Question

Both the numerator and the denominator of \( \frac{3}{4} \) are increased by a positive integer, \( x \), and those of \( \frac{15}{17} \) are decreased by the same integer. This operation results in the same value for both the fractions. What is the value of \( x \)?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

When solving problems involving fractions and operations on their numerators and denominators, remember to set the fractions equal to each other after performing the operations, and then solve for the unknown variable.

Answer 1

The Correct Option is C

Let the new value of both fractions after performing the operations be denoted as \( y \). - For the first fraction \( \frac{3}{4} \), when both the numerator and denominator are increased by \( x \), the new fraction becomes: \[ \frac{3 + x}{4 + x} \] - For the second fraction \( \frac{15}{17} \), when both the numerator and denominator are decreased by \( x \), the new fraction becomes: \[ \frac{15 - x}{17 - x} \] Since the operation results in the same value for both fractions, we can set these two fractions equal to each other: \[ \frac{3 + x}{4 + x} = \frac{15 - x}{17 - x} \] Now, cross-multiply to solve for \( x \): \[ (3 + x)(17 - x) = (15 - x)(4 + x) \] Expanding both sides: \[ (3)(17) - (3)(x) + (x)(17) - (x^2) = (15)(4) + (15)(x) - (x)(4) - (x^2) \] \[ 51 - 3x + 17x - x^2 = 60 + 15x - 4x - x^2 \] Simplify the equation: \[ 51 + 14x - x^2 = 60 + 11x - x^2 \] Cancel out \( x^2 \) from both sides: \[ 51 + 14x = 60 + 11x \] Now, solve for \( x \): \[ 14x - 11x = 60 - 51 \] \[ 3x = 9 \] \[ x = 3 \] Thus, the value of \( x \) is \( \boxed{3} \).