Question 826140
let brother's age be {{{x}}} and your age {{{y}}}

if brother is {{{one}{{ year older than {{{twice}}} your age, then we have

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


if the product of your ages is {{{55}}}, then we have 

{{{x*y=55}}}...............eq.2

now solve the system

{{{x=2y+1}}}...........eq.1
{{{x*y=55}}}...............eq.2
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{{{x*y=55}}}...............eq.2...substitute {{{x}}} from eq.1

{{{(2y+1)*y=55}}}...............solve for {{{y}}}

{{{2y^2+y=55}}}

{{{2y^2+y-55=0}}}..........use quadratic formula

 {{{y = (-1 +- sqrt( 1^2-4*2*(-55) ))/(2*2) }}} 


{{{y = (-1 +- sqrt( 1+440 ))/4 }}} 

{{{y = (-1 +- sqrt( 441 ))/4 }}} 

{{{y = (-1 +- 21)/4 }}}..............we will need only positive solution because age cannot be negative number

{{{y = (-1 +21)/4 }}}

{{{y = 20/4 }}}

{{{y = 5 }}}....................your age 

brother's age will be {{{x=2y+1}}} which is {{{x=2*5+1}}} or {{{x=11}}}

so, your brother is {{{11}}} years old