6. If the numerator and denominator of a fraction are both decreased by 1 , the new fraction is equivalent to \( \frac{2}{3} \). But, if the numerator and denominator are both increased by 1 , the fraction is equivalent to \( \frac{3}{4} \). Find the original fraction.

Let the original fraction be \( \frac{x}{y} \). According to the first condition, if both the numerator and denominator are decreased by 1, we have: \[ \frac{x-1}{y-1} = \frac{2}{3} \] Cross-multiplying gives: \[ 3(x - 1) = 2(y - 1) \implies 3x - 3 = 2y - 2 \implies 3x - 2y = 1 \quad \text{(1)} \] According to the second condition, if both the numerator and denominator are increased by 1, we have: \[ \frac{x + 1}{y + 1} = \frac{3}{4} \] Cross-multiplying gives: \[ 4(x + 1) = 3(y + 1) \implies 4x + 4 = 3y + 3 \implies 4x - 3y = -1 \quad \text{(2)} \] Now we have a system of equations to solve: 1. \( 3x - 2y = 1 \) 2. \( 4x - 3y = -1 \) We can solve these equations using the substitution or elimination method. Let's use elimination. Multiply the first equation by 3: \[ 9x - 6y = 3 \quad \text{(3)} \] Multiply the second equation by 2: \[ 8x - 6y = -2 \quad \text{(4)} \] Now, subtract equation (4) from equation (3): \[ (9x - 6y) - (8x - 6y) = 3 - (-2) \] \[ 9x - 8x = 3 + 2 \implies x = 5 \] Now substitute \( x = 5 \) back into equation (1): \[ 3(5) - 2y = 1 \] \[ 15 - 2y = 1 \implies 2y = 14 \implies y = 7 \] So, the original fraction is: \[ \frac{x}{y} = \frac{5}{7} \] The original fraction is \( \frac{5}{7} \).