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Matt71
The numerator of an improper fraction is 4 more than the denominator. When the numerator is decreased by 3 and the denominator is decreased by 1, the fraction is decreased by 3/10. Find the original improper fraction.
Let the denominator of original fraction be x numerator of original fraction = x+4 new numerator = (x+4)-3 = x+1 new denominator = x-1 new fraction = original fraction - 3/10 \[\frac{x+1}{x-1} = \frac{x+4}{x} - \frac{3}{10}\] can you solve this now?
Let the following be the original improper fraction:\[\frac{x+4}{x} \]Solve the following for x:\[\frac{x+4-3}{x-1}\text{=}\frac{x+4}{x}-\frac{3}{10}\]\[\left\{x=\frac{8}{3},x=5\right\} \]Calculate the value of the original fraction when x=8/3\[\frac{4+x}{x}\text{=}\frac{4+\frac{8}{3}}{\frac{8}{3}}\text{=}\frac{5}{2}\]and x = 5\[\frac{4+5}{5}=\frac{9}{5} \]Assume that the answer is \[\frac{9}{5} \]and verify it's validity:\[\frac{9}{5}-\frac{3}{10}\text{ = }\frac{9-3}{5-1}\]\[\frac{3}{2}\text{ = }\frac{3}{2}\]Although 5/2 is another solution to the equation, it will not pass the validity test.