Here are the questions from the image, along with their solutions and detailed workings:

1. Solve for $x$:

a) $x^2 = 4$

$x = \pm \sqrt{4} = \pm 2$ Solution: $x = 2$ or $x = -2$

b) $\frac{2x + 1}{3} = 10$

Multiply both sides by 3: $2x + 1 = 30$ Subtract 1 from both sides: $2x = 29$ Divide by 2: $x = \frac{29}{2} = 14.5$ Solution: $x = 14.5$

c) $\frac{3}{x} + 1 = 12x$

First, subtract 1 from both sides: $\frac{3}{x} = 12x - 1$ Multiply both sides by $x$ to eliminate the fraction: $3 = x(12x - 1)$ Expand: $3 = 12x^2 - x$ Rearrange to form a quadratic equation: $12x^2 - x - 3 = 0$ Solve using the quadratic formula: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(12)(-3)}}{2(12)} = \frac{1 \pm \sqrt{1 + 144}}{24} = \frac{1 \pm \sqrt{145}}{24}$ Solution: $x = \frac{1 \pm \sqrt{145}}{24}$

d) $4x + 2 = 8$

Subtract 2 from both sides: $4x = 6$ Divide by 4: $x = \frac{6}{4} = 1.5$ Solution: $x = 1.5$

e) $2x - 3 = 3b$

This equation includes an unknown $b$. Solving for $x$: $2x = 3b + 3$ $x = \frac{3b + 3}{2}$ Solution: $x = \frac{3b + 3}{2}$ (in terms of $b$)

f) $x - 4 = 1$

Add 4 to both sides: $x = 5$ Solution: $x = 5$

g) $3c + 2x = 4b$

Solve for $x$: $2x = 4b - 3c$ $x = \frac{4b - 3c}{2}$ Solution: $x = \frac{4b - 3c}{2}$ (in terms of $b$ and $c$)

2. Solve the following equations:

w) $x^3 = 27$

Take the cube root of both sides: $x = \sqrt[3]{27} = 3$ Solution: $x = 3$

i) $x^2 + x^2 = 16$

Combine like terms: $2x^2 = 16$ Divide by 2: $x^2 = 8$ Take the square root: $x = \pm \sqrt{8} = \pm 2\sqrt{2}$ Solution: $x = 2\sqrt{2}$ or $x = -2\sqrt{2}$

j) $\frac{x}{2} = 4$

Multiply both sides by 2: $x = 8$ Solution: $x = 8$

k) $\frac{2}{x + 4y} = 1$

Multiply both sides by $x + 4y$: $2 = x + 4y$ Solving for $x$: $x = 2 - 4y$ Solution: $x = 2 - 4y$

Would you like more details on any of these solutions, or do you have further questions?

Here are 5 related problems to explore:

1. Solve $x^2 = 9$.
2. Solve $\frac{4x + 2}{5} = 3$.
3. Solve $\frac{6}{x} + 2 = 5x$.
4. Solve $5x - 3 = 7b$.
5. Solve $2x^3 + 1 = 9$.

Tip: Always check your solutions by substituting them back into the original equation to ensure they satisfy the equation.