QuestionJuly 7, 2025

If the number 5 is added to both numerator and denominator of a cettain faction, the result 2/3 when the number 3 is subtracted from both numerator and denominator of the fraction.The result is 2/5 ,hence, find the fraction.

If the number 5 is added to both numerator and denominator of a cettain faction, the result 2/3 when the number 3 is subtracted from both numerator and denominator of the fraction.The result is 2/5 ,hence, find the fraction.

Solution
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Answer

Let's denote the original fraction as \( \frac{a}{b} \). According to the problem, if we add 5 to both the numerator and denominator of this fraction, the result is \( \frac{2}{3} \). This can be written as: \[ \frac{a + 5}{b + 5} = \frac{2}{3} \] Similarly, if we subtract 3 from both the numerator and denominator, the result is \( \frac{2}{5} \). This can be written as: \[ \frac{a - 3}{b - 3} = \frac{2}{5} \] Now we have two equations: 1. \( \frac{a + 5}{b + 5} = \frac{2}{3} \) 2. \( \frac{a - 3}{b - 3} = \frac{2}{5} \) We can solve these equations to find the values of ( a ) and ( b ). From the first equation, we can cross-multiply to get: \[ 3(a + 5) = 2(b + 5) \] \[ 3a + 15 = 2b + 10 \] \[ 3a - 2b = -5 \quad \text{(Equation 1)} \] From the second equation, we can cross-multiply to get: \[ 5(a - 3) = 2(b - 3) \] \[ 5a - 15 = 2b - 6 \] \[ 5a - 2b = 9 \quad \text{(Equation 2)} \] Now we have a system of two linear equations: 1. ( 3a - 2b = -5 ) 2. ( 5a - 2b = 9 ) We can solve this system by elimination. Subtract Equation 1 from Equation 2: \[ (5a - 2b) - (3a - 2b) = 9 - (-5) \] \[ 5a - 3a = 9 + 5 \] \[ 2a = 14 \] \[ a = 7 \] Now substitute ( a = 7 ) back into Equation 1: \[ 3(7) - 2b = -5 \] \[ 21 - 2b = -5 \] \[ -2b = -26 \] \[ b = 13 \] So, the original fraction is: \[ \frac{a}{b} = \frac{7}{13} \] Therefore, the original fraction is \( \frac{7}{13} \).
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