Two integers differ by four, with the property that the sum of their reciprocals is the same as 14 times the product of their reciprocals. What are the two integers?
Smaller --- x
larger ----- x+4
1/x + 1/(x+4) = 14(1/x)(1/(x+4)
multiply by x(x+4)
x+4 + x = 14
2x = 10
x = 5
the two integers are 5 and 9
check:
1/5 +1/9 = 14/45
14 times product = 14(1/5)(1/9) = 14/45
my answers are correct
To solve this problem, let's assume the two integers are x and y. We are given that "Two integers differ by four," so we can write the equation:
x - y = 4 --(Equation 1)
We are also given that "the sum of their reciprocals is the same as 14 times the product of their reciprocals." The sum of their reciprocals can be written as:
1/x + 1/y
The product of their reciprocals can be written as:
(1/x) * (1/y) = 1/(xy)
According to the problem, the sum of their reciprocals is the same as 14 times the product of their reciprocals, so we can write the equation:
1/x + 1/y = 14 * (1/(xy)) --(Equation 2)
To eliminate the fractions in Equation 2, we can multiply both sides by xy to get:
y + x = 14 --(Equation 3)
Now we have a system of equations (Equations 1 and 3) that we can solve to find the values of x and y.
From Equation 1, we have:
x - y = 4 --> x = 4 + y
Substituting this value of x into Equation 3, we get:
y + (4 + y) = 14
Simplifying:
2y + 4 = 14
Subtracting 4 from both sides:
2y = 10
Dividing both sides by 2:
y = 5
Now substitute the value of y back into x = 4 + y:
x = 4 + 5
x = 9
So, the two integers are 9 and 5.