The problem asks us to solve for $x$ in two different questions. In the first question, we have an equation $\frac{x}{2} + \frac{x}{3} = 5$. In the second question, we have an inequality $4 - x \le 13$. We are given that $x$ must be an integer.

1. Problem Description

The problem asks us to solve for $x$ in two different questions. In the first question, we have an equation $\frac{x}{2} + \frac{x}{3} = 5$. In the second question, we have an inequality $4 - x \le 13$. We are given that $x$ must be an integer.

2. Solution Steps

(i) To solve the equation $\frac{x}{2} + \frac{x}{3} = 5$, we first find a common denominator for the fractions, which is

6. Then we rewrite the equation as:

$\frac{3x}{6} + \frac{2x}{6} = 5$

Combining the terms on the left side, we have:

$\frac{5x}{6} = 5$

Multiply both sides by 6:

$5x = 30$

Divide both sides by 5:

$x = 6$

Since $x=6$ is an integer, it satisfies the condition $x \in Z$.

(ii) To solve the inequality $4 - x \le 13$, we subtract 4 from both sides:

$-x \le 9$

Multiply both sides by -

1. Remember to flip the inequality sign when multiplying by a negative number:

$x \ge -9$

Since $x \in Z$, $x$ must be an integer greater than or equal to -

9.

3. Final Answer

(i) $x = 6$

(ii) $x \ge -9$