Is the method I used an efficient way of getting the answers to my titled question? From... $$\vert x-3 \vert + \vert x-4 \vert = 9$$ The way I solved this was to assume that each abs value was positive first giving: $$x-3 + x- 4 = 9 $$ $$2x- 7 = 9$$ $$x = -1$$ Then assuming they are both negative: $$-(x-3) + -(x-4) = 9$$ $$-x+3-x+4= 9$$ $$-2x+7= 9$$ $$x = 8$$
Solutions: $x = -1$ and $x= 8$
On
This function is continuous and piecewise linear, with the pieces delimited by the values that cancel the arguments of the absolute value.
You can write this table of variations:
$$\begin{matrix}-\infty&3&4&\infty\\\hline\infty&1&1&\infty,\end{matrix}$$ which shows two solutions, in $(-\infty,3)$ and $(4,\infty)$.
These are obtained by replacing the absolute values by the appropriate signs,
$$-(x-3)-(x-4)=9$$ and $$+(x-3)+(x-4)=9.$$
On
User35508 has given you the general algebraic method. But if you want some geometric intuition, your equation just says
$$\text{the distance from $x$ to $3$ plus the distance from $x$ to $4$ is $9$}$$
where distances are measured along the real number line.
This immediately explains why there are no solutions when $x$ is between $3$ and $4$. And with just a little more thought you can easily see the two solutions are $-1$ and $8$.
Of course, this geometric approach isn't very helpful for more complicated equations, e.g. $|x^2-5x+6|+|x-4|=3$, but it's a useful tool to have in your repertoire nonetheless.
Take three intervals $$-\infty<x<3$$ $$3\leq x < 4$$ $$4\leq x<\infty$$ Then for first interval $$-(x-3)-(x-4)=9$$ $$2x-7=-9 \Longleftrightarrow x=-1$$
For second interval $$(x-3)-x+4=9$$ $$1=9$$ which is absurd
Finally for last interval $$2x-7=9\Longleftrightarrow x=8$$
NOTE:Your answers are wrong..You are not preserving equality when adding or subtracting numbers from your equations.