Results 1 to 10 of 10

Math Help - Algebra question

  1. #1
    Junior Member
    Joined
    Jul 2009
    Posts
    47

    Algebra question

    Solve for x:

    (x^2 - 4x + 4x - 16) - (x^2 - 2x - 3x + 6) = 0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by marie7 View Post
    Solve for x:

    (x^2 - 4x + 4x - 16) - (x^2 - 2x - 3x + 6) = 0
    Have you tried to simply that?

    Note that its the same as x^2-4x+4x-16-x^2+2x+3x-6=0.

    Can you try to simplify the left side first and then solve for x?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2009
    Posts
    47
    So it would be -5x -10 = 0 right? Then where would I go from there? Because the question I'm doing are quadratic equations and then there's this one which is random and doesn't have the format to do a quadratic calculation on it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2009
    Posts
    47
    -5x = 10
    x = -2 is the answer I've gotten.

    The original equation is [(x + 4) / (x +2)] - [(x - 3) / (x - 4)] = 0 and I simplified it to the equation I mentioned in my first post. -2 doesn't fit into the original equation. Maybe I've done something wrong.
    Last edited by marie7; July 31st 2009 at 05:38 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by marie7 View Post
    So it would be -5x -10 = 0 right? Then where would I go from there? Because the question I'm doing are quadratic equations and then there's this one which is random and doesn't have the format to do a quadratic calculation on it.
    I get 5x-22=0

    From there, solve it like you would solve a linear equation!

    Don't let this mess with you. Sometimes when you deal with this kind of stuff, you may not get a quadratic equation to solve!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jul 2009
    Posts
    47
    I thought -16 + 6 is -10? not -22
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by marie7 View Post
    I thought -16 + 6 is -10? not -22
    In the original expression, it was x^2+4x-4x-16-(x^2-2x-3x+6)=0\implies x^2-16-x^2+5x-6=0 \implies 5x-16-6=0\implies 5x-22=0

    So it should be x=-\frac{22}{5}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jul 2009
    Posts
    47
    I'll copy and paste this just in case you didn't see it in my above post:

    "The original equation is [(x + 4) / (x +2)] - [(x - 3) / (x - 4)] = 0 and I simplified it to the equation I mentioned in my first post. your answer doesn't fit into the original equation. Maybe I've done something wrong in my calculations."

    Note that I've just edited the first (x - 4) to (x +4) in the equation above. It was a typo, so sorry!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by marie7 View Post
    The original equation is [(x - 4) / (x +2)] - [(x - 3) / (x - 4)] = 0 and I simplified it to the equation I mentioned in my first post. -2 doesn't fit into the original equation. Maybe I've done something wrong.

    Note that I've just edited the first (x - 4) to (x +4) in the equation above. It was a typo, so sorry!

    If that's the case then, you should get

    (x-4)(x+4)-(x+2)(x-3)=0\implies x^2-16-(x^2-x-6)=0 \implies x^2-16-x^2+x+6=0\implies x-10=0\implies x=10
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Jul 2009
    Posts
    47
    Thank you very much for your help! *gives you cookies*
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Question on algebra
    Posted in the Algebra Forum
    Replies: 5
    Last Post: February 12th 2011, 07:46 AM
  2. An algebra question
    Posted in the Algebra Forum
    Replies: 7
    Last Post: June 17th 2010, 02:03 PM
  3. Algebra question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 31st 2010, 02:50 PM
  4. algebra 2 question
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 8th 2009, 08:20 PM
  5. Algebra question
    Posted in the Business Math Forum
    Replies: 1
    Last Post: July 31st 2005, 09:55 PM

Search Tags


/mathhelpforum @mathhelpforum