Absolute Value Equations #2

Case 1: For $x < - 1$

$\begin{aligned} |3x-2|-|x+1| & = x+2 \\ -3x + 2 + x+1 & = x+2 \\ \implies x & = \frac 13 \not < -1 & \small \color{#D61F06} \text{Rejected} \end{aligned}$

Case 2: For $-1 \le x < \dfrac 23$

$\begin{aligned} |3x-2|-|x+1| & = x+2 \\ -3x + 2 - x-1 & = x+2 \\ \implies x & = - \frac 15 \in \left[-1, \frac 23 \right) & \small \color{#3D99F6} \text{Accepted} \end{aligned}$

Case 3: For $x \ge \dfrac 23$

$\begin{aligned} |3x-2|-|x+1| & = x+2 \\ 3x - 2 - x-1 & = x+2 \\ \implies x & = 5 \ge \frac 23 & \small \color{#3D99F6} \text{Accepted} \end{aligned}$

Therefore, the sum of solution is $5-\dfrac 15 = \dfrac {24}5$ $\implies m+n = 24+5 = \boxed{29}$ .

For these kind of problems, we need to divide the number line denoting values of $x$ into multiple parts such that the modulus sign can be removed and the equation be transformed into a linear equation in one variable for each region of $x$ . We follow a particular algorithm for such problems. First, we need to compute the roots of the expressions inside the modulus. So,

$3x-2=0\implies x=\frac{2}{3}\quad\textrm{and}\quad x+1=0\implies x=(-1)$

Now, let's divide the domain into three regions, namely,

$x\geq \frac{2}{3}~,~x\in \left[-1,\frac{2}{3}\right]~,~x\leq (-1)$

Now, we evaluate the expression for each region as follows:

$\forall x\geq \frac{2}{3}~,~~3x-2-x-1=x+2\implies x=5\\ \forall x\in \left[-1,\frac{2}{3}\right]~,~~2-3x-x-1=x+2\implies x=\left(-\frac{1}{5}\right)\\ \forall x\leq (-1)~,~~2-3x+x+1=x+2\implies x=~\frac{1}{3}$

Check the cases with the values found. Only the first two cases yield acceptable solutions since the third case $x\leq (-1)$ gives a value $x=\frac{1}{3}\not\leq (-1)$ . So, the solutions are $x=5,\left(-\dfrac{1}{5}\right)$ .

Sum of the solutions $=5-\dfrac{1}{5}=\boxed{\dfrac{24}{5}}$

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Here, the diagram shows the nature of the expressions (positive or negative) for the domain in the three regions. I and II refers to the first and second expressions on LHS respectively.

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P.s - For better view of the picture, I recommend that you open it in a new tab and save the image first. Then open it from your downloads folder.