# Simultaneous equations

• Oct 4th 2007, 08:48 AM
Helloworld
Simultaneous equations
Hello, I've a problem with equations. It's very hard to me. Could u give the solutions with some steps?10(b^3)=8c+2(a^5)10(c^3)=8a+2(b^5)10(a^3)=8b +2(c^5)I think, that:x,y,z=0x,y,z=1but what's next?x=? y=? z=?I need help.Thanks in advance ;)
• Oct 4th 2007, 08:58 AM
topsquark
Quote:

Originally Posted by Helloworld
Hello, I've a problem with equations. It's very hard to me. Could u give the solutions with some steps?10(b^3)=8c+2(a^5)10(c^3)=8a+2(b^5)10(a^3)=8b +2(c^5)I think, that:x,y,z=0x,y,z=1but what's next?x=? y=? z=?I need help.Thanks in advance ;)

Let's write this out a little more intelligibly:
$10b^3=8c+20a^5c^3=8a+20a^3b^5=8b+2c^5$

I suppose as a first step I'd write out:
$10b^3 = 8c + 20a^5c^3 \implies 5b^3 = 4c + 10a^5c^3$

$10b^3 = 8a + 20a^3b^5 \implies 5b^3 = 4a + 10a^3b^5$

$10b^3 = 8b + 2c^5 \implies 5b^3 = 4b + c^5$

-Dan
• Oct 4th 2007, 09:06 AM
topsquark
Quote:

Originally Posted by topsquark
$5b^3 = 4c + 10a^5c^3$

$5b^3 = 4a + 10a^3b^5$

$5b^3 = 4b + c^5$

I presume a, b, and c are to be positive integers.

I can't finish this for you, but I will note a couple of points that you might be able to put together.

The first equation says that
$5b^3 = c(4 + 10a^5c^2)$

1) So $5b^3$ must be divisible by c. Similarly from the second equation $5b^3$ must be divisible by a.

2) The RHS of both the first and second equations is divisible by 2. Thus b is divisible by 2.

3) Using 2), we see from the third equation that c is also divisible by 2.

(So note that your solution of a = 1, b = 1, and c = 1 must be incorrect.)

See if these observations get you anywhere.

-Dan
• Oct 4th 2007, 09:13 AM
earboth
Quote:

Originally Posted by Helloworld
Hello, I've a problem with equations. It's very hard to me. Could u give the solutions with some steps?10(b^3)=8c+2(a^5)10(c^3)=8a+2(b^5)10(a^3)=8b +2(c^5)I think, that:x,y,z=0x,y,z=1but what's next?x=? y=? z=?I need help.Thanks in advance ;)

Hello,

do you mean

$\begin{array}{l}10(b^3)=8c+2(a^5) \\ 10(c^3)=8a+2(b^5) \\ 10(a^3)=8b+2(c^5) \end{array}$

Where does the x, y, z come from?
• Oct 4th 2007, 09:15 AM
topsquark
Quote:

Originally Posted by earboth
Hello,

do you mean

$\begin{array}{l}10(b^3)=8c+2(a^5) \\ 10(c^3)=8a+2(b^5) \\ 10(a^3)=8b+2(c^5) \end{array}$

Where does the x, y, z come from?

Ahhhh...That makes more sense...

-Dan
• Oct 4th 2007, 09:30 AM
topsquark
Quote:

Originally Posted by earboth
Hello,

do you mean

$\begin{array}{l}10(b^3)=8c+2(a^5) \\ 10(c^3)=8a+2(b^5) \\ 10(a^3)=8b+2(c^5) \end{array}$

So again we may divide each of these by 2:
$\begin{array}{l}5b^3=4c+a^5 \\ 5c^3=4a+b^5 \\ 5a^3=4b+c^5 \end{array}$
I guess this would explain your (now valid) a = 1, b = 1, c = 1 solution.

It looks to me like:
$5b^3=4c+a^5 \implies 4c = 5b^3 - a^5$
implies that a and b are either both even or both odd.

Similarly the second equation implies that b and c are either both even or both odd.

Similarly the third equation, etc.

So it looks like either all of a, b, and c are even or all of a, b, and c are odd.

-Dan
• Oct 4th 2007, 11:19 AM
Helloworld
Quote:

Originally Posted by topsquark
So again we may divide each of these by 2:
$\begin{array}{l}5b^3=4c+a^5 \\ 5c^3=4a+b^5 \\ 5a^3=4b+c^5 \end{array}$
I guess this would explain your (now valid) a = 1, b = 1, c = 1 solution.

It looks to me like:
$5b^3=4c+a^5 \implies 4c = 5b^3 - a^5$
implies that a and b are either both even or both odd.

Similarly the second equation implies that b and c are either both even or both odd.

Similarly the third equation, etc.

So it looks like either all of a, b, and c are even or all of a, b, and c are odd.

-Dan

OK, u are right. now is ok, I should divided it by 2.And what I should do now? I know, that a=b=c, but how can I check that?
• Oct 5th 2007, 08:32 AM
Helloworld
ok, now I know how can I do that ;)Topic should be closed ;)
• Oct 5th 2007, 02:51 PM
Jhevon
Quote:

Originally Posted by Helloworld
;)Topic should be closed ;)