Answer:
The original fraction is [tex]\frac{6}{14}[/tex].
Step-by-step explanation:
Solution,
Let the fraction be'[tex]\frac{x}{y}[/tex]'.
Where 'x' is the numerator and 'y' is the denominator.
Now according to question, if 2 is added to both the numerator and denominator the value of the fraction becomes 1/2.
So, framing the above sentence in equation form, we get;
[tex]\frac{x+2}{y+2}=\frac{1}{2}[/tex]
On using cross multiplication method, we get;
[tex]2(x+2)=y+2\\\\2x+4=y+2\\\\2x+4-2=y\\\\\therefore\ y=2x+2\ \ \ \ equation\ 1[/tex]
Now according to question, if 2 is Subtracted to both the numerator and denominator the value of the fraction becomes 1/3.
So, framing the above sentence in equation form, we get;
[tex]\frac{x-2}{y-2}=\frac{1}{3}[/tex]
On using cross multiplication method, we get;
[tex]3(x-2)=y-2\\\\3x-6=y-2\\\\3x-y=-2+6\\\\3x-y=4 \ \ \ \ equation\ 2[/tex]
Now Substituting the equation 1 in equation 1 we get;
[tex]3x-y=4\\\\3x-(2x+2)=4\\\\3x-2x-2=4\\\\3x-2x=4+2\\\\x=6[/tex]
Now Substituting the value of x in equation 1 we get;
[tex]y=2x+2\\\\y=2\times6+2 = 14[/tex]
Hence The original fraction is [tex]\frac{6}{14}[/tex].