You can either do it with 1 unknown or with 2 unknowns.
I'll do it both ways.
With 1 unknown
Let the number of nickels be x
Then the number of quarters, using
ONE PART = TOTAL MINUS OTHER PART,
is 11-x.
Value Value
Type Number of of
of of EACH ALL
coin coins coin coins
-------------------------------------------
nickels x $0.05 $0.05x
quarters 11-x $0.25 $0.25(11-x)
-------------------------------------------
TOTALS 11 ----- $1.55
The equation comes from the column on the right
0.05x + 0.25(11-x) = 1.55
Get rid of decimals by multiplying every term by 100:
5x + 25(11-x) = 155
5x + 275 - 25x = 155
-20x + 275 = 155
-20x = -120
x = 6 = the number of nickels.
The number of quarters is 11-x or 11-6 or 5 quarters.
Checking: 6 nickels is $0.30 and 5 quarters is $1.25
That's 11 coins.
And indeed $0.30 + $1.25 = $1.55
---------------------
With two unknowns
Let the number of nickels be x
Let the number of quarters be y
Value Value
Type Number of of
of of EACH ALL
coin coins coin coins
-------------------------------------------
nickels x $0.05 $0.05x
quarters y $0.25 $0.25y
-------------------------------------------
TOTALS 11 ----- $1.55
The first equation comes from the second column.
x + y = 11
The second equation comes from the last column.
0.05x + 0.25y = 1.55
Get rid of decimals by multiplying every term by 100:
5x + 25y = 155
So we have the system of equations:
.
We solve by substitution. Solve the first equation for y:
x + y = 11
y = 11 - x
Substitute (11 - x) for y in 5x + 25y = 155
5x + 25(11 - x) = 155
5x + 275 - 25x = 155
-20x + 275 = 155
-20x = -120
x = 6 = the number of nickels.
Substitute in y = 11 - x
y = 11 - (6
y = 5 quarters.
The number of quarters is 11-x or 11-6 or 5 quarters.
Checking: 6 nickels is $0.30 and 5 quarters is $1.25
That's 11 coins.
And indeed $0.30 + $1.25 = $1.55dwin
Edwin