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 $\boxed{\textbf{A}}$ .

1973 AHSME (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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