If the linear equation (2/3) * (6x/5 − 1/4) + 1/3 = 9x/5 holds for a real number x, what is the value of x that satisfies this equality?

Introduction / Context:
This question checks your ability to solve a linear equation involving fractions and brackets. Such questions are common in aptitude and school mathematics, and they test algebraic manipulation skills, especially dealing with fractional coefficients and clearing denominators in a neat way. The goal is to isolate x and express it as a simple rational number.

Given Data / Assumptions:

• The equation is (2/3) * (6x/5 − 1/4) + 1/3 = 9x/5.
• x is a real number.
• All denominators are non zero, so the equation is valid.
• We must find the exact value of x that makes the equality true.

Concept / Approach:
The standard approach is to first remove fractions by multiplying through by a common multiple of the denominators. Here, denominators are 3, 5, and 4. However, you can also simplify step by step. First expand the bracket, then combine like terms, and finally solve for x by isolating the variable. Careful handling of fractions is essential to avoid arithmetic mistakes.

Step-by-Step Solution:
Start with (2/3) * (6x/5 − 1/4) + 1/3 = 9x/5.Multiply both sides by 3 to clear the denominator 3: 2 * (6x/5 − 1/4) + 1 = 27x/5.Expand the left side: 2 * (6x/5) = 12x/5 and 2 * (−1/4) = −1/2, so the left side is 12x/5 − 1/2 + 1.Combine constants: −1/2 + 1 = 1/2, so we have 12x/5 + 1/2 = 27x/5.Subtract 12x/5 from both sides: 1/2 = 27x/5 − 12x/5 = 15x/5 = 3x.So 3x = 1/2, and dividing both sides by 3 gives x = 1/6.

Verification / Alternative check:
Substitute x = 1/6 into the original equation. Compute 6x/5 = 6 * (1/6) / 5 = 1/5 and 1/4 remains 1/4. So 6x/5 − 1/4 = 1/5 − 1/4 = (4 − 5) / 20 = −1/20. Then (2/3) * (−1/20) = −1/30. Add 1/3 to this: −1/30 + 1/3 = −1/30 + 10/30 = 9/30 = 3/10. Now compute the right side: 9x/5 = 9 * (1/6) / 5 = 9/6 * 1/5 = 3/10. Both sides are equal, so x = 1/6 is correct.

Why Other Options Are Wrong:

• −1/6, 1/5, and −1/5 do not satisfy the original equation when substituted back.
• Choosing 0 gives left side 1/3 and right side 0, so the equality fails.
• Only x = 1/6 balances the equation exactly.

Common Pitfalls:

• Making sign errors when distributing 2/3 across the bracket.
• Incorrectly combining fractions like 1/5 and 1/4 or 1/2 and 1.
• Forgetting to subtract the 12x/5 term from both sides, which can lead to an incorrect coefficient of x.

Final Answer:
1/6