If $x$ and $y$ are acute angles such that $x+y=\pi/4$ and $\tan y=1/6$, find the value of $\tan x$.

$\textbf{(A) }\dfrac{27\sqrt2-18}{71}\hspace{11.5em}\textbf{(B) }\dfrac{35\sqrt2-6}{71}\hspace{11.2em}\textbf{(C) }\dfrac{35\sqrt3+12}{33}$

$\textbf{(D) }\dfrac{37\sqrt3+24}{33}\hspace{11.5em}\textbf{(E) }1\hspace{15em}\textbf{(F) }\dfrac57$

$\textbf{(G) }\dfrac37\hspace{15.4em}\textbf{(H) }6\hspace{15em}\textbf{(I) }\dfrac16$

$\textbf{(J) }\dfrac12\hspace{15.7em}\textbf{(K) }\dfrac67\hspace{14.8em}\textbf{(L) }\dfrac47$

$\textbf{(M) }\sqrt3\hspace{14.5em}\textbf{(N) }\dfrac{\sqrt3}3\hspace{14em}\textbf{(O) }\dfrac56$

$\textbf{(P) }\dfrac23\hspace{15.4em}\textbf{(Q) }\dfrac1{2007}$
1. Given that \( x \) and \( y \) are acute angles such that \( x + y = \frac{\pi}{4} \) and \( \tan y = \frac{1}{6} \), we need to find the value of \( \tan x \).

2. We know that:
   \[
   \tan(x + y) = \tan\left(\frac{\pi}{4}\right) = 1
   \]

3. Using the tangent addition formula:
   \[
   \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}
   \]
   Substituting \( \tan(x + y) = 1 \) and \( \tan y = \frac{1}{6} \):
   \[
   1 = \frac{\tan x + \frac{1}{6}}{1 - \tan x \cdot \frac{1}{6}}
   \]

4. Simplify the equation:
   \[
   1 = \frac{\tan x + \frac{1}{6}}{1 - \frac{\tan x}{6}}
   \]

5. Cross-multiply to solve for \( \tan x \):
   \[
   1 - \frac{\tan x}{6} = \tan x + \frac{1}{6}
   \]

6. Clear the fraction by multiplying through by 6:
   \[
   6 - \tan x = 6 \tan x + 1
   \]

7. Combine like terms:
   \[
   6 - 1 = 6 \tan x + \tan x
   \]
   \[
   5 = 7 \tan x
   \]

8. Solve for \( \tan x \):
   \[
   \tan x = \frac{5}{7}
   \]

The final answer is \(\boxed{\frac{5}{7}}\).