Solve the linear equation [4(2x/5 − 3/2)]/3 + 7/5 = 37/5 and find the value of x.

Introduction / Context:
This question involves solving a linear equation with fractional coefficients. Such problems test comfort with fractions, distributive law, and careful use of arithmetic operations. They appear frequently in aptitude tests, where accurate handling of fractions is important for both algebra and word problems.

Given Data / Assumptions:

• The equation is [4(2x/5 − 3/2)]/3 + 7/5 = 37/5.
• x is a real number.
• We need to isolate x and find its numerical value.

Concept / Approach:
We simplify step by step: first handle the inner bracket 2x/5 − 3/2, then multiply by 4, divide by 3, and finally combine with 7/5. To remove fractions efficiently, we can multiply the entire equation by the least common multiple of all denominators. This reduces the equation to one with integer coefficients, which is easier to solve.

Step-by-Step Solution:
Start with [4(2x/5 − 3/2)]/3 + 7/5 = 37/5. First simplify inside the brackets: 2x/5 − 3/2. Multiply this by 4: 4 * (2x/5 − 3/2) = 8x/5 − 6. Now divide by 3: [4(2x/5 − 3/2)]/3 = (8x/5 − 6)/3. So the equation becomes (8x/5 − 6)/3 + 7/5 = 37/5. Multiply every term by 15 (the least common multiple of 3 and 5) to clear denominators. 15 * (8x/5 − 6)/3 = 5 * (8x/5 − 6) = 8x − 30. 15 * (7/5) = 3 * 7 = 21, and 15 * (37/5) = 3 * 37 = 111. Thus we get 8x − 30 + 21 = 111. Simplify the left side: 8x − 9 = 111. Add 9 to both sides: 8x = 120. Divide by 8: x = 120 / 8 = 15.

Verification / Alternative check:
Substitute x = 15 into the original equation. Compute 2x/5 = 2 * 15 / 5 = 6 and 6 − 3/2 = 6 − 1.5 = 4.5. Multiply by 4 to get 18, then divide by 3 to get 6. Now add 7/5, which is 1.4, so the left side is 6 + 1.4 = 7.4. The right side 37/5 is 7.4 as well. Since both sides match, x = 15 is confirmed as correct.

Why Other Options Are Wrong:
Values such as −15 or −7/5 come from sign errors when expanding or from incorrect handling of the denominators. The value 7/5 appears if one confuses x with one of the intermediate fractional terms. The value 5 would not satisfy the original equation when checked by substitution. None of these produce equal left and right sides.

Common Pitfalls:
The main sources of error are skipping steps when clearing fractions, forgetting to multiply all terms by the same factor, or mishandling negative signs. A structured approach, clearing denominators early and checking by substitution, greatly reduces these mistakes.

Final Answer:
The solution of the equation is x = 15.