Asked by Dasia
one positive number is 32 less than another number. if the reciprocal of the smaller number is added to five times the reciprocal of the larger number, the sum is 1/6. find the two numbers
Answers
Answered by
Damon
n and n+32
1/n + 5/(n+32) = 1/6
multiply all terms by 6n(n+32)
6(n+32) + 5(6n) = n(n+32)
6n + 192 + 30 n = n^2 + 32 n
n^2 - 4n - 192 = 0
(n-16)(n+12) = 0
n = 16
n+32 = 48
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check
1/16 + 5/48 = 1/6 ???
3/48 + 5/48
8/48
1/6 sure enough
1/n + 5/(n+32) = 1/6
multiply all terms by 6n(n+32)
6(n+32) + 5(6n) = n(n+32)
6n + 192 + 30 n = n^2 + 32 n
n^2 - 4n - 192 = 0
(n-16)(n+12) = 0
n = 16
n+32 = 48
================
================
check
1/16 + 5/48 = 1/6 ???
3/48 + 5/48
8/48
1/6 sure enough
Answered by
Reiny
one number --- x
the other ------ x-32
1/(x-32) + 5(1/x) = 1/6
times 6x(x-32)
6x + 30(x-32) = x(x-32)
6x + 30x - 960 = x^2 - 32x
x^2 - 68x + 960 = 0
(x-20)(x-48) = 0
x = 20 or x = 48
if x = 20 then the other is negative, but both our numbers are to be positive
so x = 48
and the two number are 48 and 16
check:
"the reciprocal of the smaller number is added to five times the reciprocal of the larger number"
= 1/16 + 5(1/48)
= 1/6 , as required
the other ------ x-32
1/(x-32) + 5(1/x) = 1/6
times 6x(x-32)
6x + 30(x-32) = x(x-32)
6x + 30x - 960 = x^2 - 32x
x^2 - 68x + 960 = 0
(x-20)(x-48) = 0
x = 20 or x = 48
if x = 20 then the other is negative, but both our numbers are to be positive
so x = 48
and the two number are 48 and 16
check:
"the reciprocal of the smaller number is added to five times the reciprocal of the larger number"
= 1/16 + 5(1/48)
= 1/6 , as required
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