Question 909098

two numbers, {{{x}}} and {{{y}}}, such that their difference,{{{x-y}}} is equal to their product {{{x*y}}}

 so,{{{x-y=xy}}}...eq.1

and the sum of their reciprocal,{{{1/x}}} and {{{1/y}}}, is {{{5}}}.

so, {{{1/x+1/y=5}}} ....eq.2

start with {{{x-y=xy}}}...eq.1 and solve for {{{x}}}

 {{{x-xy=y}}}
{{{x(1-y)=y}}}
{{{x= y/(1-y) }}}

plug it in eq.2


{{{1/x+1/y=5}}} ....eq.2

{{{1/(y/(1-y))+1/y=5}}} ...solve for {{{y}}}

{{{(1-y)/y+1/y=5}}}

{{{((1-y)+1)/y=5}}}

{{{(2-y)/y=5}}}

{{{2-y=5y}}}

{{{2=5y+y}}}

{{{2=6y}}}

{{{2/6=y}}}

{{{highlight(y=1/3)}}}

now find {{{x}}}


{{{x= y/(1-y) }}}

{{{x= (1/3)/(1-1/3) }}}

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

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

{{{x= (1*cross(3))/(cross(3)*2) }}}

{{{highlight(x= 1/2) }}}

check:

{{{x-y=xy}}}...eq.1

{{{1/2-1/3=(1/2)*(1/3)}}}

{{{1*3/2*3-1*2/3*2=1/6}}}

{{{3/6-2/6=1/6}}}

{{{(3-2)/6=1/6}}}

{{{1/6=1/6}}} ...