Question

You have 50 coin that worth $8. 25. You have dime, quarter and nickel. The number of quarter i 5 more than twice the number of nickel. How many dime, quarter and nickel do you have individually

Answer 1

Create a system of equations
d + q + n = 50
0.1d + 0.25q + 0.05n = 8.25
q = 2n + 5

Substitute q into both equations
d + (2n + 5) + n = 50
d + 3n + 5 = 50
d + 3n = 45

0.1d + 0.25(2n + 5) + 0.05n = 8.25
0.1d + 0.5n + 1.25 + 0.05n = 8.25
0.1d + 0.55n = 7

Solve for d to find n
d + 3n = 45
d = -3n + 45

0.1(-3n + 45) + 0.55n = 7
-0.3n + 4.5 + 0.55n = 7
0.25n = 2.5
n = 10

10 nickels

d + 3n = 45
d + 3(10) = 45
d + 30 = 45
d = 15

15 dimes

d + q + n = 50
15 + q + 10 = 50
q + 25 = 50
q = 25

25 quarters

Check:
0.1d + 0.25q + 0.05n = 8.25
0.1(15) + 0.25(25) + 0.05(10) = 8.25
1.5 + 6.25 + 0.5 = 8.25
8.25 = 8.25 :)