Graphical method of finding solutions.

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.

Substitute both points into the options. For \\((-1,7)\\), \\(2(-1)+7=5\\), and for \\( (2,1)\\), \\(2(2)+1=5\\). Thus both points satisfy \\(2x+y=5\\). Alternatively, the slope is \\(m=(1-7)/(2-(-1))=-2\\), giving \\(y=-2x+5\\), or \\(2x+y=5\\). Exam tip: To verify a line equation, substitute both given points and check that each one satisfies it.

Dividing the second equation by (2) gives (x+2y=10). Same left side with different constants gives parallel lines.

Fractional coordinates need more care when read from a graph. In such questions, the scale must be very clear.

Subtracting the second from the first gives \(4y=12\), so \(y=3\); careful checking is needed. The correct solution is \(\left(5,3\right)\).

Substituting \(\left(5,3\right)\) gives \(2\left(5\right)+3\left(3\right)=19\) and \(2\left(5\right)-3=7\). If both are true, this is the solution.

The point \((2,6)\) satisfies the first equation because \(2+6=8\). Substituting \(x=2\) and \(y=6\) into the second equation gives \(2k+2(6)=14\), or \(2k+12=14\). Thus, \(2k=2\) and \(k=1\). Therefore, option A is correct. Exam tip: To test whether a point lies on a line, substitute its coordinates directly into the equation.

The intersection point \((2,5)\) is a common solution of both lines. Substituting \(x=2\) and \(y=5\) in the first equation gives \(2(2)+a(5)=16\), or \(4+5a=16\). Hence, \(5a=12\) and \(a=\frac{12}{5}\). Option A is incorrect because it makes the left-hand side equal to \(14\), not \(16\). Exam tip: When a graphical solution is given, substitute the point into both equations to verify it.

In option A, dividing every term of \(2x+4y=18\) by 2 gives \(x+2y=9\). Hence both equations have exactly the same graph and represent coincident lines. Options B and C produce distinct lines with the same slope, while option D has a different slope. Exam tip: for coincident lines, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Dividing (4x-6y=18) by (2) gives (2x-3y=9). Same left side with different constants gives distinct parallel lines.

Subtracting the equations gives \(2x=8\), so \(x=4\) and \(y=3\). In a real situation this is the meeting point.

Subtracting the equations gives \(x=8\), then \(8+5y=21\) gives \(y=\frac{13}{5}\). This is the graphical intersection.

Solving both equations gives \(x=\frac{21}{5}\) and \(y=\frac{23}{5}\). This point lies on both lines.

The intersection point \((3,1)\) lies on both lines. Substituting \(x=3\) and \(y=1\) into the first line gives \(3+a(1)=10\), so \(a=7\). The second equation also checks correctly: \(2(3)-1=5\). Exam tip: Substitute the given point into the equation containing the unknown parameter.

The intersection point must satisfy both line equations. The second equation is verified because \(2+4=6\). Substituting \(x=2\) and \(y=4\) into the first equation gives \(2k+2(4)=14\), so \(2k+8=14\), \(2k=6\), and hence \(k=3\). Exam tip: Substitute the coordinates of the given intersection point into the equation containing the unknown parameter.

In \(\left(6,-2\right)\), \(x=6\) and \(y=-2\). Reversing coordinates and changing sign makes the answer wrong.

Subtracting the equations gives \(2x=8\), so \(x=4\) and \(y=8\). Whatever the context, the intersection point is the solution.

For two lines to be coincident, the ratios of their corresponding coefficients must be equal. Writing both equations in standard form gives \(\frac{2}{2}=\frac{a}{3}=\frac{18}{18}\). Hence \(\frac{a}{3}=1\), so \(a=3\). Option 2 refers only to the coefficient of \(x\), so it is not correct. Exam tip: for coincident lines, use \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

The direct answer is A: \(b=3\). First simplify the second line: divide \(2x+6y=18\) by 2 to obtain \(x+3y=9\). The first line is \(x+by=7\). For the two lines to be parallel, the coefficients of x and y must have the same ratio, so b must be 3. With b=3, the equations become \(x+3y=7\) and \(x+3y=9\). They have identical left sides but different constants, so they cannot be the same line and never meet; hence they are parallel and distinct. Option A is correct. Option B, b=2, gives a different coefficient ratio and generally produces intersecting lines. Option C, b=6, also gives a different slope. Option D, b=1, likewise does not match the required direction. Notice that b=3 does not make the lines coincident because 7 is not 9. Exam cue: same x-y coefficient ratio but unequal constants means parallel distinct lines.

Substituting \(\left(0,-3\right)\) gives \(3\left(0\right)-4\left(-3\right)=12\). A negative (y)-intercept is plotted downward on the graph.

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.