Difference between revisions of "1959 AHSME Problems/Problem 25"
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<math>\textbf{(A)}\ x^2<49 \qquad\textbf{(B)}\ x^2>1 \qquad\textbf{(C)}\ 1<x^2<49\qquad\textbf{(D)}\ -1<x<7\qquad\textbf{(E)}\ -7<x<1</math> | <math>\textbf{(A)}\ x^2<49 \qquad\textbf{(B)}\ x^2>1 \qquad\textbf{(C)}\ 1<x^2<49\qquad\textbf{(D)}\ -1<x<7\qquad\textbf{(E)}\ -7<x<1</math> | ||
== Solution == | == Solution == | ||
− | The equation <math>|3-x| < 4</math> can be solved by splitting it into two inequalities: <math>3-x<4</math> and <math>3-x<-4</math>. The solutions to those inequalities are <math>x<-1</math> and <math>x>7</math>, respectively. The common interval of those two inequalities is <math>\textbf{(D)}\ -1<x<7</math> | + | The equation <math>|3-x| < 4</math> can be solved by splitting it into two inequalities: <math>3-x<4</math> and <math>3-x<-4</math>. The solutions to those inequalities are <math>x<-1</math> and <math>x>7</math>, respectively. The common interval of those two inequalities is <math>\boxed{\textbf{(D)}\ -1<x<7}</math>. |
== See also == | == See also == | ||
{{AHSME 50p box|year=1959|num-b=24|num-a=26}} | {{AHSME 50p box|year=1959|num-b=24|num-a=26}} | ||
{{MAA Notice} | {{MAA Notice} |
Revision as of 12:25, 21 July 2024
Problem 25
The symbol means if is greater than or equal to zero, and if a is less than or equal to zero; the symbol means "less than"; the symbol means "greater than." The set of values satisfying the inequality consists of all such that:
Solution
The equation can be solved by splitting it into two inequalities: and . The solutions to those inequalities are and , respectively. The common interval of those two inequalities is .
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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All AHSME Problems and Solutions |
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