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