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
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
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
To solve this problem, let's assign variables to the unknowns.
Let's say the larger number is 'x' and the smaller number is 'y'.
According to the given information, one positive number is 32 less than another number, so we can write the equation:
x = y + 32
We are also given that the reciprocal of the smaller number (1/y) when added to five times the reciprocal of the larger number (5/x), gives us 1/6:
1/y + 5/x = 1/6
Now, we have a system of two equations:
x = y + 32
1/y + 5/x = 1/6
To solve this system, we can use the method of substitution. Rearrange the first equation to solve for 'y':
y = x - 32
Now substitute this expression for 'y' in the second equation:
1/(x - 32) + 5/x = 1/6
To eliminate the denominators, multiply the entire equation by 6x(x - 32):
6x + 30(x - 32) = x(x - 32)
Simplify the equation:
6x + 30x - 960 = x^2 - 32x
Combine like terms:
36x - 960 = x^2 - 32x
Rearrange the equation in standard quadratic form:
x^2 - 68x + 960 = 0
Solve this quadratic equation. The factors of 960 that have a difference of 68 are 40 and 28. Therefore, we can factor the equation as:
(x - 40)(x - 28) = 0
Setting each factor equal to zero:
x - 40 = 0 -> x = 40
x - 28 = 0 -> x = 28
So, we have two possible values for 'x': x = 40 and x = 28.
Now substitute these values back into the first equation to find the corresponding values of 'y':
For x = 40:
y = 40 - 32 -> y = 8
For x = 28:
y = 28 - 32 -> y = -4
Since we are looking for positive numbers, we disregard the solution (x = 28, y = -4).
Therefore, the two numbers are x = 40 and y = 8.