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

• Dec 11th 2009, 03:08 PM
heldrida
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!
• Dec 11th 2009, 03:55 PM
Gusbob
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 $J_1 = 0.04C_1$
• Dec 11th 2009, 04:41 PM
bigwave
3750 - 3000 = 750
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...
• Dec 12th 2009, 02:59 PM
heldrida
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!