Simultaneous equations

Question

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?

Answer 1

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.

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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$$\boxed{(x,y)=\left(3,\frac 12\right)}$$

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

Answer 2

Look at the second equation : $3x-6=2+2y$. If we subtract $2$ and multiply with $2$, we get $6x-16=4y$

Answer 3

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?