The addition of numerator and denominator of a fraction is three less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Find the fraction.
Answers
Step-by-step explanation:
Given:-
The addition of numerator and denominator of a fraction is three less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator.
To find:-
Find the fraction.?
Solution:-
Let the numerator be X
Let the denominator be Y
Then the Fraction = X/Y
Given that:
Condition -1:-
The addition of numerator and denominator of a fraction is three less than twice the denominator.
=>X+Y = 2Y -3
=>X +Y -2Y = -3
=>X -Y = -3 --------------(1)
Condition-2:-
If the numerator and denominator are decreased by 1, the numerator becomes half the denominator.
The numerator us decreased by 1 =X-1
The denominator is decreased by 1 = Y-1
(X-1) = (Y-1)/2
=>2(X-1) = (Y-1)
=>2X-2 = Y-1
=>2X-2 -Y = -1
=>2X-Y = -1+2
=>2X -Y = 1 ---------------(2)
On solving (1)&(2)
on subtracting (1) from (2)
2X-Y = 1
X-Y = -3
(-)
_______
X+0 = 4
______
Therefore, X = 4
on Substituting the value of X in (1) then
X-Y = -3
=>4-Y = -3
=>4+3 = Y
=>7 = Y
=>Y = 7
Therefore, Y = 7
The required fraction = 4/7
Answer:-
The required fraction for the given problem is 4/7
Check:-
The fraction = 4/7
Condition -1:-
The addition of numerator and denominator of a fraction is three less than twice the denominator.
=>4+7 = 11
and 2(7)-3
=>14-3
=>11
Verified
Condition-2:-
If the numerator and denominator are decreased by 1, the numerator becomes half the denominator.
numerator = 4-1 = 3
Denominator=7-1 =6
3 = 6/2
3=3
Verified the given relations
They are true for the fracation 4/7
Answer:
let the fraction be x/y
(x+y)=2y-3-(1)
x-1=y-1/2-(2)
solve (1)
x=y-3-(3)
solve (2)
2x-2=y-1
2x-y=1--(4)