Algebra equation, is this possible to solve ? Where to start?

**heldrida** · Dec 11th 2009, 03:08 PM

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!

**Gusbob** · Dec 11th 2009, 03:55 PM

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 $J_1 = 0.04C_1$

**bigwave** · Dec 11th 2009, 04:41 PM

just building off of Gusbob

$ J_1 = .04C_1 $
$J_2 = .02C_2$

and

$ J_1 = .05C_2 $

from
$C_1 - C_2 =750$

$ \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

$ 3750 - 3000 = 750 $

well one way to solve it anyway...

**heldrida** · Dec 12th 2009, 02:59 PM

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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Algebra equation, is this possible to solve ? Where to start?

3750 - 3000 = 750

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