Equation involving modulus Solve the equation , 
$|x+1| + |x-2| = |2x-1|$ 
This was the solution given in my book , 
$|x+1| + |x-2| = |2x-1|$ 
$(x+1)(x+2) ≥ 0$
Hence , $x≤-1 , x≥2$ 
However I couldn’t understand the second step. How did they just factorise the equation that was in the modulus ? I initially thought of squaring but I believe that would be too long. So , what how can we solve such modulus equations by factorization ?

If you feel yourself lost you can always try cases:
$$\begin{align*}\bullet\;&x<-1&\implies& -(x+1)-(x-2)=-(2x-1)\implies1=1\implies \color{red}{x<-1}\\{}\\
\bullet\;&-1\le x<\frac12&\implies& (x+1)-(x-2)=-(2x-1)\implies3=-2x+1\implies \color{red}{x=-1}\\{}\\
\bullet\;&\frac12\le x<2&\implies&(x+1)-(x-2)=(2x-1)\implies 3=2x-1\implies x=2...\text{no solution}\\{}\\
\bullet\;&2\le x&\implies& (x+1)+(x-2)=(2x-1)\implies 0=0\implies \color{red}{x\ge2}\end{align*}$$
Thus the solution set is $\;\color{red}{(-\infty,-1]\cup[2,\infty)}\;$
