Fred's current age is (x-1) years. i) What was he 4 years ago? ii) What will he be 8 years from now? iii) How old is he now, if his age in 8 years time will be three times his age 4 years ago?

$$
(x-1) - 4 = x-5
$$
Fred's age 4 years ago was $\boxed{(x-5) \text{ years}}$.

• ii) Will he be 8 years from now?

$$
(x-1) + 8 = x+7
$$
Fred's age 8 years from now will be $\boxed{(x+7) \text{ years}}$.

• iii) Is he now, if his age in 8 years time will be three times his age 4 years ago?

$$
\text{Current age} = 11-1 = 10
$$
Fred's current age is $\boxed{10 \text{ years}}$.

b)
The perimeter of a rectangular cocoa farm is 497 km. The length of the farm is $2\frac{1}{2}$ times the width.
Let $W$ be the width and $L$ be the length.
Given $L = 2\frac{1}{2}W = \frac{5}{2}W$.
The perimeter $P = 2(L+W)$.

• i) Find the width.

$$
W = \frac{497}{7}
$$
$$
W = 71
$$
The width of the farm is $\boxed{71 \text{ km}}$.

• ii) Find the length of the farm.

$$
L = \frac{5}{2} \times 71
$$
$$
L = \frac{355}{2}
$$
$$
L = 177.5
$$
The length of the farm is $\boxed{177.5 \text{ km}}$.

5. a) Factorize: $(x-y)(3m+n) - (x-y)(m-2n)$.

The factored expression is $\boxed{(x-y)(2m+3n)}$.

b)
Given $p = \begin{pmatrix} 2-3x \\ 5-2y \end{pmatrix}$, $q = \begin{pmatrix} -1 \\ 5 \end{pmatrix}$, and $p-q = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$. Find the value of $(x+y)$.

Equation 1: $2-3x = 5$
Equation 2: $5-2y = 13$

The value of $(x+y)$ is $\boxed{-5}$.

c)

• i) Find the truth set of $\frac{x-1}{2} \le \frac{1}{2} + x$.

$$
-1-1 \le x
$$
$$
-2 \le x
$$
The truth set is $\boxed{{x \mid x \ge -2}} $.

• ii) Illustrate the answer in (i) on the number line.

$$
\begin{array}{cccccccc}
\text{<---} & \bullet & \text{------------------------------------->} \\
& -2 & & & & & & x
\end{array}
$$

6. a) Copy and complete the table for the relation $y = 5-2x$ for $-3 \le x \le 4$.

For $x=-3$: $y = 5-2(-3) = 5+6 = 11$
For $x=-2$: $y = 5-2(-2) = 5+4 = 9$
For $x=-1$: $y = 5-2(-1) = 5+2 = 7$
For $x=0$: $y = 5-2(0) = 5-0 = 5$
For $x=1$: $y = 5-2(1) = 5-2 = 3$
For $x=2$: $y = 5-2(2) = 5-4 = 1$
For $x=3$: $y = 5-2(3) = 5-6 = -1$
For $x=4$: $y = 5-2(4) = 5-8 = -3$

The completed table is:

b)
Using a scale of 2 cm to 1 unit on x-axis and 2 cm to 2 units on the y-axis, draw on a graph sheet two perpendicular axes ox and oy for $-5 \le x \le 5$ and $-12 \le y \le 12$.

c)

• i) Using the table, plot all the points of the relation $y=5-2x$.

$(-3, 11), (-2, 9), (-1, 7), (0, 5), (1, 3), (2, 1), (3, -1), (4, -3)$ on the graph drawn in 6(b).

• ii) Draw a straight line through all the points.

d)
Using the graph, find the:
(Since I cannot draw a graph, I will calculate these values using the equation $y=5-2x$.)

• i) value of $y$ when $x=-2.6$.

$$
y = 5 - 2(-2.6)
$$
$$
y = 5 + 5.2
$$
$$
y = 10.2
$$
The value of $y$ when $x=-2.6$ is $\boxed{10.2}$.

• ii) value of $x$ when $y=-2.8$.

$$
x = \frac{-7.8}{-2}
$$
$$
x = 3.9
$$
The value of $x$ when $y=-2.8$ is $\boxed{3.9}$.

• iii) gradient of the line.

This is in the form $y = mx+c$, where $m$ is the gradient.