Question 1012836
let one number be {{{x}}} and another number {{{y}}}

if one is {{{11}}} less than {{{6}}} times another number we have:


{{{x=6y-11}}}....eq.1

 if their product is {{{7}}} we have:

{{{x*y=7}}}....eq.2....substitute {{{x}}} from eq.1

{{{(6y-11)*y=7}}}.....solve for {{{y}}}

{{{6y^2-11y=7}}}

{{{6y^2-11y-7=0}}}

{{{y= (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{y= (-(-11) +- sqrt((-11)^2-4*6*(-7) ))/(2*6) }}} 

{{{y= (11 +- sqrt(121+168 ))/12 }}} 

{{{y= (11 +- sqrt(289 ))/12 }}} 

{{{y= (11 +- 17)/12 }}} 

solutions:

{{{y= (11 + 17)/12 }}} 
{{{y=28/12 }}}
{{{highlight(y= 7/3) }}}
or  
{{{y= (11 - 17)/12 }}} 
{{{y=-6/12 }}}
{{{highlight(y= -1/2) }}}

the number {{{x}}} will be:

{{{x*y=7}}}....eq.2

{{{x=7/y}}}

{{{x=7/(7/3)}}}

{{{x=(3*7)/7}}}

{{{highlight(x=3)}}}

or

{{{x=7/y}}}

{{{x=7/(-1/2)}}}

{{{x=(2*7)/(-1)}}}

{{{highlight(x=-14)}}}

so, there is two pairs of numbers and they are:

{{{highlight(x=3)}}} and {{{highlight(y= 7/3) }}}

or

{{{highlight(x=-14)}}} and {{{highlight(y= -1/2) }}}