Let's tackle these problems one by one:

1. Writing the fraction as a single fraction

You have the expression: $\frac{1}{y-1} + \frac{2}{y+2}$

To add these fractions, you need a common denominator. The common denominator is $(y-1)(y+2)$.

So, we rewrite each fraction: $\frac{1}{y-1} = \frac{y+2}{(y-1)(y+2)}$ $\frac{2}{y+2} = \frac{2(y-1)}{(y-1)(y+2)}$

Now, add the fractions: $\frac{y+2 + 2(y-1)}{(y-1)(y+2)}$ Simplify the numerator: $y + 2 + 2y - 2 = 3y$

So the combined fraction is: $\frac{3y}{(y-1)(y+2)}$

2. Dividing $x^3 - x^2 + x - 21$ by $x - 3$

We perform polynomial long division:

1. Divide the leading term of the dividend by the leading term of the divisor: $\frac{x^3}{x} = x^2$
2. Multiply $x^2$ by $x - 3$: $x^2(x - 3) = x^3 - 3x^2$
3. Subtract this from the original polynomial: $(x^3 - x^2 + x - 21) - (x^3 - 3x^2) = 2x^2 + x - 21$
4. Divide the new leading term $2x^2$ by $x$: $\frac{2x^2}{x} = 2x$
5. Multiply $2x$ by $x - 3$: $2x(x - 3) = 2x^2 - 6x$
6. Subtract this from the current remainder: $(2x^2 + x - 21) - (2x^2 - 6x) = 7x - 21$
7. Divide the new leading term $7x$ by $x$: $\frac{7x}{x} = 7$
8. Multiply $7$ by $x - 3$: $7(x - 3) = 7x - 21$
9. Subtract this from the current remainder: $(7x - 21) - (7x - 21) = 0$

Thus, the quotient is $x^2 + 2x + 7$ and the remainder is $0$.

3. Solving the simultaneous equations

(i) $x + y = 9$ and $x - y = 20$

Add the two equations: $(x + y) + (x - y) = 9 + 20$ $2x = 29 \quad \Rightarrow \quad x = \frac{29}{2} = 14.5$

Substitute $x = 14.5$ into $x + y = 9$: $14.5 + y = 9 \quad \Rightarrow \quad y = 9 - 14.5 = -5.5$

Solution: $x = 14.5$, $y = -5.5$

(ii) $2x + 3y = 15$ and $2x - 4y = 20$

Subtract the second equation from the first: $(2x + 3y) - (2x - 4y) = 15 - 20$ $7y = -5 \quad \Rightarrow \quad y = -\frac{5}{7}$

Substitute $y = -\frac{5}{7}$ into $2x + 3y = 15$: $2x + 3\left(-\frac{5}{7}\right) = 15 \quad \Rightarrow \quad 2x - \frac{15}{7} = 15$ $2x = 15 + \frac{15}{7} = \frac{120}{7} \quad \Rightarrow \quad x = \frac{60}{7}$

Solution: $x = \frac{60}{7}$, $y = -\frac{5}{7}$

(iii) $4y + 3x = 25$ and $-4y - 6x = 30$

Add the two equations: $(4y + 3x) + (-4y - 6x) = 25 + 30$ $-3x = 55 \quad \Rightarrow \quad x = -\frac{55}{3}$

Substitute $x = -\frac{55}{3}$ into $4y + 3x = 25$: $4y + 3\left(-\frac{55}{3}\right) = 25 \quad \Rightarrow \quad 4y - 55 = 25$ $4y = 80 \quad \Rightarrow \quad y = 20$

Solution: $x = -\frac{55}{3}$, $y = 20$

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Would you like more details or have any questions?

Here are some related questions you might consider:

1. How can we find the least common denominator when adding fractions?
2. What are the steps involved in polynomial long division?
3. Can you provide an example of solving simultaneous equations by substitution?
4. How does the elimination method work for solving systems of equations?
5. What happens when the remainder is non-zero in polynomial division?

Tip: When solving simultaneous equations, it’s crucial to choose the method that seems easiest based on the structure of the equations.