# 1973 AHSME Problems/Problem 22

## Problem

The set of all real solutions of the inequality $$|x - 1| + |x + 2| < 5$$

is $\textbf{(A)}\ x \in ( - 3,2) \qquad \textbf{(B)}\ x \in ( - 1,2) \qquad \textbf{(C)}\ x \in ( - 2,1) \qquad$ $\textbf{(D)}\ x \in \left( - \frac32,\frac72\right) \qquad \textbf{(E)}\ \O \text{ (empty})$

## Solution

We can do casework upon the value of $x$. First, consider the case where both absolute values are positive, which is when $x \geq 1$. In this case, the equation becomes $2x + 1 < 5$. This turns into $x < 2$. Combining this with our original assumption, we get the solutions $1 \leq x < 2$.

Next, consider the case when both absolute values are negative, which is when $x < -2$. This yields $-1 - 2x < 5$, or $x > -3$. Combining this with our original assumption, we get $-3 < x < -2$

The next case is when the first absolute value is positive and the second is negative. This occurs when $x \geq 1$ and $x < -2$. Obviously, this has no solutions, since the inequalities do not overlap.

The final case is when the first absolute value is negative and the second is positive. This occurs when $x < 1$ and $x \geq -2$. This yields $3 < 5$, which is always true. Therefore, we also get the solutions $-2 \leq x < 1$.

Therefore, after combining all of our solutions, we get the range $-3 < x < 2$, which is $\boxed{\textbf{A}}$.