Let the price of a foot of redwood cost $r and a foot of pine cost $p
60r + 90p = $315
100r + 60p = $408
This forms simultaneous equation and solving it by substitution:
100(60r + 90p = $315) ... (1)
60(100r + 60p = $408) ... (2)
6000r + 9000p = $31500 ... (1)
6000r + 3600p = $24480 ... (2)
Subtracting (2) from (1):
5400p = $7020
p = $1.3
and r = ($315 - (90 x 1.3)) / 60 = $198/60 = $3.3
A foot of redwood costs $1.3
A foot of pine costs $3.3