Question 1130449

let a number be {{{x}}} and its reciprocal {{{1/x}}}


if the sum of a number and its reciprocal is {{{13 / 6}}}, we have

{{{x+1/x=13/6}}}

{{{(x^2+1)/x=13/6}}}...........cross multiply

{{{6(x^2+1)=13x}}}

{{{6x^2+6=13x}}}

{{{6x^2-4x-9x+6=0}}}

{{{(6x^2-4x)-(9x-6)=0}}}

{{{2x(3x-2)-3(3x-2)=0}}}

{{{(2x - 3) (3x - 2) = 0}}}

solutions:


if {{{(2x - 3) = 0}}}=>{{{2x=3}}}=>{{{x=3/2}}}

if {{{ (3x - 2) = 0}}}=>{{{3x=2}}}=>{{{x=2/3}}}


so, your numbers are:

{{{x=3/2}}} and its reciprocal {{{x=2/3}}}

or vice versa 

{{{x=2/3}}}  and its reciprocal {{{x=3/2}}}