substituting this into the first equation gives \(2x+3(0)=18\), hence \(2x=18\) and \(x=9\). Therefore, the intersection point is \((9,0)\). Exam tip: To find where a line meets the \(x\)-axis, put \(y=0\).

On the line \\(x=0\\), the x-coordinate of the intersection point must be 0. Substituting \\(x=0\\) in \\(4x-5y=20\\) gives \\(4(0)-5y=20\\), so \\(-5y=20\\) and \\(y=-4\\). Hence, the intersection point is \\((0,-4)\\). Option A has the wrong sign for \\(y\\). Exam tip: the line \\(x=0\\) is the y-axis, so substitute \\(x=0\\) directly into the other equation.

For the pair of linear equations a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0, the graphical classification uses the ratios of corresponding coefficients. Here a₁/a₂=5/1=5, b₁/b₂=10/2=5, and c₁/c₂=15/3=5. Since all three ratios are equal, the second equation is obtained by multiplying the first equation by a nonzero constant, so both equations represent the same geometric line. Consequently, the pair has infinitely many common solutions. Parallel distinct lines would require the first two ratios to be equal but different from the third, so option A is not suitable.

(\frac{a_1}{a_2}=\frac{b_1}{b_2}=2), but (\frac{c_1}{c_2}=\frac{9}{4}). Hence the lines are parallel and inconsistent.

In the third pair, (\frac{a_1}{a_2}\ne\frac{b_1}{b_2}). Hence the lines intersect at one point and the pair is consistent independent.

At \(x=0\), \(y=-4\), and at \(y=0\), \(x=12\). While finding intercepts, note which variable is kept zero.

\(2.25=\frac{9}{4}\) and \(-1.5=-\frac{3}{2}\). It is better to convert decimal coordinates into simplified fractions.

Substituting \(\left(3,4\right)\) gives \(2\left(3\right)+4=10\) and \(3+2\left(4\right)=11\). This is the common point of both lines.

From the first equation, \(y=5x-13\). Substituting this in the second equation gives \(x+2(5x-13)=16\), so \(11x=42\) and \(x=\frac{42}{11}\). Therefore, \(y=5\left(\frac{42}{11}\right)-13=\frac{67}{11}\). Hence, the graphs intersect at \(\left(\frac{42}{11},\frac{67}{11}\right)\). Option B incorrectly interchanges the \(x\)- and \(y\)-coordinates. Exam tip: Always substitute the intersection coordinates into both original equations to verify them.