If $|x+5|-|3x-6|=0$, find the largest possible value of $x$. Express your answer as an improper fraction.
I notice that the equation has absolute value signs, which means that there are two possible cases: either the expressions inside the signs have the same sign or opposite signs. If they have the same sign, then I can drop the absolute value signs and just solve for $x$. So I get $x+5-3x+6=0$, which simplifies to $-2x+11=0$. Solving for $x$, I get $x=\frac{11}{2}$, which is an improper fraction. This is a possible value of $x$, but is it the largest? I need to check the other case. If the expressions inside the absolute value signs have opposite signs, then I need to change the sign of one of them and then solve for $x$. There are two subcases: either $x+5$ is positive and $3x-6$ is negative, or the other way around. If $x+5$ is positive and $3x-6$ is negative, then I get $x+5+3x-6=0$, which simplifies to $4x-1=0$. Solving for $x$, I get $x=\frac{1}{4}$, which is also an improper fraction, but smaller than $\frac{11}{2}$. If $x+5$ is negative and $3x-6$ is positive, then I get $-x-5+3x-6=0$, which simplifies to $2x-11=0$. Solving for $x$, I get $x=\frac{11}{2}$, which is the same as before. So I have found two possible values of $x$: $\frac{11}{2}$ and $\frac{1}{4}$, and the largest one is $\frac{11}{2}$. I need to make sure that these values actually satisfy the original equation, by plugging them back in and verifying that they make the equation true. If I plug in $x=\frac{11}{2}$, I get $|\frac{11}{2}+5|-|3\cdot \frac{11}{2}-6|=0$, which simplifies to $|\frac{21}{2}|-|\frac{15}{2}|=0$, which is true. If I plug in $x=\frac{1}{4}$, I get $|\frac{1}{4}+5|-|3\cdot \frac{1}{4}-6|=0$, which simplifies to $|\frac{21}{4}|-|-\frac{23}{4}|=0$, which is also true. So both values are valid solutions, but the largest one is $\frac{11}{2}$. 