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### (a) Simplify the following expressions:

#### (i) \( \frac{ab^2c}{a^2bc^{-3}} \)
\[
\frac{ab^2c}{a^2bc^{-3}} = \frac{a}{a^2} \cdot \frac{b^2}{b} \cdot \frac{c}{c^{-3}}
= a^{1-2} \cdot b^{2-1} \cdot c^{1+3} = a^{-1}b^1c^4
= \frac{bc^4}{a}
\]
Thus, the simplified expression is \( \frac{bc^4}{a} \).

#### (ii) \( \left(\sqrt{x} \cdot \sqrt{y} \cdot \sqrt{z}\right) \left(\sqrt{x} \cdot y \cdot \sqrt{z}\right) \)
\[
\left(\sqrt{x} \cdot \sqrt{y} \cdot \sqrt{z}\right) \left(\sqrt{x} \cdot y \cdot \sqrt{z}\right) = \sqrt{x} \cdot \sqrt{x} \cdot \sqrt{y} \cdot y \cdot \sqrt{z} \cdot \sqrt{z}
= x \cdot \sqrt{y} \cdot y \cdot z
= x \cdot y^{3/2} \cdot z
\]
Thus, the simplified expression is \( x \cdot y^{3/2} \cdot z \).

---

### (b) Solve the equations:

#### (i) \( 2x - 3 + 4x = 7 + x \)
\[
2x + 4x - x = 7 + 3
\]
\[
5x = 10 \implies x = 2
\]
The solution is \( x = 2 \).

#### (ii) \( \log 2x^4 + \log x = \log 16 + \log x^4 \)
Using the property \( \log ab = \log a + \log b \), we combine terms:
\[
\log (2x^4 \cdot x) = \log (16 \cdot x^4)
\]
\[
\log (2x^5) = \log (16x^4)
\]
Equating the arguments:
\[
2x^5 = 16x^4
\]
Dividing by \( x^4 \) (assuming \( x \neq 0 \)):
\[
2x = 16 \implies x = 8
\]
The solution is \( x = 8 \).

---

### (c) Factorize the expressions:

#### (i) \( 5x^2y - 30x^3y^2 \)
\[
5x^2y - 30x^3y^2 = 5x^2y (1 - 6xy)
\]
Thus, the factorized form is \( 5x^2y(1 - 6xy) \).

#### (ii) \( x^2 - 100 \)
\[
x^2 - 100 = (x - 10)(x + 10)
\]
Thus, the factorized form is \( (x - 10)(x + 10) \).

---

### (d) Sag length equation:
The equation for sag length \( l \) is:
\[
l = \frac{12}{x} + x
\]
Given \( l = 18 \), we solve:
\[
18 = \frac{12}{x} + x
\]
Multiplying the equation by \( x \) to eliminate the denominator:
\[
18x = 12 + x^2
\]
Rearranging:
\[
x^2 - 18x + 12 = 0
\]
We can solve this quadratic equation using the quadratic formula:
\[
x = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(1)(12)}}{2(1)} = \frac{18 \pm \sqrt{324 - 48}}{2}
= \frac{18 \pm \sqrt{276}}{2} = \frac{18 \pm 2\sqrt{69}}{2}
\]
\[
x = 9 \pm \sqrt{69}
\]
Thus, the distance between the supports is \( x = 9 + \sqrt{69} \) or \( x = 9 - \sqrt{69} \) (taking the positive root makes physical sense).

---

### (e) Simultaneous equations:
\[
y = x^2 + 3x - 1
\]
\[
y = 2x + 1
\]
Equating the two expressions for \( y \):
\[
x^2 + 3x - 1 = 2x + 1
\]
Rearranging:
\[
x^2 + 3x - 2x - 1 - 1 = 0
\]
\[
x^2 + x - 2 = 0
\]
Factoring:
\[
(x - 1)(x + 2) = 0
\]
Thus, \( x = 1 \) or \( x = -2 \).

Substitute into \( y = 2x + 1 \):
For \( x = 1 \):
\[
y = 2(1) + 1 = 3
\]
For \( x = -2 \):
\[
y = 2(-2) + 1 = -3
\]
Thus, the solutions are \( (x, y) = (1, 3) \) and \( (x, y) = (-2, -3) \).

---

Would you like more details on any step? Here are 5 related questions you can explore:

1. How do logarithmic properties simplify equations with logs?
2. What other methods can solve quadratic equations besides factoring?
3. How do you factor more complex polynomials efficiently?
4. What is the significance of real and complex roots in sag length equations?
5. How do you graph the simultaneous equations to find solutions visually?

**Tip**: Always check for extraneous solutions when working with logarithmic and square root equations!

Math problems covering simplification, equation solving, factorization, calculating sag length in cables, and solving simultaneous equations.

This guide covers a range of high school math problems including simplification, solving linear and quadratic equations, factorization techniques, and the properties of logarithms. Suitable for students in grades 9-12, this resource offers step-by-step solutions to enhance understanding and mastery of these topics.

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