Difference between revisions of "2011 AMC 10A Problems/Problem 7"
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Now we can analyze and we see <math>-x</math> can become <math>x</math> if <math>x=-y</math> and absolute value inequalities cannot be negative, so the answer is <math>\boxed{\mathrm{(B)}}</math> | Now we can analyze and we see <math>-x</math> can become <math>x</math> if <math>x=-y</math> and absolute value inequalities cannot be negative, so the answer is <math>\boxed{\mathrm{(B)}}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/9pG49ACG5k8 | ||
| + | |||
| + | ~savannahsolver | ||
== 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 10:53, 1 November 2020
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.
Section 1: This contains A,C,D as they are all squares or square roots. From skimming, we can get an answer as maybe C
Section 2: This contains B and E From skimming we can get we can get answer as maybe B
Now we can analyze and we see 
can become 
if 
and absolute value inequalities cannot be negative, so the answer is
Video Solution
~savannahsolver
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. 
