Question

The sum of the reciprocal of the two consecutive positive number is 23 upon 132. Find the numbers

Answer 1

we can say that if the numbers are n and n+1, then the reciprocals are 1/n and 1/(n+1). So we're saying  
1/n + 1/(n+1) = 23/132; multiply the whole equation by 132n(n+1) to get rid of the denominators. 
<=> 132(n+1) + 132(n) = 23n(n+1) 
<=> 264n + 132 = 23n^2 + 23n 
<=> 23n^2 - 241n - 132 = 0 
<=> (23n + 12)(n - 11) = 0 
so n must be 11 or -12/23. But since n has to be an integer, n=11 is the only valid solution. Then the two numbers are n and n+1, i.e. 11 and 12, as before.

Answer 2

Let the two consecutive number be x and x+1
Therefore ATQ
1/x+1/x+1=23/132
x+1+x/x2+x=23/132
2x+1/x2+x=23/132
264x+132=23x2+ 23
241x= 23x2-132
23x2-241x-132=0
23x2-253x+12x-132=0
23x(x-11)+12(x-11)=0
(23x+12) (x-11)=0
X=-12/23 , x=11