Graphical method of finding solutions.

Here \(x+y=\frac{5}{2}+\frac{7}{2}=\frac{12}{2}=6\). Values of (x) and (y) are read directly from the intersection point.

For parallel lines, (\frac{k}{2}=\frac{3}{6}), so (k=1). The constants ratio is different, so the lines are distinct parallel lines.

Using \(x=y+1\) from \(x-y=1\) gives \(y=\frac{16}{5}\) and \(x=\frac{21}{5}\). On the graph this is the intersection point.

From the first equation, \(y=4x-9\). Substituting this in the second equation gives \(2x+3(4x-9)=23\), so \(14x=50\) and \(x=\frac{25}{7}\). Therefore, \(y=4\left(\frac{25}{7}\right)-9=\frac{37}{7}\). Hence, the intersection point of the two lines is \(\left(\frac{25}{7},\frac{37}{7}\right)\). In option B, the coordinates are interchanged, so the point does not satisfy the equations. Exam tip: Always substitute the intersection point into both original equations to verify it.

Substituting \(\left(5,2\right)\) gives \(5\left(5\right)+3\left(2\right)=31\) and \(5+2=7\). If both equations are true, that is the intersection.

Elimination gives \(17y=50\) and \(x=\frac{91}{17}\). Fraction coordinates can also be graphical solutions.

(\frac{a_1}{a_2}=\frac{b_1}{b_2}=3), but (\frac{c_1}{c_2}=\frac{12}{5}). Hence the lines are parallel and inconsistent.

Dividing the first equation by (4) gives (2x-3y=5). Therefore both are the same line and have infinitely many solutions.

To find the x-intercept, put y=0: 9x=45, giving x=5 and the point \\( (5,0) \\). To find the y-intercept, put x=0: -5y=45, giving y=-9 and the point \\( (0,-9) \\). Therefore, option A is correct. Option C incorrectly treats 9 and 5 as the coordinates instead of calculating the intercepts. Exam tip: set y=0 for the x-intercept and x=0 for the y-intercept.

Putting \(x=-5\) gives \(4\left(-5\right)-3y=7\), so \(y=-9\). In a vertical line, (x) is already fixed.

The intersection point must satisfy both line equations. From the first line, \(y=6\). Substituting this into the second equation gives \(5x-2(6)=23\), so \(5x=35\) and \(x=7\). Therefore, the intersection point is \((7,6)\). Exam tip: for a horizontal line \(y=c\), the y-coordinate of every point on it is immediately known to be \(c\).

The intersection point on the graph must satisfy both equations simultaneously. Adding the two equations gives \(5x=25\), so \(x=5\). Substituting this into the first equation gives \(5+3y=14\), hence \(y=3\). Therefore, the intersection point is \((5,3)\). Exam tip: verify a suspected ordered pair in both equations before selecting it.

\(\left(\frac{25}{7},\frac{23}{7}\right)\) satisfies both equations. Read fraction coordinates carefully using the graph scale.

At \(\left(2,5\right)\), \(3\left(2\right)+4\left(5\right)=26\), but \(2+5=7\) also, so check fully. The correct non-common point is \(\left(4,\frac{7}{2}\right)\).

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 (10x-6y=20), while the second is (10x-6y=25). Hence the lines are parallel and distinct.

Every point on coincident lines satisfies both equations, so the system has infinitely many solutions. Check \(a_1/a_2=b_1/b_2=c_1/c_2\). Distinct parallel lines have no solution. Exam tip: identify the line relationship before selecting the answer.

The line \(x=0\) represents the y-axis, so substitute \(x=0\) into the first equation: \(6(0)-7y=42\), giving \(-7y=42\) and hence \(y=-6\). Therefore, the intersection point is \((0,-6)\). Option A has the wrong sign for the y-coordinate, while option C is the x-intercept of the first line and does not lie on \(x=0\). Exam tip: when \(x=0\), the first coordinate of the point must be zero.

(\frac{a_1}{a_2}=\frac{b_1}{b_2}=3), but (\frac{c_1}{c_2}=\frac{11}{4}). Hence the lines are parallel and inconsistent.

At \(x=0\), \(y=-4\), and at \(y=0\), \(x=16\). While finding intercepts, note which variable is kept zero.