Sketch on a single diagram the graphs of y = 5 - |x| and y = |2x – 3|. Hence, or otherwise, solve the inequality |2x-3|<5-|x|.

a) (i) Sketch on a single diagram the graphs of $y = 5 - |x|$ and $y = |2x - 3|$.

To sketch the graphs, we identify key points and shapes for each function:

For $y = 5 - |x|$:

• If $x \ge 0$, $y = 5 - x$. This is a line segment from $(0,5)$ to $(5,0)$.
• If $x < 0$, $y = 5 - (-x) = 5 + x$. This is a line segment from $(0,5)$ to $(-5,0)$.
• The graph is a V-shape opening downwards, with its vertex at $(0,5)$. It passes through $(-5,0)$, $(0,5)$, and $(5,0)$.

For $y = |2x - 3|$:

• The critical point is where $2x - 3 = 0$, which is $x = \frac{3}{2}$.
• If $x \ge \frac{3}{2}$, $y = 2x - 3$. This is a line segment starting from $(\frac{3}{2},0)$. For example, it passes through $(2,1)$ and $(3,3)$.
• If $x < \frac{3}{2}$, $y = -(2x - 3) = -2x + 3$. This is a line segment ending at $(\frac{3}{2},0)$. For example, it passes through $(0,3)$ and $(1,1)$.
• The graph is a V-shape opening upwards, with its vertex at $(\frac{3}{2},0)$. It passes through $(0,3)$, $(\frac{3}{2},0)$, and $(3,3)$.

The intersection points of the two graphs are found by setting $5 - |x| = |2x - 3|$.
Case 1: $x < 0$
$5 + x = -(2x - 3)$
$5 + x = -2x + 3$
$3x = -2 \implies x = -\frac{2}{3}$.
$y = 5 - |-\frac{2}{3}| = 5 - \frac{2}{3} = \frac{13}{3}$. Intersection point: $(-\frac{2}{3}, \frac{13}{3})$.

Case 2: $0 \le x < \frac{3}{2}$
$5 - x = -(2x - 3)$
$5 - x = -2x + 3$
$x = -2$. This is outside the interval $0 \le x < \frac{3}{2}$, so no intersection here.

Case 3: $x \ge \frac{3}{2}$
$5 - x = 2x - 3$
$8 = 3x \implies x = \frac{8}{3}$.
$y = 5 - |\frac{8}{3}| = 5 - \frac{8}{3} = \frac{7}{3}$. Intersection point: $(\frac{8}{3}, \frac{7}{3})$.

a) (ii) Hence, or otherwise, solve the inequality $|2x - 3| < 5 - |x|$.

We need to solve the inequality $|2x - 3| < 5 - |x|$. We consider the same cases as for the intersection points.

Case 1: $x < 0$
The inequality becomes $-(2x - 3) < 5 - (-x)$:
$3 - 2x < 5 + x$
$-2 < 3x$
$x > -\frac{2}{3}$
Combining with $x < 0$, the solution for this case is $-\frac{2}{3} < x < 0$.

Case 2: $0 \le x < \frac{3}{2}$
The inequality becomes $-(2x - 3) < 5 - x$:
$3 - 2x < 5 - x$
$-2 < x$
Combining with $0 \le x < \frac{3}{2}$, the solution for this case is $0 \le x < \frac{3}{2}$.

Case 3: $x \ge \frac{3}{2}$
The inequality becomes $2x - 3 < 5 - x$:
$3x < 8$
$x < \frac{8}{3}$
Combining with $x \ge \frac{3}{2}$, the solution for this case is $\frac{3}{2} \le x < \frac{8}{3}$.

Combining the solutions from all three cases:
$(-\frac{2}{3} < x < 0) \cup (0 \le x < \frac{3}{2}) \cup (\frac{3}{2} \le x < \frac{8}{3})$
This simplifies to the interval between the two intersection points:

$$-\frac{2}{3} < x < \frac{8}{3}$$

The solution to the inequality is $\boxed{-\frac{2}{3} < x < \frac{8}{3}}$.