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Math Help - Algebra problem NEED HELP!

  1. #1
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    Algebra problem NEED HELP!

    Can anyone help me with these two algebra problems??

    1.) A broker made 2 seperate investments on behalf of a client. The first investment earned 12% and the second lost 5% for a total gain of $1040. If the amounts in these two investments had been switched then the result would have been a gain of $360. Determine how much was actually invested at each rate

    2.) While it is a common knowledge that Martians are green and have seven heads, few people realize that a Martian has eleven eyes. At a recent party, while I was conversing with a Martian, I innocently remarked that if one-half of the Martians at the party suddenly became Earthlings (while the other half remained Martians), there would be 295 eyes at the party. Somewhat offended, the Martian noted that if one-half of theEarthlings at the party became Martians (while the other half stayed Earthlings), there would be 410 heads at the party. How many Martians and how many Earthlings were at
    the party?

    ANS: This is what me and my friend worked out, but the red bolded parts is the parts i dont know how my friend go it.

    let x = number of martians
    y = number of earthling

    eyes
    295 = (x/2)*11 + (x/2)*2 + 2y <----where did those numbers come from
    295 = 13x/2 + 2y
    590 = 13x + 4y

    heads
    410 = (y/2)*1 + (y/2)*7 + 7x <----and these numbers??
    410 = 4y + 7x
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  2. #2
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    Hello, lemontea!

    1) A broker made 2 seperate investments on behalf of a client.
    The first investment earned 12% and the second lost 5% for a net gain of $1040.
    If the amounts in these two investments had been switched,
    then the result would have been a net gain of $360.
    Determine how much was actually invested at each rate

    Let x = amount invested in plan A.
    Let y = amount inveted in plan B.
    . . The net gain was $1040.

    Then we have: . {\color{red}0.12x - 0.05y \:=\:1040}

    If the amounts are switched, we have:
    y = amount invested in plan A.
    x = amount invested in plan B.
    . . The net gain was $360.

    Then we have: . {\color{red}0.12y - 0.05x \:=\:360}


    Multiply the equations by 100.

    Then we have: . {\color{blue}\begin{array}{ccc}12x - 5y & = & 104,000 \\ -5x + 12y & = & 36,000\end{array}}

    . . and we can solve the system of equations.


    I got: . \boxed{\begin{array}{ccc}x & = & \$12,000 \\ y & = & \$8,000\end{array}}


    Edit: I made a terrible blunder in my arithmetic . . . sorry!
    Last edited by Soroban; October 23rd 2007 at 05:32 AM.
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  3. #3
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    Hello again, lemontea

    2) While it is a common knowledge that Martians are green and have 7 heads, few people
    realize that a Martian has 11 eyes. At a recent party, while I was conversing with a Martian,
    I innocently remarked that if one-half of the Martians became Earthlings (while the other
    half remained Martians), there would be 295 eyes at the party. Somewhat offended, the
    Martian noted that if one-half of the Earthlings at the party became Martians (while the
    other half stayed Earthlings), there would be 410 heads at the party.
    How many Martians and how many Earthlings were at the party?
    I'll modify your solution slightly . . .

    Let . \begin{array}{ccc}M & = &\text{number of Martians} \\ E & = & \text{number of Earthlings}\end{array}


    Eyes
    . . .
    normal Martians . . . . . . . . Normal
    . . . .
    with 11 eyes . . . . . . . . .Earthlings
    295 \:= \:\overbrace{\left(\frac{M}{2}\right)\cdot 11}  \;+ \;\underbrace{\left(\frac{M}{2}\right)\cdot 2} \;+ \;\overbrace{2E}
    . . . . . . . . . . . .
    Martians with
    . . . . . . . . . . . . .
    only 2 eyes

    . . 295 \:= \:\frac{13}{2}M + 2E\quad\Rightarrow\quad{\color{blue}13M + 4E \:=\:590}



    Heads
    . . . .
    Normal . . . . . . . Earthlings
    . . . .
    Martians. . . . . . .w/ 7 heads
    410 \:= \:\overbrace{7M} + \underbrace{\left(\frac{E}{2}\right)\cdot1} + \overbrace{\left(\frac{E}{2}\right)\cdot7}
    . . . . . . . . .
    Normal
    . . . . . . . .
    Earthlings

    . . {\color{blue}7M + 4E \:=\:410}



    Solve the system of equations: . \begin{Bmatrix}M & = & 30 \\ E & = & 50\end{Bmatrix}

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  4. #4
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    Thanks for explaining it soo well..i just got a small question..for the first question...

    After this step:



    Do I do substitution or elimination..I tried substitution but the answer was different than yours...elimination would be kinda hard to do....
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  5. #5
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    Hello, lemontea!

    Do I do substitution or elimination? . . . . Use your favorite method.

    I tried substitution but the answer was different than yours.
    . .
    My answers were wrong (*blush*) . . . I've corrected them.

    Elimination would be kinda hard to do. .
    . . . It's not too bad.
    We have: . \begin{array}{cccc}12x - 5y & = & 104,000 & {\color{blue}[1]} \\ -5x + 12y & = & 36,000 & {\color{blue}[2]} \end{array}

    \begin{array}{cccc}\text{Multiply }{\color{blue}[1]}\text{ by 12:} & 144x - 60y & = & 1,248,000 \\ \text{Multiply }{\color{blue}[2]}\text{ by 5:} & -25x + 60y & = & 180,000 \end{array}

    Add the equations: . 119x \:=\:1,428,000\quad\Rightarrow\quad\boxed{ x \:=\:12,000}

    Substitute into [2]: . -5(12,000) + 12y \:=\:36,000\quad\Rightarrow\quad\boxed{y \:=\:8,000}

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