Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

kissme Group Title

8x-y+5z from the sum of 2x+3y - 6z and 5x-3y

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. tishara429 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    13331

    • 2 years ago
  2. kissme Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    pls complete solution :)

    • 2 years ago
  3. tishara429 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    6971

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

    the sum of 2x+3y - 6z and 5x-3y = (2x + 3y -6z) + (5x - 3y) = 2x + 5x +3y - 3y -6z = 7x - 6z To get 8x-y+5z from the sum of 2x+3y - 6z and 5x-3y, we need to add something. Let that something be A [ (2x + 3y -6z) + (5x - 3y) ] + A = 8x-y+5z ( 7x - 6z ) + A = 8x - y + 5z Subtract both sides by 7x ( 7x - 6z ) + A - 7x = 8x - y + 5z -7x A - 6z = x - y + 5z Add 6z to both sides A - 6z + 6z = x -y + 5z + 6z A = x - y + 11z So, to get 8x-y+5z from the sum of 2x+3y - 6z and 5x-3y, we need to add x - y + 11z.

    • 2 years ago
  5. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I hope I didn't misinterpret the question :(

    • 2 years ago
  6. tishara429 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

    • 2 years ago
  7. tishara429 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    you did fine

    • 2 years ago
  8. kissme Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    THANKS

    • 2 years ago
  9. kissme Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so whats the answer..?

    • 2 years ago
  10. tishara429 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    15

    • 2 years ago
  11. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @tishara429 Can you explain your answer? I don't understand how you got your answer.

    • 2 years ago
  12. tishara429 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    add

    • 2 years ago
  13. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Do you mind explaining it detailedly? Sorry that I'm slow!

    • 2 years ago
  14. tishara429 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ok you add the number which equals the answer u got and then you will have the correct answer so thats how you solve the question.

    • 2 years ago
  15. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    But there are some unknowns there, you can't add them directly because they are NOT the like terms, right?

    • 2 years ago
  16. tishara429 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    oh i see then u have to divide

    • 2 years ago
    • Attachments:

See more questions >>>

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.