Joe has a collection of nickels and dimes that is worth $3.95. If the number of dimes was doubled and the number of nickels was increased by 21, the value of the coins would be $8.40. How many nickels and dimes does he have?

Question

anonymous · Answer

will this one disappear too?

akshay_budhkar · Answer

lol

anonymous · Answer

$$10x+5y=395$$
$$20x+5(y+21)=840$$

anonymous · Answer

do you know how to solve the system?

anonymous · Answer

no

anonymous · Answer

not you silly !

anonymous · Answer

i would rewrite 
$$20x+5(y+21)=840$$ as 
$$20x+5y=365$$ first

anonymous · Answer

no i don't

anonymous · Answer

satellite73 ...step by step explanation ... use elimination ...easier

anonymous · Answer

then subtract the first equation from the second 
$$20x+5y=735$$
$$10x+5y=395$$ subtract to get 
$$10x=735-395=340$$ making 
$$x=34$$

anonymous · Answer

is that the anwser?

anonymous · Answer

?

anonymous · Answer

if you are noot sure ask for more help in the chat

anonymous · Answer

x=34 means there are 34 dimes. Now, you need to find y to find number of nickels.

anonymous · Answer

10x+5y=395
x = 34
=> 340 + 5y = 395
=> 5y = 395-340
=> 5y = 55
=> y = 11

So, 34 dimes and 11 nickels.

anonymous · Answer

{5 n + 10 d = 395, 5 (n + 21) + 10 (2 d) = 840}
{n = 11, d = 34}