Difference between revisions of "2011 AMC 10A Problems/Problem 7"
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− | == Solution == | + | == Solution 1 == |
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Therefore, the answer is <math> \boxed{\mathrm{(B)}} </math>. | Therefore, the answer is <math> \boxed{\mathrm{(B)}} </math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Instead of solving, we can just categorize and solve. | ||
+ | |||
+ | [hide=Section 1] | ||
+ | This contains $\mathrm{(A),(C),(D)} | ||
+ | [/hide] | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2011|ab=A|num-b=6|num-a=8}} | {{AMC10 box|year=2011|ab=A|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:34, 10 January 2018
Contents
Problem 7
Which of the following equations does NOT have a solution?
Solution 1
has no solution because absolute values output positives and this equation implies that the absolute value could output a negative.
Further: is true for
is true for
is true for
is true for
Therefore, the answer is .
Solution 2
Instead of solving, we can just categorize and solve.
[hide=Section 1] This contains $\mathrm{(A),(C),(D)} [/hide]
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.