Adding $2$ absolute values together: $|x+2| + |x-3| =5.$

Question

I came across a very basic absolute value question

$|x+2| + |x-3| =5.$

Initially, I thought the answer was $x=-2$ and $x=3$ because I let each absolute values be either positive and negative and that's what you get.

But the correct answer was an inequality; $-2\le x \le 3.$ Now, If you try subbing values in or solve it graphically, this inequality is correct.

But how am I supposed to know if the solution is to be an inequality or equality? Is there anyway to tell by looking at the expression?

Thanks,

Answer 1

Methodology/ Algorithm:

You start by checking when the expressions within the absolute value signs are positive or negative. This is because according to the definition of $|f(x)| = f(x)$ when $f(x) \ge 0$ and $|f(x)| = - f(x)$ when $f(x) \lt 0$. So first differentiate when the expressions will change signs.

Then you will arrive at a number of points where the expressions change signs. Here since you have two expressions inside absolute value signs you will have two such points. Divide the real line at these points and solve the resulting equations subject to the inequality which arises from restricting $x$ to a portion of the real line.

For an example, here the line will be split at $x = -2$ and $x = 3$. So suppose, $-2 \le x \lt3 $ then, the resulting equation , due to the inequality is,

$$ x + 2 - (x - 3) = 5 \iff 5 = 5 $$

This means the equality is satisfied for every $x$ in the given interval. Now check the other regions, namely $x \ge 3$ and $x \lt -2$ and solve the equation within this regions and you will find the equality is not satisfied for any $x$ in the stipulated conditions.

Answer 2

The triangle inequality states that for $x,y,z \in \mathbb{R}$, we have $|x-y|+|x-z| \le |y-z|$ with equality occurring if and only if $x$ is in between $y$ and $z$ inclusive. Apply that here for $y = -2$ and $z = 3$.

Answer 3

just consider all the following 5 cases:

1) $ x+2>0 $ and $ x-3>0$

2) $ x+2>0 $ and $ x-3<0$

3) $ x+2<0 $ and $ x-3>0$

4) $ x+2<0 $ and $ x-3<0$

5) $ x+2=0 $ and $ x-3=0$

Answer 4

A useful fact in absolute value equations: $|a| + |b| = |a -b| \implies ab \le 0$.

If we apply the above result here, then we are left with $(x+2)(x-3)\le 0$, the solution set to which is $x \in \left[-2,3\right]$.

Answer 5

For a more intuitive approach, plotting a graph of $|x+2|$ and $|x-3|$ would help. These are two V-shaped graphs of $\pm45^o $ gradient, touching the x-axis at $-2$ and $3$ respectively.

Then you can clearly see that for $-2\leq x\leq3$, the graphs are $$\begin{align} &y_1&&=\ \ \ x+2\\ &y_2=-(x-3)&&=-x+3\end{align}$$

giving $y_1+y_2=5$.

Thus the solution to $|x+2|+|x-3|=5$ is $-2\leq x\leq3$.