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We have the equation:

\[
\frac{1}{5}x + \frac{1}{2} = 3 \left( \frac{3}{4}x - 5 \right)
\]

Let's solve this step-by-step:

### Step 1: Distribute the 3 on the right-hand side.
Distribute the 3 to both terms inside the parentheses:

\[
3 \left( \frac{3}{4}x - 5 \right) = \frac{9}{4}x - 15
\]

Now the equation becomes:

\[
\frac{1}{5}x + \frac{1}{2} = \frac{9}{4}x - 15
\]

### Step 2: Eliminate the fractions.
To clear the fractions, find the least common denominator (LCD) of 5, 2, and 4, which is 20. Multiply the entire equation by 20:

\[
20 \left( \frac{1}{5}x \right) + 20 \left( \frac{1}{2} \right) = 20 \left( \frac{9}{4}x - 15 \right)
\]

This simplifies to:

\[
4x + 10 = 5 \times \frac{9}{4}x - 300
\]

\[
4x + 10 = 45x - 300
\]

### Step 3: Solve for \( x \).
Now, bring all the \(x\)-terms to one side and constants to the other side:

\[
4x - 45x = -300 - 10
\]

\[
-41x = -310
\]

Now, divide by \(-41\):

\[
x = \frac{-310}{-41} =  \frac{310}{41}
\]

Thus, the solution is:

\[
x = \frac{310}{41}
\]

Would you like further details or have any questions? 

Here are some related questions to expand on this topic:

1. How can you solve similar equations with decimals instead of fractions?
2. What happens if both sides of the equation have multiple variable terms?
3. How do you solve systems of equations where both equations contain fractions?
4. What methods can be used to check if your solution is correct?
5. Can this problem be solved using substitution or elimination methods?

**Tip:** When solving equations involving fractions, clearing the fractions early can simplify the process significantly.

Learn how to solve the linear equation with fractions: 1/5x + 1/2 = 3(3/4x - 5). Follow a detailed step-by-step breakdown using algebraic concepts and clearing fractions by multiplying through the least common denominator. Ideal for students in Grades 9-10.

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