Draw labeled graph of: $|x-1|-|3x-2|$. where $|\cdot|$ denoted Modulus or Absolute value function.
I tried braking graphs between intervals: $(-\infty,\frac{2}{3}),\ \big[\frac{2}{3},1),\ \big[1,\infty)$. Obtained three different graphs then combined them into one to obtain $|x-1|-|3x-2|$.
But that too cumbersome process, can this be done easier manner, than this?
Such functions are piecewise linear, so that it suffices to evaluate the function at the "corners" and draw segments in between. A corner occurs where the argument of an absolute value goes to zero.
We have
$$3x-2=0\to f\left(\frac23\right)=\frac13,\\x-1=0\to f(1)=-1$$ and this is enough to draw in $\left[\dfrac23,1\right]$.
Outside this range, we have two half-lines and it is sufficient to know their slopes, $2$ on the left and $-2$ on the right. This is because the slope of $|x|$ is $-1$ then $1$ from left to right.
This is confirmed by a grapher:
Summary of the method:
For a linear combination of absolute values of linear binomials,
evaluate the sum at all the corner points, which occur at the zeroes of the binomials,
evaluate the left and right slopes, which are minus and plus the sum of the coefficients respectively,
draw the polyline.
For $$\sum a_k|b_kx+c_k|,$$
evaluate all
$$x_n=-\frac{b_n}{a_n}$$ then
$$y_n\sum a_k|b_kx_n+c_k|,$$ and link $(x_n,y_n)$, sorted on $x$.
The slopes of the extreme half-lines are $$\mp\sum a_k.$$