# [SOLVED] Algebra indices simultaneous equation

• Jun 26th 2009, 11:03 AM
greghunter
[SOLVED] Algebra indices simultaneous equation
Solve the simultaneous equation $x-y=1$ and $(2^x)(3^y)=432$

Do you make $x=1+y$ first and then put it into the other equation?

I don't know what to do once I've substituted or if that's what you have to do.

Thank you in advance for any help
• Jun 26th 2009, 11:09 AM
Prove It
Quote:

Originally Posted by greghunter
Solve the simultaneous equation $x-y=1$ and $(2^x)(3^y)=432$

Do you make $x=1+y$ first and then put it into the other equation?

I don't know what to do once I've substituted or if that's what you have to do.

Thank you in advance for any help

Yes you're on the right track.

So you should end up with $2^{1 + y}\cdot 3^y = 432$

$2^1 \cdot 2^y \cdot 3^y = 432$

$2\cdot\left(2\cdot 3\right)^y = 432$

$2\cdot 6^y = 432$

$6^y = 216$

$6^y = 6^3$

$y = 3$.

Since $x = 1 + y$ we find that $x = 4$.
• Jun 26th 2009, 11:10 AM
earboth
Quote:

Originally Posted by greghunter
Solve the simultaneous equation $x-y=1$ and $(2^x)(3^y)=432$

Do you make $x=1+y$ first and then put it into the other equation?

I don't know what to do once I've substituted or if that's what you have to do.

Thank you in advance for any help

1. $x-y=1~\implies~y = x-1$

2. $(2^x)(3^y)=432~\implies~2^x\cdot 3^{x-1}=432$

$2^x\cdot 3^x=1296 = 6^4$

$6^x = 6^4~\implies~ x= 4~\wedge~y=3$
• Jun 26th 2009, 11:26 AM
greghunter
Cheers guys!