Solving $3|2-x| + |2x-4| = 8$ I'm not sure how to solve $3|2-x| + |2x-4| = 8$. I don't exactly need an answer, the method is what I'm interested in. I've done equations with one absolute value element. >_&lt;
$ 3|2-x| + |2x-4| = 8 \iff 5|x-2|=8 \iff 5(x-2)= \pm 8.$

Can you proceed ?
A generic approach:

A function that is a linear combination of absolute values of linear monomials, like yours, is piecewise linear and continuous. (General form of the equation: $f(x):=\sum_{k}a_k|b_k x+c_k|+d=0$.)

So it suffices to evaluate it at the "corners" (i.e. where the argument of some absolute value vanishes, $x=-c_k/b_k$) to detect changes of sign (you compute the $f(x_k)=\sum_{j}a_j|-b_j c_k/b_k+c_j|+d$).

We also need to look at the values at $\pm\infty$, the sign of which is given by the sign of $\sum_{k}a_k$.

Here we observe that there is a single corner, at $x=2$, and 

$$3|2-2|+|2\cdot2-4|-8&lt;0.$$

At infinity, the function is positive, so we have a root in $(-\infty,2)$ and another in $(2,\infty)$. To find the exact position of the roots, we remove the absolute values and supply the correct sign, to obtain ordinary linear equations. (I.e.$\sum_{k}\pm a_k(b_k x+c_k)-d=0$.)

Thus,


for the root in $(-\infty,2)$, $3(2-x)+(4-2x)-8=0$,
for the root in $(2, \infty)$, $3(x-2)+(2x-4)-8=0$.




Note that we didn't merge $|x-2|$ and $|4-2x|=2|x-2|$ for the sake of exposition of the method.