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The sum of the two numbers is 15 and the sum of their reciprocals is 3/10. Find the numbers.
x+y=15 y+x/xy=3/10 1/x + 1/15-x=3/10 150=45x-3x2 3x2-45x-150=0 x2-15-50=0 Therefore no.are 5 and 10
x+y=15
y+x/xy=3/10
1/x + 1/15-x=3/10
150=45x-3x2
3x2-45x-150=0
x2-15-50=0
Therefore no.are 5 and 10
let the numbers be x and y so x+y = 15 -------------------------(i) and 1/x + 1/y = 3/10 ==> (y + x)/xy = 3/10 (putting (i)) 15/ xy = 3/10 5/xy = 1/10 xy = 50 --------------------(ii) x = 50/y -----------------------(iii) 50/y + y = 15 50 + y2 = 15y y2 - 15y + 50 = 0 (y-10)(y-5) = 0 y = 10 and y = 5 now there will be 2 numbers for 'x' substituting y as 10 we get x = 5 ; substituting y as 5 we get x = 10; so the numbers are x = 10 and y = 5 and viceversa ! PLEASE APPROVE !
let the numbers be x and y
so
x+y = 15 -------------------------(i)
and 1/x + 1/y = 3/10
==> (y + x)/xy = 3/10 (putting (i))
15/ xy = 3/10
5/xy = 1/10
xy = 50 --------------------(ii)
x = 50/y -----------------------(iii)
50/y + y = 15
50 + y2 = 15y
y2 - 15y + 50 = 0
(y-10)(y-5) = 0
y = 10 and y = 5
now there will be 2 numbers for 'x'
substituting y as 10 we get
x = 5 ;
substituting y as 5 we get
x = 10;
so the numbers are x = 10 and y = 5 and viceversa !
PLEASE APPROVE !
let the numbers be x & y. x+y=15............1 1/x + 1/y = 3/10...........2 y+x /xy = 3.10 10(x+y)= 3xy. 10(15)= 3(15-y)y. 50 = 15y-y2. y2-15y+50 = 0. y2-10y-5y+50 = 0. y(y-10)-5(y-10)=0. (y-5)(y-10)=0. y = 5 or 10. Similarly x = 10 or 5. The numbers are 5 and 10.
let the numbers be x & y.
x+y=15............1
1/x + 1/y = 3/10...........2
y+x /xy = 3.10
10(x+y)= 3xy.
10(15)= 3(15-y)y.
50 = 15y-y2.
y2-15y+50 = 0.
y2-10y-5y+50 = 0.
y(y-10)-5(y-10)=0.
(y-5)(y-10)=0.
y = 5 or 10.
Similarly x = 10 or 5.
The numbers are 5 and 10.
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