Question 36424
Let the dimes be x
Let the nickels be y
Let the quarters be z
Equation 1 :
"twice the number of quarters is 13 more than the number of nickels"
2(z)=13+y
y=2z-13 (subsitution 1)
Equation 2:
x+y+z=210 (subsitute for y)
x+(2z-13)+z=210
x=210+13-2z-z
x=223-3z (subsitution 2)
Equation 3:
25z+5y+10x=2405
Subsitute for y:
25z+5(2z-13)+10x=2405
subsitute for x:
25z+10z-65+10(223-3z)=2405
35z-30z=2405-2230+65
5z=240
z=48
(subsitution 1)-- y=2(48)-13 
y=83
(subsitution 2)-- x=223-3(48)
x=79


Hence, there are 48 quarters, 79 dimes and 83 nickels.
Paul.