[SOLVED] Algebra indices simultaneous equation

**greghunter** · June 26th 2009, 10:03 AM

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

**Prove It** · June 26th 2009, 10:09 AM

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 $x = 4$ .

**earboth** · June 26th 2009, 10:10 AM

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