Write an equation that can be used to find two consecutive integers such that the sum of their
reciprocals is
13
84
I thought it was 13/84, but that does not work.
However, 1/6 + 1/7 = 13/42
1/x + 1/(x+1)=13 0r 84
x=integer
To find two consecutive integers such that the sum of their reciprocals is 13/84, we can set up the equation as follows:
1/n + 1/(n+1) = 13/84
To solve this equation, we need to find a common denominator. In this case, the common denominator is given by multiplying (n)(n+1). The equation becomes:
(n+1)/(n(n+1)) + n/(n(n+1)) = 13/84
Next, we can combine the fractions on the left side by adding the numerators:
(n+1 + n)/(n(n+1)) = 13/84
Simplifying further, we have:
(2n + 1)/(n(n+1)) = 13/84
To get rid of the denominators, we can cross-multiply:
(2n + 1) * 84 = 13 * (n(n+1))
Simplifying both sides of the equation:
168n + 84 = 13n^2 + 13n
Rearranging the equation and setting it equal to zero:
13n^2 + 13n - 168n - 84 = 0
Combining like terms:
13n^2 - 155n - 84 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Once we find the values of n, we can substitute it back into the equation to find the consecutive integers.