Graphical method of finding solutions.

The intersection point must satisfy both equations. Substituting \(x=-2\) into \(3x+y=1\) gives \(3(-2)+y=1\), or \(-6+y=1\), so \(y=7\). Therefore, the intersection point is \((-2,7)\). Option C has the wrong sign for the \(y\)-coordinate. Exam tip: Every point on the vertical line \(x=-2\) has \(x\)-coordinate \(-2\).

Every point on the line \(y=4\) has y-coordinate 4. Substituting this into the other equation gives \(2x+3(4)=22\), so \(2x=10\) and \(x=5\). Therefore, the intersection point is \((5,4)\). Option B incorrectly swaps the x- and y-coordinates. Exam tip: for a horizontal line \(y=c\), the y-coordinate is fixed at \(c\).

Subtracting the equations gives (2x=8), so (x=4) and (y=8). In real life, the meeting point is the intersection point.

Substituting \(\left(5,3\right)\) makes both equations true. In graphical method this common point is the solution.

At \(\left(4,3\right)\), \(4+3\left(3\right)=13\), so check options carefully. The correct intersection is \(\left(\frac{24}{7},\frac{27}{7}\right)\).

From \(2x-y=3\), \(y=2x-3\), and substituting in the first equation gives \(x=\frac{24}{7}\). Then \(y=\frac{27}{7}\).

At \(\left(4,2\right)\), \(3\left(4\right)+2\left(2\right)=16\) and \(4+2=6\). If both are true, the point is the intersection.

Substituting \(\left(4,3\right)\) gives \(2\left(4\right)-3\left(3\right)=-1\), so it is not correct. The correct solution is \(\left(\frac{22}{5},\frac{13}{5}\right)\).

Using \(y=7-x\) from \(x+y=7\), we get \(2x-3\left(7-x\right)=1\). Thus \(x=\frac{22}{5}\) and \(y=\frac{13}{5}\).

The intersection point must satisfy both line equations. Subtracting the second equation from the first gives \(3x=9\), so \(x=3\). Substituting this into \(x+y=9\) gives \(y=6\). Therefore, the intersection point is \((3,6)\). For example, option C satisfies \(x+y=9\), but \(4x+y=21\), so it is incorrect. Exam tip: always verify the obtained point in both equations.

At \(\left(3,2\right)\), \(3-2\left(2\right)=-1\), so it is not correct. The correct solution is \(\left(\frac{18}{7},\frac{23}{7}\right)\).

The first equation gives \(x=2y-4\). Substituting in \(3x+y=11\) gives \(y=\frac{23}{7}\) and \(x=\frac{18}{7}\).

At \(\left(3,3\right)\), \(2\left(3\right)+5\left(3\right)=21\) and \(3+3=6\). This is the common point of both lines.

To find the x-intercept, put y=0: 3x=30, so x=10 and the point is (10, 0). To find the y-intercept, put x=0: 5y=30, so y=6 and the point is (0, 6). Option B reverses the two intercept values. Exam tip: set y=0 for the x-intercept and x=0 for the y-intercept.

To find the \(y\)-intercept, set \(x=0\), because every point on the \(y\)-axis has an \(x\)-coordinate of zero. Thus, \(4(0)-3y=24\), giving \(y=-8\). Therefore, the \(y\)-intercept is \((0,-8)\). Option C is the \(x\)-intercept because it is obtained by setting \(y=0\). Exam tip: set \(x=0\) for the \(y\)-intercept and \(y=0\) for the \(x\)-intercept.

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.