Two positive integers differ by 4 and sum of their reciprocals is 10/21. Then, one of the numbers is ?
Answers
Answer:
Let two positive integers be x and y respectively.
So by statement ,
x-y = 4.........................eq (1)
1/x +1/y = 10/21
(x + y)/xy = 10/21
(x + y)21 = 10xy
21x +21y = 10xy...........eq (2)
So , from eq (1)
x = y+4
Put above equation in eq (2)
21(y+4) +21y = 10(y+4)y
21y +84 +21y = 40y + 10y^2
-10y^2+2y+84= 0
10y^2 -2y -84 =0
Or 5y^2 -y -42 =0
By solving the quadratic equation we will get the value of y as
y = 3 , -2.8
Therefore , x=7 , 1.2
Bt since it is given in the question that integers should be positive
Therefore y = 3 and x = 7 ( neglect y = -2.8 as it is negative )
Answer
Let two positive integers be x and y respectively.
So by statement ,
x-y = 4.........................eq (1)
1/x +1/y = 10/21
(x + y)/xy = 10/21
(x + y)21 = 10xy
21x +21y = 10xy...........eq (2)
So , from eq (1)
x = y+4
Put above equation in eq (2)
21(y+4) +21y = 10(y+4)y
21y +84 +21y = 40y + 10y^2
-10y^2+2y+84= 0
10y^2 -2y -84 =0
Or 5y^2 -y -42 =0
By solving the quadratic equation we will get the value of y as
y = 3 , -2.8
Therefore , x=7 , 1.2
Bt since it is given in the question that integers should be positive
Therefore y = 3 and x = 7 ( neglect y = -2.8 as it is negative )