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What is the sum of all the solutions of <math>x = \left|2x-|60-2x|\right|</math>? | What is the sum of all the solutions of <math>x = \left|2x-|60-2x|\right|</math>? | ||
− | <math> | + | <math>\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 92 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 124</math> |
− | \ | ||
− | \qquad | ||
− | \ | ||
− | \qquad | ||
− | \ | ||
− | \qquad | ||
− | \ | ||
− | \qquad | ||
− | \ | ||
− | </math> | ||
− | == Solution == | + | == Solution 1== |
− | + | We evaluate this in cases: | |
+ | |||
+ | ''Case 1'' | ||
+ | <math>x<30</math> | ||
+ | |||
+ | When <math>x<30</math> we are going to have <math>60-2x>0</math>. When <math>x>0</math> we are going to have <math>|x|>0\implies x>0</math> and when <math>-x>0</math> we are going to have <math>|x|>0\implies -x>0</math>. Therefore we have <math>x=|2x-(60-2x)|</math>. | ||
+ | <math>x=|2x-60+2x|\implies x=|4x-60|</math> | ||
+ | |||
+ | ''Subcase 1 ''<math>30>x>15</math> | ||
+ | |||
+ | When <math>30>x>15</math> we are going to have <math>4x-60>0</math>. When this happens, we can express <math>|4x-60|</math> as <math>4x-60</math>. | ||
+ | Therefore we get <math>x=4x-60\implies -3x=-60\implies x=20</math>. We check if <math>x=20</math> is in the domain of the numbers that we put into this subcase, and it is, since <math>30>20>15</math>. Therefore <math>20</math> is one possible solution. | ||
+ | |||
+ | '' Subcase 2 '' <math>x<15</math> | ||
+ | |||
+ | When <math>x<15</math> we are going to have <math>4x-60<0</math>, therefore <math>|4x-60|</math> can be expressed in the form <math>60-4x</math>. | ||
+ | We have the equation <math>x=60-4x\implies 5x=60\implies x=12</math>. Since <math>12</math> is less than <math>15</math>, <math>12</math> is another possible solution. | ||
+ | <math>x=|2x-|60-2x||</math> | ||
+ | |||
+ | ''Case 2 '': <math>x>30</math> | ||
+ | |||
+ | When <math>x>30</math>, <math>60-2x<0</math>. When <math>x<0</math> we can express this in the form <math>-x</math>. Therefore we have <math>-(60-2x)=2x-60</math>. This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we have | ||
+ | <math>x=|2x-(2x-60)|</math> | ||
+ | |||
+ | <math>x=|2x-2x+60|</math> | ||
+ | |||
+ | <math>x=|60|</math> | ||
+ | |||
+ | <math>x=60</math> | ||
+ | |||
+ | We have now evaluated all the cases, and found the solution to be <math>\{60,12,20\}</math> which have a sum of <math>\boxed{\textbf{(C)}\ 92}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | From the equation <math>x = \left|2x-|60-2x|\right|</math> , we have <math>x = 2x-|60-2x|</math> , or <math>-x = 2x-|60-2x|</math>. Therefore, <math>x=|60-2x|</math> , or <math>3x=|60-2x|</math>. From here we have four possible cases: | ||
+ | |||
+ | 1. <math>x=60-2x</math>; this simplifies to <math>3x=60</math>, so <math>x=20</math>. | ||
+ | |||
+ | 2. <math>-x=60-2x</math>; this simplifies to <math>x=60</math>. | ||
+ | |||
+ | 3. <math>3x=60-2x</math>; this simplifies to <math>5x=60</math>, so <math>x=12</math>. | ||
+ | |||
+ | 4. <math>-3x=60-2x</math>; this simplifies to <math>-x=60</math>, so <math>x=-60</math>. However, this solution is extraneous because the absolute value of <math>2x-|60-2x|</math> cannot be negative. | ||
+ | |||
+ | The sum of all of the solutions of <math>x</math> is <math>20+60+12=\boxed{\textbf{(C)}\ 92}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/vYXz4wStBUU?t=272 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | https://youtu.be/1DjO74VEr3E | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2010|ab=B|num-b=12|num-a=14}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:00, 22 March 2024
Problem
What is the sum of all the solutions of ?
Solution 1
We evaluate this in cases:
Case 1
When we are going to have
. When
we are going to have
and when
we are going to have
. Therefore we have
.
Subcase 1
When we are going to have
. When this happens, we can express
as
.
Therefore we get
. We check if
is in the domain of the numbers that we put into this subcase, and it is, since
. Therefore
is one possible solution.
Subcase 2
When we are going to have
, therefore
can be expressed in the form
.
We have the equation
. Since
is less than
,
is another possible solution.
Case 2 :
When ,
. When
we can express this in the form
. Therefore we have
. This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we have
We have now evaluated all the cases, and found the solution to be which have a sum of
Solution 2
From the equation , we have
, or
. Therefore,
, or
. From here we have four possible cases:
1. ; this simplifies to
, so
.
2. ; this simplifies to
.
3. ; this simplifies to
, so
.
4. ; this simplifies to
, so
. However, this solution is extraneous because the absolute value of
cannot be negative.
The sum of all of the solutions of is
Video Solution
https://youtu.be/vYXz4wStBUU?t=272
~IceMatrix
~savannahsolver
See Also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 |
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