10. The sum of the numerator and denominator of a certain positive fraction is 8. If 2 is
added to both the numerator and denominator, the fraction is increased by 4/35. Find
the fraction by quadratic equation
Find
Answers
Step-by-step explanation:
Given :-
The sum of the numerator and denominator of a certain positive fraction is 8. If 2 is added to both the numerator and denominator, the fraction is increased by 4/35.
To find :-
Find the fraction by quadratic equation
Solution:-
Let the numerator of a fraction be X
Let the denominator of the fraction be Y
Then the fraction = X/Y
Given that
The sum of the numerator and denominator = 8
=> X+Y = 8 -----(1)
and
=> Y = 8-X ------(2)
If 2 is added to both the numerator and denominator, the fraction is increased by 4/35
=> (X+2)/(Y+2) =(X/Y) + (4/35)
=> (X+2)/(Y+2) - (X/Y) = 4/35
=> [(X+2)(Y)-(Y+2)(X)]/[(Y+2)Y]=4/35
=> (XY+2Y-XY-2X)/(Y^2+2Y ) = 4/35
=> (2Y-2X)/(Y^2+2Y) = 4/35
=> 2(Y-X)/(Y^2+2Y) = 4/35
=> (Y-X)/(Y^2+2Y) = 2/35
On applying cross multiplication then
=> 35(Y-X) = 2(Y^2+2Y)
=> 35Y-35X = 2Y^2+4Y
=> 35Y-35X-2Y^2-4Y = 0
=> -2Y^2-35X+31Y = 0
=> 2Y^2+35X-31Y = 0
From (1)
=> 2(8-X)^2+35X-31(8-X) = 0
=> 2(64-16X+X^2)+35X-248+31X=0
=> 128-32X+2X^2+35X-248+31X = 0
=> -120+34X+2X^2 = 0
=> 2X^2+34X-120 = 0
=> 2(X^2+17X-60)=0
=>X^2+17X-60 = 0
=>X^2-3X+20X-60=0
=>X(X-3)+20(X-3) =0
=> (X-3)(X+20)=0
=>>X-3 = 0 or X+20 = 0
=> X = 3 or X= -20
But given that
The fraction is a positive
So, X = 3
Then Y = 8-3 = 5
X = 3
and Y = 5
Fraction = 3/5
Answer:-
The required fraction for the given problem is 3/&
Check :-
X = 3 and Y = 5 then
Their sum = 3 +5 = 8
If 2 is added to 3 and 5 they becomes 5 and 7
=> (5/7)-(3/5)
=(25-21)/35
= 4/35
Verified the given relations in the given problem
Answer:
here you go
Step-by-step explanation:
I HOPE IT WILL HELPFUL TO YOU
