To solve the system of inequalities \( \frac{1}{3}(8x - 3) \leq 7 \) and \( \frac{1}{6}(x + 5) \leq 2 \), we need to solve each inequality separately and then find the intersection of the solutions.

### Step 1: Solve the first inequality

\[
\frac{1}{3}(8x - 3) \leq 7
\]

Multiply both sides by 3 to eliminate the fraction:

\[
8x - 3 \leq 21
\]

Add 3 to both sides:

\[
8x \leq 24
\]

Divide by 8:

\[
x \leq 3
\]

### Step 2: Solve the second inequality

\[
\frac{1}{6}(x + 5) \leq 2
\]

Multiply both sides by 6 to eliminate the fraction:

\[
x + 5 \leq 12
\]

Subtract 5 from both sides:

\[
x \leq 7
\]

### Step 3: Find the intersection of the solutions

From the first inequality, we have \( x \leq 3 \), and from the second inequality, we have \( x \leq 7 \).

The solution to the system is the intersection of these two intervals:

\[
x \leq 3
\]

So, the final solution is \( x \leq 3 \).

Would you like more details or have any questions? Here are some related problems to try:

1. Solve the inequality \( \frac{1}{4}(12x - 8) \leq 6 \).
2. Solve \( 2(x + 1) \geq 3x - 4 \).
3. Determine the solution to \( \frac{1}{5}(3x - 7) \leq 4 \).
4. Solve the system: \( \frac{2}{3}(x + 2) \leq 5 \) and \( \frac{1}{2}(x - 1) \leq 4 \).
5. Solve \( 3x + 4 \leq 2x + 5 \).

**Tip:** When solving inequalities, always remember to reverse the inequality sign when multiplying or dividing by a negative number.

Learn how to solve the system of fractional inequalities 1/3(8x - 3) ≤ 7 and 1/6(x + 5) ≤ 2 with detailed step-by-step solutions. This problem covers key algebraic concepts, including solving fractional inequalities and finding intersections of solutions. Suitable for students in Grades 7-9.

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