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John saved 65 coins that are nickels and dimes only. If he had $4.90 in total, how many nickels did he have?

_____nickels

Respuesta :

This is a classic algebra problem with 2 equations and 2 unknowns. Let N = #nickles and D=#dimes.

Then N+D=65 the number of coins he has.

Also 5N +10D=490 working the amount of money in pennies. Put equation 1 into 2 .

5N+10(65-N)=490 .

5N+650-10N=490 .

-5N=-160 .

N=32 and D=33

To check 32 + 33 = 65

160 + 330 = 490 (in 1c increments)