transforming from absolute sign to plus minus sign
$begingroup$
I have recently encountered the following algebra transformation from an absolute sign to plus minus sign.
I am unable to get my head around on how it really works? What is the underlying principle that justify such transformation?
linear-algebra exponential-function absolute-value
asked Dec 4 '18 at 2:50
ThorThor
23217
$endgroup$
add a comment |
$begingroup$
I have recently encountered the following algebra transformation from an absolute sign to plus minus sign.
I am unable to get my head around on how it really works? What is the underlying principle that justify such transformation?
linear-algebra exponential-function absolute-value
asked Dec 4 '18 at 2:50
ThorThor
23217
$endgroup$
$begingroup$
Substitute in the $y = ...$ equation into the original one with the absolute values. Try both the plus and minus versions, and see what happens.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 2:53
$begingroup$
That make sense. Thank you
$endgroup$
– Thor
Dec 4 '18 at 2:57
add a comment |
$begingroup$
I have recently encountered the following algebra transformation from an absolute sign to plus minus sign.
I am unable to get my head around on how it really works? What is the underlying principle that justify such transformation?
linear-algebra exponential-function absolute-value
asked Dec 4 '18 at 2:50
ThorThor
23217
$endgroup$
I have recently encountered the following algebra transformation from an absolute sign to plus minus sign.
I am unable to get my head around on how it really works? What is the underlying principle that justify such transformation?
linear-algebra exponential-function absolute-value
linear-algebra exponential-function absolute-value
asked Dec 4 '18 at 2:50
ThorThor
23217
asked Dec 4 '18 at 2:50
ThorThor
23217
asked Dec 4 '18 at 2:50
ThorThor
23217
asked Dec 4 '18 at 2:50
ThorThor
23217
asked Dec 4 '18 at 2:50
ThorThor
23217
23217
$begingroup$
Substitute in the $y = ...$ equation into the original one with the absolute values. Try both the plus and minus versions, and see what happens.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 2:53
$begingroup$
That make sense. Thank you
$endgroup$
– Thor
Dec 4 '18 at 2:57
add a comment |
$begingroup$
Substitute in the $y = ...$ equation into the original one with the absolute values. Try both the plus and minus versions, and see what happens.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 2:53
$begingroup$
That make sense. Thank you
$endgroup$
– Thor
Dec 4 '18 at 2:57
$begingroup$
Substitute in the $y = ...$ equation into the original one with the absolute values. Try both the plus and minus versions, and see what happens.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 2:53
$begingroup$
Substitute in the $y = ...$ equation into the original one with the absolute values. Try both the plus and minus versions, and see what happens.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 2:53
$begingroup$
That make sense. Thank you
$endgroup$
– Thor
Dec 4 '18 at 2:57
$begingroup$
That make sense. Thank you
$endgroup$
– Thor
Dec 4 '18 at 2:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $|x|=1$, then it is either $x=1$ or $x=-1$.
Hence if $|x|=y$, then it is possible that $x=y$ or $x=-y$.
answered Dec 4 '18 at 2:52
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Thank you for the explanation. Every make sense now.
$endgroup$
– Thor
Dec 4 '18 at 2:57
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $|x|=1$, then it is either $x=1$ or $x=-1$.
Hence if $|x|=y$, then it is possible that $x=y$ or $x=-y$.
answered Dec 4 '18 at 2:52
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Thank you for the explanation. Every make sense now.
$endgroup$
– Thor
Dec 4 '18 at 2:57
add a comment |
$begingroup$
If $|x|=1$, then it is either $x=1$ or $x=-1$.
Hence if $|x|=y$, then it is possible that $x=y$ or $x=-y$.
answered Dec 4 '18 at 2:52
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Thank you for the explanation. Every make sense now.
$endgroup$
– Thor
Dec 4 '18 at 2:57
add a comment |
$begingroup$
If $|x|=1$, then it is either $x=1$ or $x=-1$.
Hence if $|x|=y$, then it is possible that $x=y$ or $x=-y$.
answered Dec 4 '18 at 2:52
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
If $|x|=1$, then it is either $x=1$ or $x=-1$.
Hence if $|x|=y$, then it is possible that $x=y$ or $x=-y$.
answered Dec 4 '18 at 2:52
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 4 '18 at 2:52
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 4 '18 at 2:52
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 4 '18 at 2:52
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
$begingroup$
Thank you for the explanation. Every make sense now.
$endgroup$
– Thor
Dec 4 '18 at 2:57
add a comment |
$begingroup$
Thank you for the explanation. Every make sense now.
$endgroup$
– Thor
Dec 4 '18 at 2:57
$begingroup$
Thank you for the explanation. Every make sense now.
$endgroup$
– Thor
Dec 4 '18 at 2:57
$begingroup$
Thank you for the explanation. Every make sense now.
$endgroup$
– Thor
Dec 4 '18 at 2:57
add a comment |
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$begingroup$
Substitute in the $y = ...$ equation into the original one with the absolute values. Try both the plus and minus versions, and see what happens.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 2:53
$begingroup$
That make sense. Thank you
$endgroup$
– Thor
Dec 4 '18 at 2:57