Results 1 to 10 of 10

Math Help - Simultaneous Equations, 2 Problems.

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    3

    Simultaneous Equations, 2 Problems.

    Could someone please explain to me how to work out the following equations, I have tried various methods but can't come up with the correct answers.

    x+y=2
    2x-y=2

    5x=8y
    4x-3y+17=0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by henriklarsson1 View Post
    Could someone please explain to me how to work out the following equations, I have tried various methods but can't come up with the correct answers.

    x+y=2
    2x-y=2
    (x+y)+(2x-y)=2+2\Rightarrow\ x+2x+y-y=2+2


    5x=8y
    4x-3y+17=0
    Get equal coefficients for y

    5x=8y\Rightarrow\ 15x=24y

    4x-3y+17=0\Rightarrow\ 8(4x-3y+17)=4(0)=0

    Try the same technique as for the first part.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    Posts
    3
    Thanks for your help but in the back of my book it says equation 1 is x= 4/3 y= 2/3.

    I'll give you an example of one I know how to work out: 2x-y=5, x-2y=4. (2,-1)
    The method I have been taught is: 2(2x-y=5) x-2y=4
    4x-2y=10 x-2y=4
    3x=6
    x=2
    -y= 5-4
    y=-1 (sorry about the mess)

    I'm still unsure how to work out the previous ones using this method.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by henriklarsson1 View Post
    Thanks for your help but in the back of my book it says equation 1 is x= 4/3 y= 2/3.
    That is exactly the answer I get when using Archie's method!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,099
    Thanks
    67
    Quote Originally Posted by henriklarsson1 View Post
    Thanks for your help but in the back of my book it says equation 1 is x= 4/3 y= 2/3.

    I'll give you an example of one I know how to work out: 2x-y=5, x-2y=4. (2,-1)
    The method I have been taught is: 2(2x-y=5) x-2y=4
    4x-2y=10 x-2y=4
    3x=6
    x=2
    -y= 5-4
    y=-1 (sorry about the mess)

    I'm still unsure how to work out the previous ones using this method.
    That's known as "elimination"
    Your 1st problem:
    x+y=2
    2x-y=2
    Add the 2 equations:
    3x = 4 ; OK?

    Your 2nd problem:
    5x=8y
    4x-3y+17=0
    Rewrite:
    5x - 8y = 0 [1]
    4x - 3y = -17 [2]
    [1] * 4, [2] * 5:
    20x - 32y = 0
    20x - 15y = -85
    Finish it...

    If you can't, you need classroom help...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by henriklarsson1 View Post
    Thanks for your help but in the back of my book it says equation 1 is x= 4/3 y= 2/3.

    I'll give you an example of one I know how to work out: 2x-y=5, x-2y=4. (2,-1)
    The method I have been taught is:

    2(2x-y=5) x-2y=4

    4x-2y=10 x-2y=4

    3x=6

    x=2

    -y= 5-4

    y=-1

    I'm still unsure how to work out the previous ones using this method.
    It's very very similar.

    You should understand why the steps are being taken.
    It just makes the process so much easier.

    Let's just examine the steps of the example you gave that you know how to do...

    2x-y=5,\;\;\;x-2y=4

    Solve for x, so to set that up, get the same amount of y in both,
    hence multiply the first by 2...

    2x-y=5\Rightarrow\ 2(2x-y)=2(5)\Rightarrow\ 4x-2y=10

    Now we have

    4x-2y=10,\;\;\;x-2y=4

    therefore subtract them to remove y

    (4x-2y)-(x-2y)=10-4\Rightarrow\ 4x-x-2y--2y=6

    \Rightarrow\ 3x-2y+2y=6\Rightarrow\ 3x=6

    3x=3(2)\Rightarrow\ x=2

    2(2)-y=5\Rightarrow\ 4-y=5=4-(-1)\Rightarrow\ y=-1

    So that's why pickslides said it's the same thing.


    However, you can choose to add or subtract.
    You subtracted above because you had the same multiple of y.

    We can do that also for the questions

    x+y=2,\;\;\;2x-y=2\Rightarrow

    2(x+y)=2(2)\Rightarrow\ 2x+2y=4

    Subtract

    (2x+2y)-(2x-y)=4-2=2

    2x-2x+2y--y=2

    2y+y=2\Rightarrow\ 3y=2\Rightarrow\ y=\frac{2}{3}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2011
    Posts
    3
    Thanks guys, really appreciate your help!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    904
    Thanks
    27
    Thanks to Wilmer for simpler solutions particularly problem 1. There is a sign error in your solution to 2.x and y must have the same signs.


    bjh
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by bjhopper View Post
    Thanks to Wilmer for simpler solutions particularly problem 1. There is a sign error in your solution to 2.x and y must have the same signs.


    bjh
    The solution was correct, the signs are opposite.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    904
    Thanks
    27

    simultaneous equations

    Sorry for my mistake.I did this differently
    5x=8y
    4x-3y+17=0
    x=8y/5
    4*8y/5 -15y/5 =-17
    32y/5 - 15y /5 = -17
    17y/5 = -17
    y =-5
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simultaneous Equations 4 variables, 4 equations
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 7th 2011, 04:06 PM
  2. simultaneous equation word problems
    Posted in the Algebra Forum
    Replies: 5
    Last Post: February 17th 2010, 07:56 PM
  3. Replies: 3
    Last Post: February 27th 2009, 07:05 PM
  4. Replies: 2
    Last Post: August 7th 2008, 11:38 PM
  5. Simultaneous Equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 1st 2008, 10:35 AM

Search Tags


/mathhelpforum @mathhelpforum