Graphical method of finding solutions.

A value table for a line is correct only when every listed ordered pair satisfies the equation. Rearrange 4x+y=16 as y=16-4x. For x=1, y=16-4(1)=12, giving the point (1,12). For x=3, y=16-4(3)=4, giving (3,4). Therefore option A contains two valid points on the line. In option B, the values of y are interchanged and do not satisfy the equation. Option C gives (0,4), for which 4x+y=4, and (4,16), for which the value is 32. Option D also fails substitution, so only A is correct.

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 \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Dividing (2x+6y=18) by (2) gives (x+3y=9). Thus (b=3) makes the lines parallel and distinct.

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.