# Ann had three less than twice as many nickels as dimes. If the total value of his coins was \$1.45 how many....?

Ann had three less than twice as many nickels as dimes. If the total value of her coins was \$1.45 how many of each coin did she have?

Relevance

Let d = dimes

Let 2d-3 = nickles

10d + 5(2d-3) = 145

10d + 10d-15 = 145

20d-15 = 145

20d = 160

d = 8

2d-3 = 13

8 dimes and 13 nickles

• the sides of this concern are: sixteen dimes and a million nickel = 17 money 0 dimes and 33 nickels = 33 money the respond would desire to be someplace in between yet on the threshold of the 1st shrink. 15 dimes and 3 nickels = 18 money 21-18=3 So, 12 dimes and 9 nickels = 21 money is the respond.

• 145 = 5n + 10d

where n = nickels and d = dimes

also,

2d - 3 = n

Using substitution,

145 = 5(2d-3) +10d

distribute the 5

145 = 10d -15 + 10d

combine like terms and add 15

160 = 20d

divide by 20

d= 8

plug that into one of the formulas

n = 2 * 8 -3

n=13

There are 13 nickels and 8 dimes.

• 8 dimes 13 nickels