Page 1 of 2 12 LastLast
Results 1 to 15 of 19

Math Help - Solving Systems

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    20

    Solving Systems

    How do I solve this using any algebraic method?

    3x-2y=10
    5x+3y=15

    2x-4y=-6
    -x+2y=3
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member integral's Avatar
    Joined
    Dec 2009
    From
    Arkansas
    Posts
    200
    solve for x on one equation and substitute it in the other
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,629
    Thanks
    430
    Quote Originally Posted by Vuong View Post
    How do I solve this using any algebraic method?

    3(3x-2y=10)
    2(5x+3y=15)

    2x-4y=-6
    2(-x+2y=3)
    solve using elimination ... distribute, add the equations and solve for the remaining variable. back substitute to find the other variable.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    3x-2y=10
    5x+3y=15

    From the second equation you know that :

    5x = 15 - 3y

    Thus 15x = 45 - 9y

    And so, 3x = 9 - 1.8y.

    Substituting back this result into the first equation gives :

    9 - 1.8y - 2y = 10

    9 - 3.8y = 10

    3.8y = -1

    y = \frac{-1}{3.8}

    Now substitute the value of y found into any of the equations to retrieve x.

    Can you do it for the second system now ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Vuong View Post

    3x-2y=10
    5x+3y=15

    You can also use the elimination method

    make

    3x-2y=10 ...(1)
    5x+3y=15 ...(2)

    To eliminate x take (1) \times 5 and (2) \times 3<br />

    gives

    15x-10y=50 ...(3)
    15x+9y=45 ...(4)

    Now taking (4) form (3) gives

    -19y = 5

    Can you take it from here?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2009
    Posts
    20
    Quote Originally Posted by Bacterius View Post
    3x-2y=10
    5x+3y=15

    From the second equation you know that :

    5x = 15 - 3y

    Thus 15x = 45 - 9y

    And so, 3x = 9 - 1.8y.

    Substituting back this result into the first equation gives :

    9 - 1.8y - 2y = 10

    9 - 3.8y = 10

    3.8y = -1

    y = \frac{-1}{3.8}

    Now substitute the value of y found into any of the equations to retrieve x.

    Can you do it for the second system now ?
    How did you get the 9 for 9 - 1.8y - 2y = 10?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2009
    Posts
    20
    Quote Originally Posted by pickslides View Post
    You can also use the elimination method

    make

    3x-2y=10 ...(1)
    5x+3y=15 ...(2)

    To eliminate x take (1) \times 5 and (2) \times 3<br />

    gives

    15x-10y=50 ...(3)
    15x+9y=45 ...(4)

    Now taking (4) form (3) gives

    -19y = 5

    Can you take it from here?
    y = -3.8 then plug it in?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    Quote Originally Posted by Vuong View Post
    How did you get the 9 for 9 - 1.8y - 2y = 10?
    Comes from the substitution of 3x = 9 - 1.8y.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Vuong View Post
    y = -3.8 then plug it in?

    To either (1),(2),(3) or (4) to find x.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Nov 2009
    Posts
    20
    Quote Originally Posted by pickslides View Post

    To either (1),(2),(3) or (4) to find x.
    Thanks, I got 5.86
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    That's not right, you should find something around x \approx 3.15
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Nov 2009
    Posts
    20
    Quote Originally Posted by Bacterius View Post
    That's not right, you should find something around x \approx 3.15
    I plugged -3.8 into 3x-2y=10 x_x
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Vuong View Post
    y = -3.8 then plug it in?
    Are you sure this is correct?

    -19y = 5 \implies y = -\frac{5}{19} \approx -0.263
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Nov 2009
    Posts
    20
    Quote Originally Posted by pickslides View Post
    Are you sure this is correct?

    -19y = 5 \implies y = -\frac{5}{19} \approx -0.263
    So I plug -0.263 into one of them?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    Quote Originally Posted by Vuong View Post
    So I plug -0.263 into one of them?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Solving for 2 systems at once
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 5th 2011, 01:32 PM
  2. Solving Systems
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: October 14th 2009, 10:02 AM
  3. Solving systems
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 12th 2009, 07:43 PM
  4. Solving Systems
    Posted in the Advanced Algebra Forum
    Replies: 22
    Last Post: September 15th 2007, 11:53 PM
  5. Replies: 3
    Last Post: October 11th 2006, 09:15 PM

Search Tags


/mathhelpforum @mathhelpforum