# Thread: [SOLVED] Algebra indices simultaneous equation

1. ## [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

2. 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 $x = 4$.

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

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

$6^x = 6^4~\implies~ x= 4~\wedge~y=3$

4. Cheers guys!

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