Question 414862
Let the smaller integer be {{{x}}}, then the larger one is {{{x + 1}}} (since they are consecutive).

Knowing that the sum of the reciprocals of these two integers is 3 / 2, we obtain an equation:

{{{(1 /x) + 1/(x + 1) = 3 /2}}}


{{{(x + 1 + x) / (x(x + 1)) = 3/ 2}}}...cross multiply


{{{2(x + 1 + x) = 3x(x + 1)}}}

{{{2x + 2 + 2x = 3x^2 + 3x}}}

{{{4x + 2  = 3x^2 + 3x}}}


{{{2  = 3x^2 + 3x-4x}}}


{{{ 3x^2 -x -2 =0}}}.........use quadratic formula

*[invoke quadratic_formula 3, 1, -2, "x"]


So our solutions are:

{{{x=2/3}}}....is not an integer, so we discard this solution

 or {{{x=-1}}}

If the smaller integer is {{{-1}}}, then the second must be {{{-1 + 1 = 0}}}

Answer: {{{-1}}}, {{{0}}}