Given that 3x + 1/(5x) = 7 for a non zero real number x, what is the simplified value of the algebraic expression 5x / (15x^2 + 15x + 1)?

Introduction / Context:
This question tests your ability to manipulate algebraic expressions using a given relation. Instead of solving directly for x, you can transform the target expression using the original equation and algebraic identities, which is a common technique in aptitude exams to avoid solving messy quadratics completely.

Given Data / Assumptions:

• We are given 3x + 1/(5x) = 7.
• x is a non zero real number, so division by x is valid.
• We must find the value of E = 5x / (15x^2 + 15x + 1).
• The answer should be a simplified rational number.

Concept / Approach:
Rather than solving the equation for x directly, we introduce a substitution to simplify the algebra. Let y = 5x, which allows us to rewrite both the given equation and the target expression in terms of y. Then we use the quadratic equation satisfied by y to simplify the denominator of E. This often leads to cancellation and a simple final value.

Step-by-Step Solution:
Let y = 5x. Then x = y/5. The given equation becomes 3x + 1/(5x) = 3(y/5) + 1/y = 7.Multiply both sides by 5y to clear denominators: 5y * (3y/5 + 1/y) = 7 * 5y, which gives 3y^2 + 5 = 35y.Rearrange to form a quadratic: 3y^2 − 35y + 5 = 0. Thus y satisfies this equation.Now rewrite E in terms of y. First note that 15x^2 + 15x + 1 with x = y/5 becomes 15(y^2/25) + 15(y/5) + 1 = (3/5)y^2 + 3y + 1.So E = 5x / (15x^2 + 15x + 1) = y / ((3/5)y^2 + 3y + 1). Multiply numerator and denominator by 5 to get E = 5y / (3y^2 + 15y + 5).Use the quadratic 3y^2 − 35y + 5 = 0 to replace 3y^2 by 35y − 5. Then the denominator becomes (35y − 5) + 15y + 5 = 50y.Thus E = 5y / (50y) = 1/10, since y is non zero.

Verification / Alternative check:
We can approximate a numerical solution. Solve 3x + 1/(5x) = 7 for x using a calculator and then substitute into the original expression 5x / (15x^2 + 15x + 1). When this is done, the value is extremely close to 0.1, which equals 1/10. This confirms the algebraic result and shows that the substitution and simplification steps are consistent.

Why Other Options Are Wrong:

• 1/5 and 2/5 are larger than the true value and would result from incomplete simplification or misuse of the quadratic relation.
• 10 and 5/2 are far too large and indicate a serious algebraic error such as inverting the fraction incorrectly.
• Only 1/10 matches the simplified result derived from consistent algebraic manipulation.

Common Pitfalls:

• Trying to solve directly for x, which leads to complicated algebra instead of a neat substitution.
• Making errors while transforming the denominator 15x^2 + 15x + 1 into an expression in y.
• Forgetting to use the quadratic relation 3y^2 − 35y + 5 = 0 to simplify the denominator.

Final Answer:
1/10