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Math Help - Algebra equation, is this possible to solve ? Where to start?

  1. #1
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    Algebra equation, is this possible to solve ? Where to start?

    Hi,

    I would like to know if this is possible to solve ? Is there enough data ?

    C1 - C2 = 750 euros
    J1 = C1 * 8/12 * 0,06
    J2 = C2 * 6/12 * 0,04
    J1 = 2,5 * J2

    What I've done so far was,

    J1 = 2,5(C2 * 6/12 * 0,04)
    J1 = 2,5( 0,24C2/12 )
    J1 = 0,6C2/12

    12J1 = 0,6C2

    C2 = 12J1/0,6


    Now I'm thinking about getting C1, by doing J1 = C1 * 8/12 * 0,06 and so on. But I dont know if this is the right way to work it.

    Thanks,

    any sugestion is appreciated!
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  2. #2
    Super Member
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    I don't know what exactly you're trying to solve but I shall give you a method that can be used to solve for all the unknowns C1, C2, J1, and J2.

    First I multiply out the fraction and decimals for J1 and J2.

    J_1 = 0.04C_1
    J_2 = 0.02C_2

    We know a constant value: C_1 - C_2 = 750. To put it into this format, try:

    J_1 - 2J_2 = 0.04(C_1 - C_2)

    Simply substitute  J_1 = 2.5J_2 and C_1 - C_2 = 750 into this equation. You can now solve for  J_2 as it's the only variable left.

    With  J_2 you can solve for  C_2 with J_2 = 0.02C_2

    With  C_2 you can solve for  C_1 with C_1 - C_2 = 750

    With  C_1 you can solve for  J_1 with J_1 = 0.04C_1
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  3. #3
    Super Member bigwave's Avatar
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    3750 - 3000 = 750

    just building off of Gusbob

     <br />
J_1 = .04C_1<br />
    J_2 = .02C_2

    and

     <br />
J_1 = .05C_2<br />

    from
    C_1 - C_2 =750

     <br />
\frac{J_1}{.04} - \frac{J_1}{.05} = 750
    so  J_1 = 15

    now

    C_1 = \frac{15}{.04} and C_2 = \frac{15}{.04}

    Therefore

     <br />
3750 - 3000 = 750<br />

    well one way to solve it anyway...
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  4. #4
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    Thanks a lot! I understood it very well, this helped me and teached me a lot!

    I tought this wouldn't be possible to solve. Thanks for your time Gusbob and bigwave!
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