Results 1 to 3 of 3

Math Help - Different questions about Equations

  1. #1
    Junior Member Fnus's Avatar
    Joined
    Nov 2006
    Posts
    48

    Different questions about Equations

    Hi, this might be really simple, but I just don't get it, so I'd like some help ^^

    Question 1:
    'Part of a garden consist of a square lawn with a path 1.5 metres wide around its perimeter. If the lawn area is two-thirds of the total area, find the lenght of a side of the lawn.'

    Question 2:
    Solve the simultanious equations:
    xy= 42
    x - 4y = 17

    Question 3:
    Solve the equation:
    (I don't know how to write fractions, but it's fractions..)

    1/x + 2 + 1/ x^2 - 4 = 2/5

    Help is very appreciated. ^^
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member TriKri's Avatar
    Joined
    Nov 2006
    Posts
    358
    Thanks
    1
    Quote Originally Posted by Fnus View Post
    Hi, this might be really simple, but I just don't get it, so I'd like some help ^^

    Question 1:
    'Part of a garden consist of a square lawn with a path 1.5 metres wide around its perimeter. If the lawn area is two-thirds of the total area, find the lenght of a side of the lawn.'

    Question 2:
    Solve the simultanious equations:
    xy= 42
    x - 4y = 17

    Question 3:
    Solve the equation:
    (I don't know how to write fractions, but it's fractions..)

    1/x + 2 + 1/ x^2 - 4 = 2/5

    Help is very appreciated. ^^

    The fisrt problem is answerd by setting up solving the equation x^2 = \frac{2}{3} * (x + 1.5 m)^2. It's a second order equation.

    To solv the second problem, you start this way:

    Allway isolate one of the unknown variables so you get an equation for it. In this case, you can isolate y by dividing the first equation by x. You'll get
    y= 42/x
    which also tells you that x isn't zero.
    By insertion into the second equation you'll get
    x - 4*42/x = 17
    x^2 - 84 = 17x
    That's a second order equation too. But I guess you can see that stuff for yourself
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member TriKri's Avatar
    Joined
    Nov 2006
    Posts
    358
    Thanks
    1
    The fisrt problem is answerd by setting up solving the equation x^2 = \frac{2}{3} * (x + 1.5 m)^2. It's a second order equation.

    To solv the second problem, you start this way:

    Allway isolate one of the unknown variables so you get an equation for it. In this case, you can isolate y by dividing the first equation by x. You'll get
    y= 42/x
    which also tells you that x isn't zero.
    By insertion into the second equation you'll get
    x - 4*42/x = 17
    x^2 - 84 = 17x
    That's a second order equation too. But I guess you can see that stuff for yourself

    The third, multiply the equation by x^2 and you'll get a second order equation here too.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Indices Equations Questions
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 10th 2010, 11:30 AM
  2. A few questions regarding Linear Equations.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 30th 2009, 08:17 PM
  3. Differential Equations questions
    Posted in the Differential Equations Forum
    Replies: 8
    Last Post: June 26th 2009, 07:47 AM
  4. Differential Equations (3 questions)
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: December 23rd 2008, 11:24 PM
  5. Finding Equations. 3 questions actually.
    Posted in the Geometry Forum
    Replies: 1
    Last Post: December 12th 2008, 05:22 AM

Search Tags


/mathhelpforum @mathhelpforum