1973 AHSME Problems/Problem 22

Problem

The set of all real solutions of the inequality

$|x - 1| + |x + 2| < 3$

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 $2x + 1 < 3$ . This turns into $x < 1$ , which is a contradiction with our original assumption of $x \geq 1$ . Therefore, this case yields no solutions.

Next, consider the case when both absolute values are negative, which is when $x < -2$ . This yields $3 - 2x < 3$ , or $x > 0$ . Again, this yields a contradiction to the assumption that $x < -2$ , so this yields no solutions.

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 also has no solutions.

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 $3 < 3$ , which also has no solutions.

Therefore, there are no solutions and the answer is $\boxed{\textbf{E}}$ .

Note

The answer key lists the answer as $\boxed{\textbf{A}}$ , but graphing the expression, there are clearly no solutions. The answer given by the answer key would be correct if the equation were $|x - 1| + |x + 2| < 5$ As a result, either the answer key is incorrect or the problem is incorrect.

To see a graph of the expression, go here: https://www.desmos.com/calculator/iqj0rtqc92

1973 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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1973 AHSME Problems/Problem 22

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