Please help with the world Problems. THank you so much

Question

demar has 68 coins in his pockets: all pennies, nickels, and dimes. the total value of the coins is $4.33. If the number of dimes is one less than twice the number of nickels , how many coins of each type does he have?

Answer 1

A substitution is applicable here

For this one you can set dimes up in terms of nickels since Dimes = 2(nickels) -1

Since you know the total number of coins 68, then 68 minus Dimes + Nickels is pennies.

Together the pennies, nickels and dimes all add up to $4.33

Let Nickels = x

Let Dimes = 2x -1

Note: Nickels + Dimes = x + 2x -1 = 3x -1

Let Pennies = 68 - (3x -1) = 68 - 3x +1= 69 -3x

Express the total cost 4.33 as a combination of coins. The coins can be expressed as values multiplied by their respective quantities.

0.05x + 0.10(2x -1) + 0.01(69 -3x)

0.05x +0.20x - 0.10 + .69 - .03x = 4.33

Combine like terms

0.20x + 0.05x - 0.03x + 0.69 - 0.10 = 4.33

0.22x + 0.59 = 4.33

Subtract 0.59 from both sides of the equation

0.22x = 3.74

Divide both sides by 0.22 to get x

x = 3.74/0.22

x = 17

So we have 17 nickels

2(17) - 1 = dimes

33 dimes

and 68-(3(17) -1) pennies

68-(51 -1)

68 - 50 = 18 pennies

We can check by counting our money

0.05(17) + 0.10(33) + 0.01(18) = 4.33

0.85 + 3.30 + 0.18 = 4.33

I hope you find this useful and please leave a message if you have any questions.