Let's solve each equation:

(i) $x + 7 = 15$

1. Subtract $7$ from both sides to isolate $x$: $x = 15 - 7$
2. Simplify: $x = 8$

So, the solution for (i) is $x = 8$.

(ii) $5(3x + 8) = 10$

1. First, divide both sides by $5$ to simplify: $3x + 8 = \frac{10}{5}$ $3x + 8 = 2$

2. Next, subtract $8$ from both sides: $3x = 2 - 8$ $3x = -6$

3. Finally, divide by $3$ to solve for $x$: $x = \frac{-6}{3}$ $x = -2$

So, the solution for (ii) is $x = -2$.

Summary of Solutions

• (i) $x = 8$
• (ii) $x = -2$

Would you like further details on these solutions or have any questions?

Additional Practice Questions:

1. Solve $x - 5 = 10$.
2. Solve $4(2x + 3) = 20$.
3. Solve $7x + 10 = 24$.
4. Solve $x/4 + 6 = 3$.
5. Solve $2(x + 7) = 6$.

Tip: When solving linear equations, aim to isolate $x$ by performing inverse operations step-by-step.