Question 279488
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Let *[tex \Large x] represent the numerator of the original fraction.  Let *[tex \Large x\ +\ 5] represent the denominator of the original fraction.  Subtracting 1 from the numerator and setting it equal to *[tex \Large \frac{1}{3}] results in:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\,-\,1}{x\,+\,5}\ =\ \frac{1}{3}]


Cross-multiply and solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ -\ 3\ =\ x\ +\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ =\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 4]


Original fraction is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4}{9}]


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4\,-\,1}{9}\ =\ \frac{3}{9}\ =\ \frac{1}{3}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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