Difference between revisions of "2011 AMC 10A Problems/Problem 7"
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− | == Solution == | + | == Solution 1 == |
− | <math>|-3x|+5=0</math> has no solution because absolute values output | + | <math>|-3x|+5=0</math> has no solution because absolute values only output nonnegative numbers. |
Further: | Further: | ||
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<math>|-3x|-4=0</math> is true for <math>x = \frac{4}{3}, -\frac{4}{3}</math> | <math>|-3x|-4=0</math> is true for <math>x = \frac{4}{3}, -\frac{4}{3}</math> | ||
+ | |||
+ | Therefore, the answer is <math> \boxed{\mathrm{(B)}} </math>. | ||
+ | |||
+ | ==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 <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}} |
Latest revision as of 20:23, 21 August 2023
Problem 7
Which of the following equations does NOT have a solution?
Solution 1
has no solution because absolute values only output nonnegative numbers.
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.