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Difference between revisions of "1973 AHSME Problems/Problem 22"

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Latest revision as of 13:02, 20 February 2020

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

See Also

1973 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions
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