A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 4 years ago

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?

  • This Question is Closed
  1. lgbasallote
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  2. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thank you for being so specific! :)

  3. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.