Question

in a certain positive fraction the denominator is greater than the numerator by 3. if 1 is subtracted from both the numerator and denominator,the fraction is decreased by 1/14.find the fraction

Answer 1

In the given Question,

Denominator is greater than the numerator by 3.

$\sf{So, let\: the \: fraction\: be \: \dfrac{x}{x+3}}$

When 1 is subtracted from numerator and denominator, the fraction becomes

$\sf{\dfrac{x-1}{x+3-1}}\\\\\\= \sf{\dfrac{x-1}{x-2}}$

According to the Question,

$\sf{\dfrac{x}{x+3}-\dfrac{1}{14}=\dfrac{x-1}{x+2}}$

$\implies \: \sf{\dfrac{14x-1(x+3)}{14(x+3)}=\dfrac{x-1}{x+2}}$

$\dashrightarrow \: \: \sf{\dfrac{14x-x-3}{14x+42}=\dfrac{x-1}{x+2}}$

$\dashrightarrow \: \: \sf{\dfrac{13x-3}{14x+42}=\dfrac{x-1}{x+2}}$

$\dashrightarrow \: \: \sf{(13x-3)(x+2)=(14x+42)(x-1)}$

$\implies \: \sf{13x^{2}+26x-3x-6=14x^{2}-14x+42x-42}$

$\implies \: \sf{13x^{2}+23x-6=14x^{2} +28x-42}$

$\implies \: \sf{14x^{2} -13x^{2} +28x-23x=42-6}$

$\implies \: \sf{x^{2} +5x=36}$

$\implies \: \sf{x^{2} +5x-36=0}$

On factorising the above equation,

$\\\implies \: \sf{x^{2} +9x-4x-36=0}$

$\implies \: \sf{x(x+9)-4(x+9)=0}$

$\implies \: \sf{(x +9)(x-4)=0}$

So, the value of x is -9 or 4

It is given that the fraction is positive.

So, x = 4

$\\\\\bf{\underline{For \: finding\: the \: fraction;}}$

$\sf{Fraction \longrightarrow \dfrac{x}{x+3}=\dfrac{4}{7}}$

$\therefore \: \boxed{\boxed{\bf{Fraction \to \dfrac{4}{7}}}}$