Equation involving modulus

Question

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 ?

Answer 1

we distinguish four cases: 1)$$x\geq 2$$ this is impossible 2)$$\frac{1}{2}\le x<2$$ then we have $$x+1-x+2=2x-1$$ 3)$$-1\le x<\frac{1}{2}$$ we get $$x+1-x+2=-2x+1$$ 4) $$x<-1$$ then we have $$-x-1-x+2=-2x+1$$

Answer 2

By the triangle inequality $$|2x-1|=|x+1|+|x-2|\geq|x+1+x-2|=|2x-1|.$$ The equality occurs, when $x+1$ and $x-2$ have same sings, which gives the answer: $$(-\infty,-1]\cup[2,+\infty).$$

Answer 3

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 $\;\color{red}{(-\infty,-1]\cup[2,\infty)}\;$