Simultaneous equations I have the question  
Solve the simultaneous equations 
$$\begin{cases}
3^{x-1} = 9^{2y} \\
8^{x-2} = 4^{1+y}
\end{cases}$$
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I know that $x-1=4y$ and $3X-6=2+2y$ .
However when I checked the solutions this should become $6X-16=4Y$ .
How is this? 
 A: Look at the second equation : $3x-6=2+2y$. If we subtract $2$ and multiply with $2$, we get $6x-16=4y$
A: write the system in the form
$$3^{x-1}=3^{4y}$$ and
$$2^{3(x-2)}=2^{2(1+y)}$$
and you will get
$$x-1=4y$$
and
$$3(x-2)=2(1+y)$$
can you proceed?
A: From $$\begin{align*} & 3^{x-1}=9^{2y}\\ & 8^{x-2}=4^{y+1}\end{align*}\tag1$$
We can rewrite $9$ as $3^2$, $8$ as $2^3$ and $4$ as $2^2$. Doing so and setting the powers equivalent, we get$$\begin{align*} & x-1=4y\\ & 3x-6=2+2y\end{align*}\tag2$$
To answer your question, look at the second equation. Moving the $2$ to the left hand side and multiplying everything by $2$, we get our desired equation$$6x-16=4y\tag3$$
Which is the equation in the book.

To actually solve, we can substitute $4y$ from $(3)$ with the first equation of $(2)$ to solve for $x$.$$\begin{align*} & 6x-16=x-1\\ & 5x=15\implies x=3\end{align*}\tag4$$
Substituting that into $x-1=4y$ gives $y$ as$$y=\frac {x-1}4=\frac 12\tag5$$
Therefore, we have $x,y$ as$$\boxed{(x,y)=\left(3,\frac 12\right)}$$

If you have any questions or confusions, you can ask me!
