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 | ||
− | <cmath> |x - 1| + |x + 2| < | + | <cmath> |x - 1| + |x + 2| < 5</cmath> |
is | is | ||
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==Solution== | ==Solution== | ||
− | 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 < | + | 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>-1 - 2x < 5</math>, or <math>x > -3</math>. Combining this with our original assumption, we get <math>-3 < x < -2</math> | |
− | + | 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. | |
− | + | 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>. | |
− | The | ||
− | < | ||
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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== | ||
− | {{AHSME | + | {{AHSME 30p box|year=1973|num-b=21|num-a=23}} |
Latest revision as of 14:02, 20 February 2020
Problem
The set of all real solutions of the inequality
is
Solution
We can do casework upon the value of . First, consider the case where both absolute values are positive, which is when
. In this case, the equation becomes
. This turns into
. Combining this with our original assumption, we get the solutions
.
Next, consider the case when both absolute values are negative, which is when . This yields
, or
. Combining this with our original assumption, we get
The next case is when the first absolute value is positive and the second is negative. This occurs when and
. 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 and
. This yields
, which is always true. Therefore, we also get the solutions
.
Therefore, after combining all of our solutions, we get the range , which is
.
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 |