(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 \...

Answer Details

(a) Finding x.

From the diagram: BA is parallel to DE, the angle at B (angle ABC) is \(52^\circ\), and the angle at D is a reflex angle of \(312^\circ\), while x is the angle at C (angle BCD) formed by the broken line B-C-D.

First find the true (non-reflex) angle at D:

Method: draw an auxiliary line through C parallel to both BA and DE. Because \(BA \parallel DE\), a line drawn through C parallel to them is parallel to each, and it splits the angle at C into two parts.

• BC is a transversal cutting \(BA\) and the line through C. The alternate angles are equal, so the part of x adjacent to CB equals \(\angle ABC = 52^\circ\).
• CD is a transversal cutting \(DE\) and the line through C. The alternate angles are equal, so the part of x adjacent to CD equals \(\angle CDE = 48^\circ\).

Adding the two parts:

x = 100\(^\circ\).

(b) Graph of the inequalities.

The three inequalities are \(y - 2x < 5\), \(2y + x \geq 4\) and \(y + 2x \leq 10\). Draw the three boundary lines, decide each shaded side with a test point, and the required region is where all three overlap.

Boundary line 1: \(y - 2x = 5\), i.e. \(y = 2x + 5\). Passes through \((0,5)\) and \((-2.5,0)\). Broken line (strict \(<\)). Test \((0,0)\): \(0-0=0<5\) true, so shade the side containing the origin (below/right of this line).

Boundary line 2: \(2y + x = 4\), i.e. \(y = \dfrac{4-x}{2}\). Passes through \((0,2)\) and \((4,0)\). Solid line (\(\geq\)). Test \((0,0)\): \(0+0=0\), not \(\geq 4\), so shade the side NOT containing the origin (above this line).

Boundary line 3: \(y + 2x = 10\), i.e. \(y = 10 - 2x\). Passes through \((0,10)\) and \((5,0)\). Solid line (\(\leq\)). Test \((0,0)\): \(0+0=0\leq 10\) true, so shade the side containing the origin (below/left of this line).

The required region is the closed polygon common to all three shaded areas: on or above \(2y+x=4\), on or below \(y+2x=10\), and below \(y=2x+5\). A convenient interior check point such as \((2,2)\) satisfies all three: \(2-4=-2<5\), \(4+2=6\geq4\), \(2+4=6\leq10\). Shade the overlapping region.