(a) In the diagram, BA is parallel to DE. Find the value of x. (b) Illustrate graphically and shade the region in which inequalities \(y - 2x < 5 ; 2y + x \...

(a)

Through C, draw the line RCE parallel to BA and DE.

Since BA is parallel to RC,

\[\angle BCR=52^\circ\]

Also, since RC is parallel to DE,

\[\angle RCD=48^\circ.\]

Hence, using angles about a point,

\[x=360^\circ-(52^\circ+48^\circ)=260^\circ.\]

Therefore, \(x=260^\circ\).

(b)

First draw the boundary lines:

\[y-2x=5\quad\Rightarrow\quad y=2x+5\]

\[2y+x=4\quad\Rightarrow\quad y=2-\frac{x}{2}\]

\[y+2x=10\quad\Rightarrow\quad y=10-2x.\]

The shaded feasible region is shown below. The broken boundary \(y=2x+5\) is not included because the inequality is strict.

The corner points of the region are obtained from intersections of the boundary lines:

\[\left(-\frac65,\frac{13}{5}\right),\qquad \left(\frac54,\frac{15}{2}\right),\qquad \left(\frac{16}{3},-\frac23\right).\]

The points \(\left(-\frac65,\frac{13}{5}\right)\) and \(\left(\frac54,\frac{15}{2}\right)\) are excluded because they lie on \(y-2x=5\).