1) My coin collection is composed of pennies, dimes, nickels and quarters. There are five less
pennies than there are dimes. The number of nickels is three more than twice as many
pennies. The number of quarters is one less than two times the number of dimes. If the total
value of the coins is $7.87, how many of each type of coin are in my collection?
Let ! x = the number of dimes
!
x – 5 = the number of pennies
!
2(x – 5) + 3 = 2x – 10 + 3 = 2x – 7 = the number of nickels
!
2x – 1 = the number of quarters
So...
!
!
!
! 10(x) = the value of the dimes
1(x – 5) = the value of the pennies
5(2x – 7) = the value of the nickels
25(2x – 1) = the value of the quarters
10x + x − 5 + 5(2x − 7) + 25(2x − 1) = 787
10x + x − 5 + 10x − 35 + 50x − 25 = 787
71x − 65 = 787
Equation:
+ 65 + 65
71x = 852
71x 852
=
71
71
x = 12
So...!
!
!
!
x = the number of dimes = 12
x – 5 = the number of pennies = 12 – 5 = 7
2(x – 5) + 3 = 2x – 10 + 3 = 2x – 7 = the number of nickels = 2(12) – 7 = 24 – 7 = 17
2x – 1 = the number of quarters = 2(12) – 1 = 24 – 1 = 23
Therefore: There are 12 dimes, 7 pennies, 17 nickels, and 23 quarters.
2) My brother has a bowl full of coins. In the bowl are pennies, dimes, nickels and quarters.
There are two less dimes than there are pennies. There are twice as many nickels as there
are dimes. There are three more quarters than twice the number of pennies. If the total
value of the coins is $9.58, how many of each type of coin are in my brother’s coin bowl?
Let ! x = the number of pennies
!
x – 2 = the number of dimes
!
2(x – 2) = 2x – 4 = the number of nickels
!
2x + 3 = the number of quarters
So...
!
!
!
! 1(x) = x = the value of the pennies
10(x – 2) = the value of the dimes
5(2x – 4) = the value of the nickels
25(2x + 3) = the value of the quarters
x + 10(x − 2) + 5(2x − 4) + 25(2x + 3) = 958
x + 10x − 20 + 10x − 20 + 50x + 75 = 958
71x + 35 = 958
Equation:
− 35 − 35
71x = 923
71x 923
=
71
71
x = 13
So...!
!
!
!
x = the number of pennies = 13
x – 2 = the number of dimes = 13 – 2 = 11
2(x – 2) = 2x – 4 = the number of nickels = 2(13) – 4 = 26 – 4 = 22
2x + 3 = the number of quarters = 2(13) + 3 = 26 + 3 = 29
Therefore: There are 13 pennies, 11 dimes, 22 nickels, and 29 quarters.
3) I was in a contest where I had to gather coins from the bottom of a pool. I gathered half
dollars, quarters, and nickels. In the end I had 7 more nickels than I had half dollars. I had
three times the number of quarters than I had half dollars. Altogether I gathered $8.15. How
many of each coin did I gather?
Let ! x = the number of half dollars
!
x + 7 = the number of nickels
!
3x = the number of quarters
So... ! 50x = the value of the half dollars
!
5(x + 7) = the value of the nickels
!
25(3x) = the value of the quarters
50x + 5(x + 7) + 25(3x) = 815
50x + 5x + 35 + 75x = 815
130x + 35 = 815
Equation:
− 35 − 35
130x = 780
130x 780
=
130 130
x=6
So...! x = the number of half dollars = 6
!
x + 7 = the number of nickels = 6 + 7 = 13
!
3x = the number of quarters = 3(6) = 18
Therefore: There are 16 half dollars, 13 nickels and 18 quarters.
Procedure
Justification
−x
− 10 = 4(3x + 5)
3
Given equation
−x
− 10 = 12x + 20
3
−x
− 10 + 10 = 12x + 20 + 10
3
−x
+ 0 = 12x + 30
3
−x
= 12x + 30
3
( −3)
−x
= (12x + 30 ) ( −3)
3
? Distributive Property
? Equality Property of Addition
Inverse Property of Addition
? Identity Property
? Equality Property
1x = −3(12x + 30 )
? Inverse Property
1x = −36x − 90
Distributive Property
x = −36x − 90
? Identity Property
36x + x = −36x − 90 + 36x
37x = 0 − 90
Equality Property of Addition
? Inverse Property (and Combining Like
Terms)
37x = −90
Identity Property of Addition
37x −90
=
37
37
? Equality Property
1x =
−90
37
? Inverse Property
x=
−90
37
? Identity Property
x = −2
16
37
Identity Property/Simplifying