Answer :
Let the denominator be x
Numerator is x-3
The fraction is (x-3)/x
when 2 is added to numerator and denominator
we have fraction as (x-1)/(x+2)
ATQ (x-3)/x +(x-1)/(x+2)= 29/20
⇒ (x-3)((x+2) +x(x-1)=x(x+2)*29/20
⇒ (x²-6-x + x²-x)20= 29x²+58x
⇒ 40x²-120-40x=29x²+58x
⇒ 11x²-98x- 120=0
⇒ 11x² - 110x +12x -120=0
⇒ (11x+12)(x-10)=0
Canceling the negative fraction value of x we have x=10
Then the fraction (x-3)/x=7/10
Numerator is x-3
The fraction is (x-3)/x
when 2 is added to numerator and denominator
we have fraction as (x-1)/(x+2)
ATQ (x-3)/x +(x-1)/(x+2)= 29/20
⇒ (x-3)((x+2) +x(x-1)=x(x+2)*29/20
⇒ (x²-6-x + x²-x)20= 29x²+58x
⇒ 40x²-120-40x=29x²+58x
⇒ 11x²-98x- 120=0
⇒ 11x² - 110x +12x -120=0
⇒ (11x+12)(x-10)=0
Canceling the negative fraction value of x we have x=10
Then the fraction (x-3)/x=7/10
Answer:
(x-3)/x=7/10
Step-by-step explanation:
Let the denominator be x
So, Numerator is x-3
We got fraction,
[tex]\frac{x}{x-3}[/tex]
According to question, when 2 is added to numerator and denominator,
Then, we get = (x-1)/(x+2)
So,
=> (x-3)/x +(x-1)/(x+2)= 29/20
=> (x-3)((x+2) +x(x-1)
=> x(x+2)*29/20
⇒ (x²-6-x + x²-x)20= 29x²+58x
⇒ 40x²-120-40x=29x²+58x
⇒ 11x²-98x- 120=0
⇒ 11x² - 110x +12x -120=0
⇒ (11x+12)(x-10)=0
We will have to cancel the negative fraction value of x.
=> X = 10
We have the fraction,
=> (x-3)/x=7/10