<s>[INST] Finding $\frac{3x+y}{6x-1} = ?$ $$2^{3x} = 18$$
$$2^y = 9$$
$$\frac{3x+y}{6x-1} = ?$$
Let me show my attempt:
$$2^y = 9, 2^y = 3^2, y = 1$$
$$2^{3x} = 18, x = 1$$
I think I've gone too wrong
 [/INST] $$2^{3x} = 18$$
$$2^y = 3^2$$
$\mapsto 2^{y+1} = 9 * 2 =18$

This means that:

$2^{y+1} = 2^{3x}$

Comparing exponents (due to same base on both sides of $=$)

$y+1=3x$
$\mapsto y = 3x-1$

We have to find:

$$\frac{3x+y}{6x-1}$$

Substituting value of $y$

$= \frac{3x + (3x-1)}{6x-1}$
$= \frac{6x-1}{6x-1}$
$=1$
$$\therefore \frac{3x+y}{6x-1} = 1$$
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