Solve the system of linear equations: 2x + 3y = 12 and 3x − 2y = 4.

Equation (1): $2x + 3y = 12$
Equation (2): $3x - 2y = 4$

To eliminate $y$, multiply Equation (1) by 2 and Equation (2) by 3:
$(2x + 3y = 12) \times 2 \implies 4x + 6y = 24$ (3)
$(3x - 2y = 4) \times 3 \implies 9x - 6y = 12$ (4)

Add Equation (3) and Equation (4):

Using Equation (1): $2x + 3y = 12$

The final answer is $\boxed{x=\frac{36}{13}, y=\frac{28}{13}}$.

Q.10

Set $p(x) = 0$:

Factor the quadratic equation:

The zeroes are $x=2$ and $x=3$.
Let $\alpha = 2$ and $\beta = 3$.

For a quadratic polynomial $ax^2 + bx + c$, the sum of zeroes is $-\frac{b}{a}$ and the product of zeroes is $\frac{c}{a}$.
For $p(x) = x^2 - 5x + 6$, we have $a=1$, $b=-5$, $c=6$.

Sum of zeroes:
$\alpha + \beta = 2 + 3 = 5$
$-\frac{b}{a} = -\frac{(-5)}{1} = 5$
The sum of zeroes matches.

Product of zeroes:
$\alpha \beta = 2 \times 3 = 6$
$\frac{c}{a} = \frac{6}{1} = 6$
The product of zeroes matches.

The final answer is $\boxed{\text{Zeroes are } 2, 3. \text{ Verified relations: sum } 5 = -\frac{(-5)}{1}, \text{ product } 6 = \frac{6}{1}}$.

Q.11

If $x=0$: $y = 2(0) + 1 = 1$. So, point A is $(0, 1)$.
If $x=1$: $y = 2(1) + 1 = 3$. So, point B is $(1, 3)$.
If $x=-1$: $y = 2(-1) + 1 = -1$. So, point C is $(-1, -1)$.

Plot the points $(0, 1)$ and $(1, 3)$ (or any other two points you found) on a coordinate plane. Draw a straight line passing through these two points. This line represents the graph of the equation $2x - y + 1 = 0$.

The final answer is $\boxed{\text{The graph is a line passing through points like } (0,1) \text{ and } (1,3)}$.

Q.12

The triangle is right-angled at Q.
Hypotenuse $PR = 13$ cm.
Leg $PQ = 5$ cm.

In a right-angled triangle, $\sin(\text{angle}) = \frac{\text{Opposite side}}{\text{Hypotenuse}}$.
For angle P, the opposite side is QR and the hypotenuse is PR.

The final answer is $\boxed{QR=12 \text{ cm}, \sin P=\frac{12}{13}}$.

Q.13

If $\sqrt{3}$ is rational, it can be expressed as a fraction $\frac{a}{b}$, where $a$ and $b$ are integers, $b \ne 0$, and $a$ and $b$ are coprime (their greatest common divisor is 1).

From equation $(*)$, $a^2$ is a multiple of 3.
If $a^2$ is divisible by 3, then $a$ must also be divisible by 3 (a property of prime numbers).
So, we can write $a = 3c$ for some integer $c$.

Divide both sides by 3:

From this equation, $b^2$ is a multiple of 3.
Therefore, $b$ must also be divisible by 3.

We have shown that both $a$ and $b$ are divisible by 3. This contradicts our initial assumption that $a$ and $b$ are coprime.
Since our initial assumption leads to a contradiction, it must be false.
Therefore, $\sqrt{3}$ is an irrational number.

The final answer is $\boxed{\text{Proof by contradiction: Assume } \sqrt{3} \text{ is rational, leading to } a \text{ and } b \text{ both being divisible by 3, contradicting their coprimality. Thus, } \sqrt{3} \text{ is irrational.}}$.

Q.14

The final answer is $\boxed{\text{Mean}=25.5}$.

Q.15

Let $A = (-1, 8)$, $B = (0, 5)$, and $C = (2, -1)$.

The slope $m_{AB}$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Since $m_{AB} = m_{BC} = -3$, the slopes of the line segments AB and BC are equal.
This means that points A, B, and C lie on the same straight line, i.e., they are collinear.

The final answer is $\boxed{\text{The slope of AB is -3 and the slope of BC is -3. Since the slopes are equal, the points are collinear.}}$.