Results 1 to 2 of 2

Thread: Hi, rather easy systems of equations problem that I would love help on

  1. February 11th 2009, 11:48 AM #1
    lahoo
    lahoo is offline
    Newbie
    Joined
    Feb 2009
    Posts
    1

    Hi, rather easy systems of equations problem that I would love help on

    thanks for looking!



    Please say what equations you used. Thanks a lot, I love this site!
    Follow Math Help Forum on Facebook and Google+
  2. February 11th 2009, 12:58 PM #2
    Soroban
    Soroban is offline
    Super Member
    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,242
    Thanks
    370
    Hello, lahoo!

    Person X invested $10,000 in three accounts.
    . . Account A pays 4.5% interest.
    . . Account B pays 6.2% interest.
    . . Account C pays 3.5% interest.

    X decides to invest into account B double the sum of the other two accounts.
    When the first dividends came, they earned $549 interest.
    How much invested into each account?

    If we organize the information, everything is clearer . . .

    . . \begin{array}{|c|c|c||c|} \hline<br /> \text{Account} & \text{Amount} & \text{Rate} & \text{Interest} \\ \hline\hline<br /> A & x & 0.045 & 0.045x \\ \hline<br /> B & 2x+2y & 0.062 & 0.062(2x+2y) \\ \hline<br /> C & y & 0.035 & 0.035y \\ \hline\hline<br /> \text{Total} & 10,\!000 & & 549 \\ \hline \end{array}


    In the second column, we have: . x + (2x + 2y) + y \:=\:10,\!000 \quad\Rightarrow\quad 3x + 3y \:=\:10,\!000\;\;{\color{blue}[1]}


    In the last column: . 0.045x + 0.062(2x+2y) + 0.35y \:=\:549 \quad\Rightarrow\quad 0.169x + 0.159y \:=\:549

    Multiply by 1000: . 169x + 159y \:=\:549,\!000\;\;{\color{blue}[2]}


    Solve the system of equations: . \begin{array}{cccc}3x + 3y &=& 10,\!000 & {\color{blue}[1]} \\<br /> 169x + 159y &=& 549,\!000 & {\color{blue}[2]} \end{array}


    . . \begin{array}{cccccc}\text{Multiply }{\color{blue}[1]}\text{ by -53:} & \text{-}159x - 159y &=& \text{-}530,\!000 \\<br /> \text{Add }{\color{blue}[2]}\!: & 169x + 159y &=& 549,\!000 \end{array}

    And we get: . 10x \:=\:19,\!000 \quad\Rightarrow\quad x \:=\:\boxed{\$1900} in account A.


    Substitute into [1]: . 3(1900) + 3y \:=\:10,\!000 \quad\Rightarrow\quad y \:=\:\frac{4300}{3} \:=\:\boxed{\$1433.33} in account C.


    Then: . 2(1900 + 1433.33) \;=\;\boxed{\$6666.67} in account B.
    Follow Math Help Forum on Facebook and Google+
« Finding Nth Term | find missing integer-algebraic division »

Similar Math Help Forum Discussions

  1. Another word problem (systems of linear equations)
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 10th 2010, 03:22 PM
  2. Systems of Equations Story Problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 6th 2009, 06:19 PM
  3. Systems of Equations word problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 25th 2009, 05:39 AM
  4. Systems of Equations application problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 19th 2009, 10:21 PM
  5. systems of equations (word problem)
    Posted in the Algebra Forum
    Replies: 6
    Last Post: January 27th 2008, 03:58 PM

Search Tags


/mathhelpforum @mathhelpforum