Fraction becomes 1/3 when 1 is …

Let the fraction be {tex}\frac{x}{y}{/tex}, Then, according to the question,
{tex}\frac{{x - 1}}{y} = \frac{1}{3}{/tex} ...(1)
{tex}\frac{x}{{y + 8}} = \frac{1}{4}{/tex} ...(2)
{tex}\Rightarrow{/tex} 3(x - 1) = y ...(3)
4x = y + 8 = ...(4)
{tex}\Rightarrow{/tex} 3x - y - 3 = 0 ...(5)
4x - y - 8 = 0 ...(6)
To solve the equation (5) and (6) by cross multiplication method
We draw the diagram below;

Then, {tex}\frac{x}{{( - 1)( - 8) - ( - 1)( - 3)}} = \frac{y}{{( - 3)(4) - ( - 8)(3)}}{/tex}{tex} = \frac{1}{{(3)( - 1) - (4)( - 1)}}{/tex}
{tex}\Rightarrow \frac{x}{{8 - 3}} = \frac{y}{{ - 12 + 24}} = \frac{1}{{ - 3 + 4}}{/tex}
{tex}\Rightarrow \frac{x}{5} = \frac{y}{{12}} = \frac{1}{1}{/tex}
{tex}\Rightarrow{/tex} x = 5 and y = 12
Hence, the required fraction is {tex}\frac{5}{{12}}{/tex}.
Verification :substituting x = 5, y = 12
We find that both the equations (1) and (2) are satisfied as shown below:
{tex}\frac{{x - 1}}{y} = \frac{{5 - 1}}{{12}} = \frac{4}{{12}} = \frac{1}{3}{/tex}
{tex}\frac{x}{{y + 8}} = \frac{5}{{12 + 8}} = \frac{5}{{20}} = \frac{1}{4}{/tex}
Hence, the solution we have got is correct.

Let x be the numerator and y be the denominator A.T.Q x-1/y=1/3 x/y+8=1/4 By cross multiplication =3 (x-1)=y &4x=y+8 =3x-3=y &4x=y+8 =3x-y=3-------(1) &4x-y=8--------(2) From (1) x=3+y/3-------(3) Put x=3+y/2 in (2) we get 4 (3+y/3)=8 =12+4y/3-y=8 =12+4y/3-y=8 =12+4y-3y/3=8 =12+4y-3y=24 =12+y=24 =y=24-12 y=12 Put y=12 in (3) x=3+12/3=15/3=5 Therefore numerator(x)=5 And denominator (y)=12 :) hope its is helpful :)