Question

Heidi and Mike have 51 dimes and nickles. If the value of the coins is $4.10, how many coins of each type were there?

Answer 1

let x = no. of nickles
y = no. of dimes

x + y = 51
5(cents)x + 10(cents)y = 410 cents

multiply the first equation by -10..

-10x - 10y = -510
5x + 10y = 410

-5x = -100
x = 20
y = 31

There are 20 nickles and 31 dimes

Answer 2

thank you for being so specific! :)