Question 949416
let's say Kei is {{{x}}} years old and Mira is {{{y}}} years old

if Kei is twice as old as Mira, then we have:

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

In three years, they will be {{{x+3}}} and {{{y+3}}} years old

the sum of their ages will be
{{{(x+3)+(y+3)}}} 
 
four times as much as Kei’s age {{{6}}} years ago, {{{4(x-6)}}}

so, the sum is:

 {{{(x+3)+(y+3)=4(x-6)}}} ......eq.2

substitute {{{x}}} from eq.1

{{{(2y+3)+(y+3)=4(2y-6)}}} ......eq.2....solve for {{{y}}}

{{{2y+3+y+3=8y-24}}} 

{{{3y+6=8y-24}}} 

{{{24+6=8y-3y}}} 

{{{30=5y}}} 

{{{30/5=y}}} 

{{{highlight(y=6)}}} 

Mira is {{{highlight(6)}}} years old.

and Kei is {{{highlight(12)}}} years old