Find all such $a$ that $x+2\lvert x-3 \rvert = 7\lvert x-a \rvert + 3 \lvert x-a-4|$ has at least one root. In the equation, $a$ is a parameter and $x$ is a variable:
$$x+2\lvert x-3 \rvert = 7\lvert x-a \rvert + 3 \lvert x-a-4|.$$
I want to find all values of $a$ that make the equation have at least one real root.
Context:
My textbook says this can be accomplished by looking at the restrictions of the functions on both sides of the equation. The only thing I can think of is to find their min/max values.
What I've done: I found some restrictions but failed to come up with a solution: 


*

*$f\left(x\right) = 7\lvert x-a \rvert + 3 \lvert x-a-4|$; min $f\left(x\right)=f\left(a\right)=12$

*$g\left(x\right) = x+2\lvert x-3 \rvert$; min $g\left(x\right)=g\left(3\right) = 3$

*$f\left(x\right)=g\left(x\right) \implies g\left(x\right) \ge 12 \implies x \in \left(-\infty;-6\right)\cup\left(6;+\infty\right)$.

 A: For big enough $x$, the RHS will outgrow the LHS, regardless of the value of $a$. This is because the RHS is asymptotic to $10x$, while the LHS is asymptotic to $3x$.
Furthermore, both sides of our equation are piecewise linear and continuous. Therefore, our question is equivalent to finding the $a$’s, such that for some $x$, the RHS does not exceed the LHS, and we need only to look at their values at $x=3,a,a+4$ to determine them:


*

*At $x=3$, the LHS equals $3$, while the RHS equals $7|3-a|+3|-a-1|$. It can be verified that no $a$ satisfy the desired inequality in this case.

*At $x=a$, the LHS equals $a+2|a-3|$, while the RHS equals $12$. In this case, $a\in\mathbb{R}\setminus(-6,6)$ will satisfy the desired inequality.

*At $x=a+4$, the LHS equals $a+4+2|a+1|$, while the RHS equals $28$. It can be verified that the $a$ that satisfy the desired inequality in this case are all contained in the solution set of the last case.


Our final answer is therefore that for precisely $\boxed{a\in\mathbb{R}\setminus(-6,6)}$ does our original equation have at least a real root.
