Question 919049


{{{10}}} years ago {{{Alfred}}}'s age was {{{3}}} times that of his son {{{Jerry}}}.

{{{Alfred-10=3(Jerry-10)}}}.........eq.1

{{{15}}} years from now {{{Alfred}}}'s age will be {{{12}}} less than twice of {{{Jerry}}} age

{{{Alfred+15=2(Jerry+15)-12}}}.........eq.2

solve the system:

{{{Alfred-10=3(Jerry-10)}}}.........eq.1
{{{Alfred+15=2(Jerry+15)-12}}}.........eq.2
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{{{Alfred-10=3(Jerry-10)}}}.........eq.1 solve for {{{Alfred}}}

{{{Alfred=3(Jerry-10)+10}}}

{{{Alfred=3Jerry-30+10}}}

 {{{Alfred=3Jerry-20}}}.......substitute in eq.2


{{{Alfred+15=2(Jerry+15)-12}}}.........eq.2..solve for {{{Jerry}}}

{{{3Jerry-5=2Jerry+30-12}}}

{{{3Jerry-2Jerry=5+18}}}

{{{highlight(Jerry=23)}}}



find {{{Alfred=3Jerry-20}}}

{{{Alfred=3*23-20}}}

{{{Alfred=69-20}}}

{{{highlight(Alfred=49)}}}


check:

{{{10}}} years ago {{{Alfred}}}'s age was {{{49-10=39}}} and {{{Jerry}}}'s age was {{{23-10=13}}}
as you can see, {{{39}}} is times times more than {{{13}}}


{{{15}}} years from now {{{Alfred}}}'s age will be {{{49+15=64}}} and {{{Jerry}}}'s age will be {{{23+15=38}}} 

 than twice of {{{Jerry}}} age is {{{2*38=76}}} and {{{76-64=12}}}