Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ineedbiohelpandquick

  • 3 years ago

solving by elimination 1) 2x-y=3 4x+3y=21 when i solve it i get x=1.2 and y= -.6 but when i check it back in to the equation it is wrong will someone help me!

  • This Question is Closed
  1. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    2x-y=3 4x+3y=21 3*(2x-y)=3*3 4x+3y=21 6x-3y=9 4x+3y=21 ----------- 10x = 30 10x = 30 x = 30/10 x = 3 Now that we know that x = 3, we can use it to find y 2x-y = 3 2(3)-y = 3 6 - y = 3 -y = 3-6 -y = -3 y = 3 So x = 3 and y = 3

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

    thank you so much! would you be willing to explain this one? x-y=2 -2x+2y=5

  3. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    x-y=2 -2x+2y=5 2(x-y)=2*2 -2x+2y=5 2x-2y=4 -2x+2y=5 ----------- 0x+0y = 9 0 = 9 ... which is FALSE So there are no solutions. This means that the system is inconsistent.

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

    okay the zero threw me off when i had done it! do you know how to solve by using back substitution by chance?

  5. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Solving by substitution: x-y=2 x = 2 + y ------------------------------------------ -2x+2y = 5 -2(2+y) + 2y = 5 -4 - 2y + 2y = 5 -4 = 5 ... which is FALSE So again, the system is inconsistent and there are no solutions.

  6. ineedbiohelpandquick
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    here is one... 2x-y+5z=24 y+2z=4 z=6

  7. ineedbiohelpandquick
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for using back-substitution

  8. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    y+2z=4 y+2(6)=4 y + 12 = 4 y = 4-12 y = -8 ------------------------------------------------------- 2x-y+5z = 24 2x - (-8) + 5(6) = 24 2x + 8 + 30 = 24 2x + 38 = 24 2x = 24-38 2x = -14 x = -14/2 x = -7 So the solutions are x = -7, y = -8 and z = 6

  9. ineedbiohelpandquick
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you are incredible thank you!

  10. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You're welcome

  11. ineedbiohelpandquick
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hey are you still on here? can you help me with solve using row-operations... x+y+z=6 2x-y+z=3 3x -z=0

  12. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    x+y+z=6 2x-y+z=3 3x -z=0 1 1 1 6 2 -1 1 3 3 0 -1 0 1 1 1 6 2 -1 1 3 0 -3 -4 -18 R3 - 3R1 1 1 1 6 0 -3 -1 -9 R2 - 2R1 0 -3 -4 -18 1 1 1 6 0 1 1/3 3 (-1/3)*R2 0 -3 -4 -18 1 1 1 6 0 1 1/3 3 0 0 -3 -9 R3 + 3R2 1 1 1 6 0 1 1/3 3 0 0 1 3 (-1/3)*R3 x+y+z = 6 y+(1/3)z = 3 z = 3 ============================================= ============================================= y+(1/3)z = 3 y+(1/3)(3) = 3 y + 1 = 3 y = 3-1 y = 2 ---------------- x+y+z = 6 x+2+3 = 6 x + 5 = 6 x = 1 So the solutions are x = 1, y = 2 and z = 3

  13. ineedbiohelpandquick
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    May i ask you what the R stands for

  14. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    R stands for the row So R1 means "row 1", R2 means "row 2", and R3 means "row 3"

  15. ineedbiohelpandquick
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay so how did you determine subtract row 3- 3 by row1

  16. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    When going from 1 1 1 6 2 -1 1 3 3 0 -1 0 to 1 1 1 6 2 -1 1 3 0 -3 -4 -18 R3 - 3R1 I did this to make that lower left corner value go from 3 to 0 Notice that 3 - 3*1 = 3-3 = 0 This basically "eliminates" the x variable in equation 3 which gets me one step closer to solving for x, y, and z

  17. ineedbiohelpandquick
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay can you tell me what you get for this one then... 6y+4z=-18 3x+3y = 9 2x -z =12

  18. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    using the same method of row reduction or some other method?

  19. ineedbiohelpandquick
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    row reduction :)

  20. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    alright thanks, one sec

  21. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    6y+4z=-18 3x+3y = 9 2x -z =12 0 6 4 -18 3 3 0 9 2 0 -1 12 0 6 4 -18 1 1 0 3 (1/3)*R2 2 0 -1 12 1 1 0 3 R1 <--> R2 0 6 4 -18 2 0 -1 12 1 1 0 3 0 6 4 -18 0 -2 -1 6 R3 - 2*R1 1 1 0 3 0 -2 -1 6 R2 <--> R3 0 6 4 -18 1 1 0 3 0 1 1/2 -3 (-1/2)*R2 0 6 4 -18 1 1 0 3 0 1 1/2 -3 0 0 1 0 R3 - 6*R2 x+y = 3 y + (1/2)z = -3 z = 0 ------------------------------------------------------- y + (1/2)z = -3 y + (1/2)(0) = -3 y = -3 -------------------- x+y = 3 x + (-3) = 3 x - 3 = 3 x = 3+3 x = 6 So the solutions are x = 6, y = -3, and z = 0

  22. ineedbiohelpandquick
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    THANK YOU! You don't know how much you have helped me! there is only one more I can't seem to get right do you have time for one more? :)

  23. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sure whats the problem

  24. ineedbiohelpandquick
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    same thing so solving by row operations :)... its x-2y+5z=2 4x -z=0

  25. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    x-2y+5z=2 4x -z=0 1 -2 5 2 4 0 -1 0 1 -2 5 2 0 8 -21 -8 R2 - 4*R1 x-2y+5z = 2 8y-21z=-8 ------------------------------------------------------- 8y-21z=-8 8y = -8 + 21z y = (21z - 8)/8 ----------------------- x-2y+5z = 2 x-2( (21z - 8)/8 )+5z = 2 x - 4(21z - 8) + 5z = 2 x - 84z + 32 + 5z = 2 x - 79z + 32 = 2 x - 79z = 2-32 x - 79z = -30 x = 79z - 30 Now if you let z = s, where s is any number, then the solutions are x = 79s - 30, y = (21s - 8)/8, z = s This means that there are an infinite number of solutions. So the system is consistent and dependent.

  26. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    A very quick way to determine if you'll have a dependent system is noticing that there are more variables than equations.

  27. ineedbiohelpandquick
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so the answer is consistent and dependent or x = 79z - 30 sorry I am a little confused haha

  28. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Well both aspects make up the total answer. The idea here is that there are an infinite number of points where the two equations x-2y+5z=2 and 4x -z=0 intersect So this makes the system dependent (since one equation technically "depends" on the other -- it changes as the other one does) Because at least one solution exists, this makes the system consistent. The answers of x = 79s-30, etc... are the more explicit forms of the solutions and are more specific than saying "there are an infinite number of solutions" This is because you can't just say that something random like (1,2,3) is a solution even though I did say "there are an infinite number of solutions". All of the solutions fit a very specific algebraic form.

  29. ineedbiohelpandquick
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That makes sense! Thank you again! I appreciate all your help!

  30. jim_thompson5910
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You're welcome, glad it's all clicking.

  31. 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.