Question 822831
Quinn  is  twice  as  old  as  Karl  
<pre>
            Q = 2K
</pre>
Quinn is ... four  years  older  than  Parker
<pre>
            Q = P + 4
</pre>
  the  sum  of  their  ages  is  21.
<pre>
    Q + K + P = 21

So we have this system of equations:

            Q = 2K
            Q = P + 4
    Q + K + P = 21

Substitute 2K for Q in the 2nd and 3rd equations:

           2K = P + 4
   2K + K + P = 21

Simplify the second one

           2K = P + 4
       3K + P = 21

Solve the first one for P

       2K - 4 = P

Substitute in 3K + P = 21

3K + (2K - 4) = 21
  3K + 2K - 4 = 21
           5K = 25
            K = 5

Substitute 5 for K in 2K - 4 = P

     2(5) - 4 = P
       10 - 4 = P
            6 = P    <----answer!

Substitute  6 for P in Q = P + 4
 
            Q = 6 + 4
            Q = 10

Checking:

Quinn, who is 10 is twice as old as Karl, who is 5, and Quinn, who is 10,
is also four years older than Parker, who is 6. The sum of their ages is 21.
Q + P + K = 10 + 6 + 5 = 21.

So it checks.

Edwin</pre>