jmc

We begin by moving the second inequality to the right side of the equation, giving us $|x+5|=|3x-6|$. From here, we can split the equation into two separate cases. For the first case, note that if $x+5$ and $3x-6$ have the same sign, then $x+5=3x-6$:

Case 1: $$\begin{array}{rl}x+5& =3x-6\\ \Rightarrow -2x& =-11\\ \Rightarrow x& =\frac{11}{2}\end{array}$$If we plug this value of $x$ back into the original equation to check our answer, we get that $\left|\frac{11}{2}+5\right|-\left|3\left(\frac{11}{2}\right)-6\right|=0$ or $0=0$. Since this is true, we can accept $x=\frac{11}{2}$ as a valid solution.

For case two, note that if $x+5$ has a different sign than $3x-6$, then $x+5=-(3x-6)$.

Case 2: $$\begin{array}{rl}x+5& =-(3x-6)\\ x+5& =-3x+6\\ \Rightarrow 4x& =1\\ \Rightarrow x& =\frac{1}{4}\end{array}$$If we plug this value of $x$ back into the original equation to check our answer, we get that $\left|\frac{1}{4}+5\right|-\left|3\left(\frac{1}{4}\right)-6\right|=0$, which also gives us $0=0$. This is always true, so we can accept $x=\frac{1}{4}$ as a valid solution as well. Thus, our two possible solutions are $\frac{1}{4}$ and $\frac{11}{2}$. Since the question asks for the largest possible value of $x$, our final solution is $\boxed{\frac{11}{2}}$.