Let the number of dimes be x
Let the number of quarters be y
Value Value
Type Number of of
of of EACH ALL
coin coins coin coins
-------------------------------------------
dimes x $0.10 $0.10x
quarters y $0.25 $0.25y
-------------------------------------------
TOTALS 23 ----- $4.55
The first equation comes from the second column.
x + y = 23
The second equation comes from the last column.
0.1x + 0.25y = 4.55
Get rid of decimals by multiplying every term by 100:
10x + 25y = 455
So we have the system of equations:
.
We solve by substitution. Solve the first equation for y:
x + y = 23
y = 23 - x
Substitute (23 - x) for y in 10x + 25y = 455
10x + 25(23 - x) = 455
10x + 575 - 25x = 455
-15x + 575 = 455
-15x = -120
x = 8 = the number of dimes.
Substitute in y = 23 - x
y = 23 - (8
y = 15 quarters.
The number of quarters is 23-x or 23-8 or 15 quarters.
Checking: 8 dimes is $0.80 and 15 quarters is $3.75
That's 23 coins.
And indeed $0.80 + $3.75 = $4.55
So it checks. 8 dimes and 15 quarters
Edwin