Write down four inequalities which fully describe the unshaded region R in the figure S below.

• Line L1 (horizontal line): This line passes through $y=2$. Its equation is $y=2$.
• Line L2 (vertical line): This line passes through $x=-2$. Its equation is $x=-2$.
• Line L3 (diagonal line): This line passes through points $(0, -1)$ and $(3, 0)$. The slope $m = \frac{0 - (-1)}{3 - 0} = \frac{1}{3}$. Using the point-slope form $y - y_1 = m(x - x_1)$ with $(3, 0)$: $y - 0 = \frac{1}{3}(x - 3)$ $y = \frac{1}{3}x - 1$ Multiply by 3: $3y = x - 3$, which can be written as $x - 3y = 3$.
• Line L4 (diagonal line): This line passes through points $(0, 4)$ and $(4, 0)$. The slope $m = \frac{0 - 4}{4 - 0} = \frac{-4}{4} = -1$. Using the point-slope form $y - y_1 = m(x - x_1)$ with $(4, 0)$: $y - 0 = -1(x - 4)$ $y = -x + 4$, which can be written as $x + y = 4$.

• For L1 ($y=2$): The unshaded region R is below or on the line. So, $y \le 2$.
• For L2 ($x=-2$): The unshaded region R is to the right of or on the line. So, $x \ge -2$.
• For L3 ($x - 3y = 3$): The unshaded region R is above or on the line. Test a point in R, e.g., $(0, 2)$. $0 - 3(2) = -6$. Since $-6 < 3$, and R is above the line, the inequality is $x - 3y \le 3$.
• For L4 ($x + y = 4$): The unshaded region R is below or on the line. Test a point in R, e.g., $(0, 0)$. $0 + 0 = 0$. Since $0 < 4$, the inequality is $x + y \le 4$.

The four inequalities are:

1. $y \le 2$
2. $x \ge -2$
3. $x - 3y \le 3$
4. $x + y \le 4$

9. A line PQ = 12.5cm. By using another line, divide PQ into nine equal parts.

This is a geometric construction problem.

Label the points $P_1, P_2, P_3, \dots, P_9$ such that $PP_1 = P_1P_2 = \dots = P_8P_9$. The length of these segments is arbitrary, but they must be equal to each other.

These parallel lines will intersect the line segment PQ at points that divide PQ into nine equal parts.

The points of intersection on PQ will divide PQ into 9 equal segments.