Results 1 to 7 of 7

Math Help - solving this equation.

  1. #1
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517

    solving this equation.

    y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})

    any ideas, just confused with the gradient
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    11
    Solve for y or x? This is a recent question that belongs to other thread, so I'll assume that you want to isolate y:

    y-2=2\sqrt{3}\left( x-\frac{\pi }{3} \right)\implies y=2\sqrt{3}x-\frac{2\sqrt{3}}{3}\pi +2=2\sqrt{3}x-\frac{2}{\sqrt{3}}\pi +2.

    Does that make sense?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by jvignacio View Post
    y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})

    any ideas, just confused with the gradient
    An equation of a line is in the form y = mx + c

    Where:
    • y, x = Variables
    • m = Gradient
    • c = y-intercept
    y - 2 = 2\sqrt{3}\left(x-\frac{\pi}{3}\right)
    y-2 = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3}
    y= 2\sqrt{3}x -  \frac{2\pi\sqrt{3}}{3} + 2
    y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}

    ^ Compare this form with y = mx + c to work out what your gradiant \left(m\right) is.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    Quote Originally Posted by Air View Post
    An equation of a line is in the form y = mx + c


    Where:
    • y, x = Variables
    • m = Gradient
    • c = y-intercept
    y - 2 = 2\sqrt{3}\left(x-\frac{\pi}{3}\right)
    y-2 = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3}
    y= 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + 2
    y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}

    ^ Compare this form with y = mx + c to work out what your gradiant \left(m\right) is.

    how did u get +6 on the numerator at the end?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    Quote Originally Posted by Krizalid View Post
    Solve for y or x? This is a recent question that belongs to other thread, so I'll assume that you want to isolate y:

    y-2=2\sqrt{3}\left( x-\frac{\pi }{3} \right)\implies y=2\sqrt{3}x-\frac{2\sqrt{3}}{3}\pi +2=2\sqrt{3}x-\frac{2}{\sqrt{3}}\pi +2.

    Does that make sense?
    yeah it was for solving y!
    makes sence now. gracias mate
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by jvignacio View Post
    how did u get +6 on the numerator at the end?
    y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}

    Is...

    y = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + \frac{6}{3}

    Which is...

    y = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + 2

    By keeping it as y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}, I though it would be easier for you to compare the form with y=mx+c.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    Quote Originally Posted by Air View Post
    y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}

    Is...

    y = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + \frac{6}{3}

    Which is...

    y = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + 2

    By keeping it as y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}, I though it would be easier for you to compare the form with y=mx+c.
    yes yes now i understand, cheers man!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. HELP on solving this equation for Q
    Posted in the Algebra Forum
    Replies: 6
    Last Post: September 19th 2010, 04:39 PM
  2. Solving an equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 30th 2009, 04:31 AM
  3. Need help solving this equation
    Posted in the Trigonometry Forum
    Replies: 0
    Last Post: May 28th 2009, 05:08 AM
  4. solving diff equation w/ bournoulli's equation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 14th 2008, 06:06 PM
  5. solving an equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: August 30th 2007, 07:03 PM

Search Tags


/mathhelpforum @mathhelpforum