Question

the denominator of a fraction exceeds its numerator by 3 if 1 is added to both numerator and denominator a difference between the new and the original fraction is 1 by 24 find the original fraction

Answer 1

Hey

Let the denominator of a fraction be =x
let the numerator of a fraction be =x+3
required fraction=x+3/x
new fraction=x+3+1/x+1
=x+4/x+1
A/Q
x+3/x - x+4/x+1 =1/24
(x+3)(x+1)-(x)(x+4)/x(x+1)=1/24
x2+x+3x+3-x2-4x/x2+x=1/24
3/x2+x=1/24
3×24=x2+x
72=x2+x
0=x2+x-72
0=x2+9x-8x-72
0=x(x+9)-8(x+9)
0=(x+9)(x-8)
let x+9=0 and x-8=0
x= -9 and x=8
• •
• denominator can not be negative 
•
• •denominator=x=8
and numerator=x+3=8+3=11
required fraction=11/8

Answer 2

Hey! ! !

Solution :-

● Let the denominator of a fraction be =x

and,

● let the numerator of a fraction be =x+3

required fraction=x+3/x

new fraction=x+3+1/x+1
=x+4/x+1

☆ A/Q

x+3/x - x+4/x+1 =1/24

(x+3)(x+1)-(x)(x+4)/x(x+1)=1/24

x2+x+3x+3-x2-4x/x2+x=1/24

3/x2+x=1/24

3×24=x2+x

72=x2+x

x2+x-72=0

x2+9x-8x-72=0

x(x+9)-8(x+9)=0

(x+9)(x-8)=0

let x+9=0 and x-8=0
x= -9 and x=8

☆ Denominator can't be negative 

●denominator=x=8

and numerator=x+3=8+3=11

:- required fraction=11/8

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☆ Regards :- ♡♡《 Nitish kr singh 》♡♡