Question 723947

let numbers be {{{x}}} and {{{y}}}

given:

The sum of two numbers is {{{10}}}.

{{{x+y=10}}}....eq.1 

The sum of their squares is {{{68}}}.

{{{x^2+y^2=68}}}....eq.2 

solve the system:

{{{x+y=10}}}....eq.1 
{{{x^2+y^2=68}}}....eq.2 
__________________________

start with eq.1

{{{x+y=10}}}....eq.1 ...solve for {{{y}}}

{{{y=10-x}}}..... substitute it in eq.2


{{{x^2+(10-x)^2=68}}}....eq.2 ...solve for {{{x}}}

{{{x^2+10^2-20x+x^2=68}}}

{{{2x^2+100-20x=68}}}

{{{2x^2-20x+100-68=0}}}

{{{2x^2-20x+32=0}}}.......simplify, divide all by {{{2}}}

{{{x^2-10x+16=0}}}.......use quadratic formula

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

{{{x = (-(-10) +- sqrt( (-10)^2-4*1*16 ))/(2*1) }}}

{{{x = (10 +- sqrt( 100-64 ))/2 }}}

{{{x = (10 +- sqrt( 36 ))/2 }}}


{{{x = (10 +- 6)/2 }}}

solutions:

{{{x = (10 + 6)/2 }}}

{{{x = 16/2 }}}

{{{x = 8 }}}

or

{{{x = (10 -6)/2 }}}

{{{x = 4/2 }}}

{{{x = 2}}}


now find {{{y}}}

{{{y=10-x}}}

{{{y=10-8}}}

{{{y=2}}}

or

{{{y=10-x}}}

{{{y=10-2}}}

{{{y=8}}}

solution pairs are: {{{x = 8 }}} and {{{y=2}}}

or {{{x = 2 }}} and {{{y=8}}}