solving linear equation with modulus

Question

here's the question $$|X-1|+|2X+3|=|3X-4|.$$ I tried to obtain the range of this question by setting three intervals -infinity to $-3/2$, $-3/2$ to $1$ and from $1$ to $4/3$

but what i obtained were simply values of $x$. and the answer given behind the text book for this question is

(-infinity,1]union[3/2,infinity). 

i don't want a straight solution....i'll have to tackle many such question so please help me with the strategy to tackle such question,so that i can grow deeper understanding for approaching such question rather than copying the repeatative steps. Thankyou

Answer 1

Let consider the cases for each interval for $|X-1|+|2X+3|=|3X-4|$

• for $x<-\frac32 \implies -X+1-2X-3=-3X+4 \implies-2=4$

• for $-\frac32\le x<1 \implies -X+1+2X+3=-3X+4\implies 4X=0\implies X=0$

• for $1\le x<\frac43\implies X-1+2X+3=-3X+4\implies 6X=2\implies X=\frac13$

• for $x\ge\frac43\implies X-1+2X+3=3X-4\implies 2=-4$

thus the only solution is $X=0$.

Answer 2

After squaring we get $$|(x-1)(2x+3)| = 2x^2-17x+3$$ So if $x\notin (-{3\over 2},1)$ we get: $$2x^2+x-3 = 2x^2-17x+3\Longrightarrow x={1\over 3}$$ which is not OK.

If $x\in (-{3\over 2},1)$ we get: $$-2x^2-x+3 = 2x^2-17x+3\Longrightarrow 4x^2-16x = 0 \Longrightarrow x\in\{0,4\}$$ So $x=0$.