The sum of the numerator and the denominator of a fraction is $12$ . If $1$ is added to both numerator and the denominator the fraction becomes $\frac{3}{4}$ find the fraction.

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Solution

$letthefractionbe(ab),thena+b=12orb==12−aand(a+1b+1)=(34)(a+112−a+1)=(34)4a+4=3(13−a)4a+4=39−3a7a=35a=(357)=5∴b=12−5=7∴fraction(57)$

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The sum of the numerator and the denominator of a fraction is 12. If 1 is added to both numerator and the denominator the fraction becomes 34 find the fraction.

letthefractionbe(ab),thena+b=12orb==12−aand(a+1b+1)=(34)(a+112−a+1)=(34)4a+4=3(13−a)4a+4=39−3a7a=35a=(357)=5∴b=12−5=7∴fraction(57)

The sum of the numerator and the denominator of a fraction is $12$ . If $1$ is added to both numerator and the denominator the fraction becomes $\frac{3}{4}$ find the fraction.

$letthefractionbe(ab),thena+b=12orb==12−aand(a+1b+1)=(34)(a+112−a+1)=(34)4a+4=3(13−a)4a+4=39−3a7a=35a=(357)=5∴b=12−5=7∴fraction(57)$