Solving for $x$ in the absolute value equation

Question

$$|x-1|+|2x-3|=|3x-4|$$

I tried to solve the cases separately for the range of values of $x$, but it did not work. Can anyone please show me the way.

Answer 1

Twisted/Confusing way to solve this problem:

$|x-1|+|2x-3|=|3x-4|$...eq(I).
Let us call the three expressions as $A, B, C$ , so we have $A+B=C$

$\begin{matrix} \text{when} & \text{expression}\\ x\geq1&A=x-1 & i\\x<1& A=-(x-1) & ii\\ x\geq\frac{3}{2} & B=2x-3 & iii\\ x<\frac{3}{2} & B=-(2x-3) & iv\\ x\geq \frac{4}{3} & C=3x-4 & v\\ x<\frac{4}{3} & C=-(3x-4) & vi \end{matrix}$

• When we say that $x\geq \frac{3}{2}$ we also say that $x\geq1$ and $x\geq \frac{4}{3}$; $(\frac{3}{2}-\frac{4}{3}=\frac{1}{6}\Rightarrow \frac{3}{2}>\frac{4}{3})$

LHS of eq (I) $= i + iii = v$ RHS of eq (I)

• When we say that $x\leq 1$ we also say that $x\leq \frac{4}{3}$ and $x\leq \frac{3}{2}$

LHS of eq (I) $= ii + iv = vi $ RHS of eq (I)

for all the values of $x \leq 1 $ and $ x \geq \frac{3}{2}$, eq (I) holds

• When $ 1 < x < \frac{4}{3} $, LHS of eq (I) $i + iv \neq vi $ RHS of eq (I), eq (i) does not hold
• When $ \frac{4}{3} < x < \frac{3}{2}$, LHS of eq (I) $ i + iv \neq v$ RHS of eq (I) ,eq (I) does not hold

So the solution is $x\leq 1$ and $x\geq \frac{3}{2}$

Answer 2

There are 4 cases to be dealt with separately: $x\leq 1$, $1<x\leq\frac43$, $\frac43 <x\leq\frac32$, and $x>\frac32$.

In each of those intervals, you have to figure out whether each absolute value expression $|Y|$ means $Y$ or $-Y$. In each case, that will give you a linear equation without absolute values which you can then solve. Whatever solution you get for $x$ in each case has to be consistent with the inequality defining that case.

Does that give you a push in the right direction?

Answer 3

$$|x-1|+|2x-3|\geq|x-1+2x-3|=|3x-4|.$$ The equality occurs, when it occurs in the triangle inequality.

$$|x|+|y|\geq|x+y|$$ or $$x^2+y^2+2|xy|\geq x^2+2xy+y^2$$ or $$|xy|\geq xy.$$ In the last inequality the equality occurs for $xy\geq0.$

Thus, our inequality is equivalent to $$(x-1)(2x-3)\geq0,$$ which gives the answer: $$(-\infty,1]\cup[1.5,+\infty)$$