Question 733591
let one girl's be {{{x}}} and another {{{y}}}
if two girls' ages, {{{11}}} years apart, then {{{x-y=11}}}...eq.1

when multiplied,{{{x*y=210}}} .......eq.2

 The sum of their ages is {{{x+y}}}

start with:

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

{{{x*y=210}}} .......eq.2
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{{{x-y=11}}}...eq.1...solve for {{{y}}}

{{{x-11=y}}}...eq.1...substitute {{{y}}} in eq.2

{{{x*(x-11)=210}}} .......eq.2.....solve for {{{y}}}

{{{x^2-11x=210}}}

{{{x^2-11x-210=0}}}......use quadratic formula



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


{{{x = (-(-11) +- sqrt((-11)^2-4*1*(-210) ))/(2*1) }}}


{{{x = (11 +- sqrt(121+840) ))/2 }}}


{{{x = (11 +- sqrt(961) ))/2 }}}


{{{x = (11 +- 31 ))/2 }}}


solutions:

{{{x = (11 + 31 )/2 }}}

{{{x = 42/2 }}}

{{{highlight(x = 21 )}}}

and

{{{x = (11 -31 )/2 }}}

{{{x = -20/2 }}}

{{{x = -10 }}}.....since we are looking for age, we do not need this solution

now find {{{y}}}

{{{x-11=y}}}

{{{21-11=y}}}

{{{highlight(10=y)}}}


so, one girl's be {{{highlight(x = 21 )}}} and another {{{highlight(10=y)}}} and the sum of their ages is {{{x+y=21+10=highlight(31)}}}