Question 919024

{{{3}}} years ago {{{A}}}'s age was {{{double}}} of {{{B}}}.

{{{A-3=2(B-3)}}}.........eq.1

{{{7}}} years later the sum of their united ages will be {{{83}}} years.

{{{(A+7)+(B+3)=83}}}.........eq.2

solve the system:

{{{A-3=2(B-3)}}}.........eq.1
{{{(A+7)+(B+7)=83}}}.........eq.2
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{{{A-3=2(B-3)}}}.........eq.1 solve for {{{A}}}

{{{A=2(B-3)+3}}}.......substitute in eq.2


{{{(2(B-3)+3+7)+(B+7)=83}}}.........eq.2..solve for {{{B}}}

{{{2B-6+3+7+B+7=83}}}

{{{3B+11=83}}}

{{{3B=83-11}}}

{{{3B=72}}}

{{{B=72/3}}}

{{{B=24}}}

find {{{A=2(B-3)+3}}}

{{{A=2(24-3)+3}}}

{{{A=2(21)+3}}}

{{{A=42+3}}}

{{{A=45}}}

check:

{{{3}}} years ago {{{A}}}'s age was {{{45-3=42}}}

{{{3}}} years ago {{{B}}}'s age was {{{24-3=21}}}


 {{{42}}} is {{{double}}} of {{{21}}}-true


{{{7}}} years later the sum of their united ages will be {{{83}}} years.


{{{A=45}}} {{{7}}} years later will be {{{A=45+7=52}}}

{{{B=24}}} {{{7}}} years later will be {{{B=24+7=31}}}

and {{{A+B=52+31=83}}}-true