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The numerator of a fraction is 3 less than the denominator. If 1 is added to both its numerator and denominator, it becomes 2/3. Find the fraction​

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konrad509

[tex]\dfrac{x-3}{x}[/tex] - initial fraction

[tex]\dfrac{x-3+1}{x+1}=\dfrac{2}{3}\\\\\dfrac{x-2}{x+1}=\dfrac{2}{3}\\\\2(x+1)=3(x-2)\\2x+2=3x-6\\x=8\\\\\dfrac{8-3}{8}=\dfrac{5}{8}[/tex]

It's [tex]\dfrac{5}{8}[/tex]

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