(a)(i) Rearrange the equations:
\[y-\frac{3x}{4}=3\quad\Rightarrow\quad y=\frac34x+3\]
\[y+2x=6\quad\Rightarrow\quad y=6-2x\]
Line Points used for plotting \(y=\frac34x+3\) \((-4,0),\ (0,3),\ (4,6)\) \(y=6-2x\) \((0,6),\ (1,4),\ (3,0)\)
Using the scale of 2 cm to 1 unit on both axes, the required graph is:
Plot both straight lines on the same axes. Shade region R on and above the solid line y = 3x/4 + 3. The boundary \(y-\frac34x=3\) is drawn as a solid line since the inequality includes equality. Region \(R\), the side above this line, is shaded.
(ii) The two lines intersect at approximately
\[\boxed{(1.1,\ 3.8)}\]
This agrees with the exact solution:
\[\frac34x+3=6-2x\]
\[\frac{11}{4}x=3\quad\Rightarrow\quad x=\frac{12}{11}\]
\[y=6-2\left(\frac{12}{11}\right)=\frac{42}{11}\]
Thus the exact coordinates are \(\left(\frac{12}{11},\frac{42}{11}\right)\), which are approximately \((1.1,3.8)\).
(iii) \[y-\frac34x\geq3\quad\Rightarrow\quad y\geq\frac34x+3.\]
Hence, the required region is the half-plane on and above \(y=\frac34x+3\), labelled \(R\) on the graph.
(b)
\[(x+2)(x+c)=x^2+(c+2)x+2c.\]
Comparing with \(x^2+bx+18\):
\[2c=18\Rightarrow c=9,\]
\[b=c+2=9+2=11.\]
\[\boxed{c=9,\quad b=11}\]