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I have opened the image you provided, and it contains several mathematical problems. Let's go through each one step by step:

### 1.1 Solve for \( x \):
#### 1.1.1: \( 2x - 3 = 6 \)
Solve the linear equation:
\[
2x - 3 = 6
\]
Add 3 to both sides:
\[
2x = 9
\]
Divide by 2:
\[
x = \frac{9}{2} = 4.5
\]

#### 1.1.2: \( 3^{2x} = 81 \)
Solve the exponential equation:
\[
3^{2x} = 81
\]
Since \( 81 = 3^4 \), equate the exponents:
\[
2x = 4 \quad \Rightarrow \quad x = \frac{4}{2} = 2
\]

#### 1.1.3: \( y^2 - 4y = 5 \)
Solve the quadratic equation:
\[
y^2 - 4y - 5 = 0
\]
Factorize:
\[
(y - 5)(y + 1) = 0
\]
Thus, \( y = 5 \) or \( y = -1 \).

#### 1.1.4: \( x(x - 3) = 0 \)
Solve this factored quadratic equation:
\[
x(x - 3) = 0
\]
This gives two possible solutions:
\[
x = 0 \quad \text{or} \quad x = 3
\]

---

### 1.2 Solve for \( x \) and \( y \) simultaneously:
\[
2y + x = 3 \quad \text{and} \quad y^2 + 3x = 7
\]
We will solve this system of equations. From the first equation, express \( x \) in terms of \( y \):
\[
x = 3 - 2y
\]
Substitute this into the second equation:
\[
y^2 + 3(3 - 2y) = 7
\]
Expand and simplify:
\[
y^2 + 9 - 6y = 7 \quad \Rightarrow \quad y^2 - 6y + 2 = 0
\]
Solve the quadratic using the quadratic formula:
\[
y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(2)}}{2(1)} = \frac{6 \pm \sqrt{36 - 8}}{2} = \frac{6 \pm \sqrt{28}}{2} = \frac{6 \pm 2\sqrt{7}}{2}
\]
Simplify:
\[
y = 3 \pm \sqrt{7}
\]
For each \( y \), substitute into \( x = 3 - 2y \) to get the corresponding \( x \).

---

### 1.3 Prove that the roots of the following equation are real for all real values of \( a \):
\[
x^2 + 2(a + 1)x + (a^2 + a) = 0
\]
The discriminant of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:
\[
\Delta = B^2 - 4AC
\]
Here, \( A = 1 \), \( B = 2(a + 1) \), and \( C = a^2 + a \). Substitute these into the discriminant formula:
\[
\Delta = [2(a + 1)]^2 - 4(1)(a^2 + a)
\]
Simplify:
\[
\Delta = 4(a + 1)^2 - 4(a^2 + a) = 4(a^2 + 2a + 1) - 4(a^2 + a)
\]
\[
\Delta = 4a^2 + 8a + 4 - 4a^2 - 4a = 4a + 4
\]
Factor:
\[
\Delta = 4(a + 1)
\]
Since \( 4(a + 1) \geq 0 \) for all real \( a \), the discriminant is non-negative, meaning the roots are real for all real values of \( a \).

---

Would you like any details clarified or have further questions?

Here are five related problems you can try:
1. Solve \( 5x - 7 = 3 \).
2. Solve \( 2^{3x} = 16 \).
3. Solve \( z^2 - 5z + 6 = 0 \).
4. Solve the simultaneous equations \( 3x + 4y = 7 \) and \( 2x - y = 3 \).
5. Prove that the roots of \( x^2 + bx + c = 0 \) are real for all real values of \( b \) and \( c \).

**Tip**: For solving quadratic equations, always check if factoring is possible before using the quadratic formula.

Solve the following mathematical problems: 1.1 Solve for x in various equations, 1.2 Solve for x and y simultaneously, and 1.3 Prove that the roots of the equation are real for all real values of a.

This page provides step-by-step solutions to a set of problems involving linear, exponential, quadratic, and simultaneous equations. It includes solving for x and y in various scenarios and proving the reality of roots using the discriminant formula, suitable for students in Grades 9-11.

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