Results 1 to 4 of 4

Math Help - urgent algebra help needed!

  1. #1
    Member
    Joined
    Nov 2008
    Posts
    87

    urgent algebra help needed!

    Linear Systems question - solve this system:

    1/x + 3/y = 3/4
    3/x - 2/y = 5/12

    Polynomials question:

    Divide 2x 1 into 4x4 5x2 + 2x + 4.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2008
    Posts
    135
    Quote Originally Posted by 14041471 View Post
    Linear Systems question - solve this system:

    1/x + 3/y = 3/4
    3/x - 2/y = 5/12

    Polynomials question:

    Divide 2x 1 into 4x4 5x2 + 2x + 4.
    For the first question, multiply the first equation by -3 and then add it to the second equation:

    -3(1/x + 3/y = 3/4)
    + 3/x - 2/y = 5/12

    -3/x - 9/y = -9/4
    +3/x - 2/y = 5/12 Find a common denominator for the right hand sides of the equations:

    -3/x - 9/y = -27/12
    +3/x - 2/y = 5/12 Multiplied the right hand side of the top equation by 3/3, which is essentially multiplying by 1.

    Now add them:
    -11/y = -22/12

    Cancel out the negatives and cross multiply to get:
    12*11 = 22*y
    132 = 22y
    y = 6

    Substitute this value back into this original equation:
    3/x - 2/y = 5/12 to find x.

    3/x - 2/6 = 5/12
    3/x = 5/12 + 4/12 got a common denominator for the constants
    3/x = 9/12

    Cross Multiply:
    3*12 = 9*x
    36 = 9x
    x = 4

    For practice, Substitute the x and y values back into both of the original equations to show that they check out.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by 14041471 View Post
    Linear Systems question - solve this system:

    1/x + 3/y = 3/4
    3/x - 2/y = 5/12

    Polynomials question:

    Divide 2x 1 into 4x4 5x2 + 2x + 4.
    Multiply the first equation by 2, the second equation by 3:

    1/x + 3/y = 3/4 ==> 2/x + 6/y = 6/4

    3/x - 2/y = 5/12 ==> 9/x-6/y = 15/12

    Add columnwise and you'll get 11/x = 11/4 ==> x = 4

    Re-substitute and calculate y: y = 6
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2008
    Posts
    135

    2nd question

    Quote Originally Posted by 14041471 View Post
    Linear Systems question - solve this system:

    1/x + 3/y = 3/4
    3/x - 2/y = 5/12

    Polynomials question:

    Divide 2x 1 into 4x4 5x2 + 2x + 4.
    For the polynomials question, you start by asking yourself how many times 2x can go into 4x^4. Well, it would be 2x^3, because 2x*2x^3 = 4x^4.

    So you multiply (2x-1) by 2x^3 and subtract it from the entire above polynomial: Looks like:

    2x^3(2x-1) = 4x^4 - 2x^3

    Now, do: (4x^4 - 5x^2 + 2x + 4) - (4x^4 - 2x^3)
    The polynomial now looks like:
    ii) 2x^3 - 5x^2 + 2x + 4

    We do the same thing as above, ask ourselves how many times 2x can go into 2x^3. it would be x^2 times, since 2x*x^2 = 2x^3

    So you multiply (2x-1) by x^2 and subtract it from the second stage polynomial (ii) above:

    (x^2)(2x-1) = 2x^3 - x^2

    Now, do (2x^3 - 5x^2 + 2x + 4) - (2x^3 - x^2)
    The polynomial now looks like:
    (iii) -4x^2 + 2x + 4

    We do the same, and se that 2x can go into -4x^2 a number of -2x times because 2x*-2x = -4x^2.

    So, you multiply (2x-1) by -2x and subtract it from the third stage polynomial (iii)

    -2x(2x - 1) = -4x^2 + 2x

    Now, do (-4x^2 + 2x + 4) - (-4x^2 + 2x)
    The polynomial now looks like:
    (iv) 4

    You can't divide 2x into 4 any further, so we are done. The ending form looks like:

    (2x-1)(2x^3 + x^2 -2x) + 4

    Expand this to check to see if you get the original polynomial at the beginning. Hope this helps
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Urgent help needed please
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: December 23rd 2008, 03:38 PM
  2. algebra urgent help needed
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 25th 2008, 05:47 AM
  3. Urgent Algebra homework help needed
    Posted in the Algebra Forum
    Replies: 5
    Last Post: May 2nd 2007, 09:43 AM
  4. Urgent Help Needed
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 27th 2006, 12:15 PM
  5. Urgent help needed with linear algebra problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: June 13th 2005, 10:32 AM

Search Tags


/mathhelpforum @mathhelpforum