Question 984049
let's a positive number be {{{x}}} and its reciprocal {{{1/x}}}

given:

{{{x+13=140(1/x)}}} ....solve for {{{x}}}

{{{x+13=140/x}}}
{{{(x+13)x=140}}}
{{{x^2+13x=140}}}
{{{x^2+13x-140=0}}}...factor completely
{{{x^2-7x+20x-140=0}}}
{{{(x^2-7x)+(20x-140)=0}}}
{{{x(x-7)+20(x-7)=0}}}
{{{(x-7)(x+20) = 0}}}

solutions:

if {{{(x-7) = 0}}}=>{{{highlight(x=7)}}}
if {{{(x+20) = 0}}}=>{{{highlight(x=-20)}}}


check:

{{{x+13=140(1/x)}}}...if {{{x=7}}}
{{{7+13=140(1/7)}}}
{{{20=140/7}}}
{{{20=20}}}........true

{{{x+13=140(1/x)}}}...if {{{x=-20}}}
{{{-20+13=140(1/-20)}}}
{{{-7=-140/20}}}
{{{-7=-7}}}........true