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

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The set of all real solutions of the inequality  
 
The set of all real solutions of the inequality  
  
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<cmath> |x - 1| + |x + 2| < 3</cmath>
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<cmath> |x - 1| + |x + 2| < 5</cmath>
  
 
is  
 
is  
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==Solution==
 
==Solution==
  
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We can do casework upon the value of <math>x</math>. First, consider the case where both absolute values are positive, which is when <math>x \geq 1</math>. In this case, the equation becomes <math>2x + 1 < 3</math>. This turns into <math>x < 1</math>, which is a contradiction with our original assumption of  <math>x \geq 1</math>. Therefore, this case yields no solutions.
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We can do casework upon the value of <math>x</math>. First, consider the case where both absolute values are positive, which is when <math>x \geq 1</math>. In this case, the equation becomes <math>2x + 1 < 5</math>. This turns into <math>x < 2</math>. Combining this with our original assumption, we get the solutions <math>1 \leq x < 2</math>.  
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Next, consider the case when both absolute values are negative, which is when <math>x < -2</math>. This yields <math>3 - 2x < 3</math>, or <math>x > 0</math>. Again, this yields a contradiction to the assumption that  <math>x < -2</math>, so this yields no solutions.
 
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The next case is when the first absolute value is positive and the second is negative. This occurs when <math>x \geq 1</math> and <math>x < -2</math>. Obviously, this also has no solutions.  
 
  
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The final case is when the first absolute value is negative and the second is positive. This occurs when <math>x < 1</math> and <math>x \geq -2</math>. This yields <math>3 < 3</math>, which also has no solutions.
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Next, consider the case when both absolute values are negative, which is when <math>x < -2</math>. This yields <math>-1 - 2x < 5</math>, or <math>x > -3</math>. Combining this with our original assumption, we get <math>-3 < x < -2</math>
  
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Therefore, there are no solutions and the answer is <math>\boxed{\textbf{E}}</math>.
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The next case is when the first absolute value is positive and the second is negative. This occurs when <math>x \geq 1</math> and <math>x < -2</math>. Obviously, this has no solutions, since the inequalities do not overlap.
  
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== Note ==
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The final case is when the first absolute value is negative and the second is positive. This occurs when <math>x < 1</math> and <math>x \geq -2</math>. This yields <math>3 < 5</math>, which is always true. Therefore, we also get the solutions <math>-2 \leq x < 1</math>.
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The answer key lists the answer as <math>\boxed{\textbf{A}}</math>, but graphing the expression, there are clearly no solutions. The answer given by the answer key would be correct if the equation were
 
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<cmath> |x - 1| + |x + 2| < 5</cmath>
 
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As a result, either the answer key is incorrect or the problem is incorrect.  
 
  
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To see a graph of the expression, go here: https://www.desmos.com/calculator/iqj0rtqc92
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Therefore, after combining all of our solutions, we get the range <math>-3 < x < 2</math>, which is <math>\boxed{\textbf{A}}</math>.
  
 
==See Also==
 
==See Also==
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{{AHSME 35p box|year=1973|num-b=21|num-a=23}}
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{{AHSME 30p box|year=1973|num-b=21|num-a=23}}

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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